Fundamentals of Engineering Calculus, Differential...

Fundamentals of EngineeringCalculus, Differential Equations & Transforms, and

Numerical Analysis

Brody Dylan Johnson

St. Louis University

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis1 / 30

Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30

Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30

Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30

Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30

Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30

Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30

Overview

Overview

Agenda:

Problem solving with Just-In-Time lectures (50 minutes)

Group work with more problems (30 minutes)

Quiz (30 minutes)

Topics:

Calculus: Differential Calculus, Integral Calculus, Centroids andMoments of Inertia, Vector Calculus.

Differential Equations and Transforms: Differential Equations, FourierSeries, Laplace Transforms, Euler’s Approximation

Numerical Analysis: Root Solving with Bisection Method and Newton’sMethod.

Acknowledgement: Many problems are taken from the Hughes-Hallett,Gleason, McCallum, et al. Calculus textbook.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis2 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:

Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:

Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 1: Find the maximum and minimum values of f (x) = x3 − 3x2 + 20on [−1, 3].

Solution:Check endpoints and critical points (f ′(x) = 0).

f ′(x) = 3x2 − 6x, f ′(x) = 0 =⇒ x = 0, 2.

f ′′(x) = 6x− 6, f ′′(0) < 0 concave down; local max, f ′′(2) >0 concave up; local min.

Compare values: f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.

Minima: f (x) = 16 at x = −1 or x = 2.

Maxima: f (x) = 20 at x = 0 or x = 3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis3 / 30

Calculus

Differential Calculus

Problem 2: Find dydx if y = xx.

Solution:

Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30

Calculus

Differential Calculus

Problem 2: Find dydx if y = xx.

Solution:

Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30

Calculus

Differential Calculus

Problem 2: Find dydx if y = xx.

Solution:Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30

Calculus

Differential Calculus

Problem 2: Find dydx if y = xx.

Solution:Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.

1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30

Calculus

Differential Calculus

Problem 2: Find dydx if y = xx.

Solution:Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30

Calculus

Differential Calculus

Problem 2: Find dydx if y = xx.

Solution:Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30

Calculus

Differential Calculus

Problem 2: Find dydx if y = xx.

Solution:Apply logarithm and then use implicit differentiation.

Differentiate ln y = x ln x w.r.t x.1y

dydx

= ln x + x1x

(product rule).

dydx

= y(ln x + 1).

dydx

= xx(ln x + 1).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis4 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:

Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:

Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 3: Find the standard form of the tangent line to the hyperbolax2 − 2y2 = 8 at the point (−4, 2).

Solution:Tangent Line to y = f (x) at x = a: y = f (a) + f ′(a)(x− a).

Use implicit differentiation to find dydx since f is not given explicitly.

2x− 4ydydx

= 0

dydx

=2x4y

.

dydx

∣∣∣(−4,2)

=−88

= −1.

Tangent line: y = 2 + (−1)(x− (−4)) or y = −x− 2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis5 / 30

Calculus

Differential Calculus

Problem 4: Evaluate the following limit.

limx→∞

xe−x.

Solution:

Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.

Recall: L’Hopital’s Rule states that the limit of an indeterminate formf (x)/g(x) can be evaluated using the ratio of the derivatives f ′(x)/g′(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30

Calculus

Differential Calculus

Problem 4: Evaluate the following limit.

limx→∞

xe−x.

Solution:

Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.

Recall: L’Hopital’s Rule states that the limit of an indeterminate formf (x)/g(x) can be evaluated using the ratio of the derivatives f ′(x)/g′(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30

Calculus

Differential Calculus

Problem 4: Evaluate the following limit.

limx→∞

xe−x.

Solution:Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.

Recall: L’Hopital’s Rule states that the limit of an indeterminate formf (x)/g(x) can be evaluated using the ratio of the derivatives f ′(x)/g′(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30

Calculus

Differential Calculus

Problem 4: Evaluate the following limit.

limx→∞

xe−x.

Solution:Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.

Recall: L’Hopital’s Rule states that the limit of an indeterminate formf (x)/g(x) can be evaluated using the ratio of the derivatives f ′(x)/g′(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30

Calculus

Differential Calculus

Problem 4: Evaluate the following limit.

limx→∞

xe−x.

Solution:Indeterminate form: 0 · ∞. Use L’Hopital’s Rule.

So,

limx→∞

xe−x = limx→∞

xex

H= lim

x→∞

1ex =

1∞

= 0.

Recall: L’Hopital’s Rule states that the limit of an indeterminate formf (x)/g(x) can be evaluated using the ratio of the derivatives f ′(x)/g′(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis6 / 30

Calculus

Differential Calculus

Problem 5: Find the partial derivative ∂m∂c if

m =m0√

1− v2/c2.

Solution:

Notation: if z = f (x, y) then ∂z∂x = ∂f

∂x = fx.

Treat other variables as constants and differentiate w.r.t. indicatedvariable.∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30

Calculus

Differential Calculus

Problem 5: Find the partial derivative ∂m∂c if

m =m0√

1− v2/c2.

Solution:

Notation: if z = f (x, y) then ∂z∂x = ∂f

∂x = fx.

Treat other variables as constants and differentiate w.r.t. indicatedvariable.∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30

Calculus

Differential Calculus

Problem 5: Find the partial derivative ∂m∂c if

m =m0√

1− v2/c2.

Solution:

Notation: if z = f (x, y) then ∂z∂x = ∂f

∂x = fx.

Treat other variables as constants and differentiate w.r.t. indicatedvariable.∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30

Calculus

Differential Calculus

Problem 5: Find the partial derivative ∂m∂c if

m =m0√

1− v2/c2.

Solution:

Notation: if z = f (x, y) then ∂z∂x = ∂f

∂x = fx.

Treat other variables as constants and differentiate w.r.t. indicatedvariable.

∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30

Calculus

Differential Calculus

Problem 5: Find the partial derivative ∂m∂c if

m =m0√

1− v2/c2.

Solution:

Notation: if z = f (x, y) then ∂z∂x = ∂f

∂x = fx.

Treat other variables as constants and differentiate w.r.t. indicatedvariable.∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30

Calculus

Differential Calculus

Problem 5: Find the partial derivative ∂m∂c if

m =m0√

1− v2/c2.

Solution:

Notation: if z = f (x, y) then ∂z∂x = ∂f

∂x = fx.

Treat other variables as constants and differentiate w.r.t. indicatedvariable.∂m∂c

=∂

∂c

[m0

(1− v2/c2)12

]= −1

2m0

(1− v2/c2)32

(−2)(−v2/c3)

Simplify.∂m∂c

=−m0v2

c3(1− v2/c2)32

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis7 / 30

Calculus

Differential Calculus

Problem 6: What is the slope of the curve y =5− 3x5 + 3x

when it crosses the

positive x-axis?

Solution:

The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5

3 . The slope is given by dydx .

Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30

Calculus

Differential Calculus

Problem 6: What is the slope of the curve y =5− 3x5 + 3x

when it crosses the

positive x-axis?

Solution:

The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5

3 . The slope is given by dydx .

Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30

Calculus

Differential Calculus

Problem 6: What is the slope of the curve y =5− 3x5 + 3x

when it crosses the

positive x-axis?

Solution:The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5

3 . The slope is given by dydx .

Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30

Calculus

Differential Calculus

Problem 6: What is the slope of the curve y =5− 3x5 + 3x

when it crosses the

positive x-axis?

Solution:The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5

3 . The slope is given by dydx .

Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30

Calculus

Differential Calculus

Problem 6: What is the slope of the curve y =5− 3x5 + 3x

when it crosses the

positive x-axis?

Solution:The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5

3 . The slope is given by dydx .

Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30

Calculus

Differential Calculus

Problem 6: What is the slope of the curve y =5− 3x5 + 3x

when it crosses the

positive x-axis?

Solution:The graph crosses the positive x-axis when y = 0 and x > 0, i.e.,5− 3x = 0 or x = 5

3 . The slope is given by dydx .

Quotient Rule:ddx

(fg

)=

f ′g− g′fg2 .

dydx

=(−3)(5 + 3x)− (3)(5− 3x)

(5 + 3x)2 .

dydx

∣∣∣x= 5

3

=(−3)(10)− 3(0)

102 = −0.3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis8 / 30

Calculus

Differential Calculus

Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:

Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

Fixed Cost: g(v) = 675 (dollars/hour) is constant w.r.t. speed.

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v it takes 1/vhours to go one mile.

Minimize: C(v) = 675/v + 0.1v2,C′(v) = −675/v2 + 0.2v = (0.2v3 − 675)/v2. Critical Pointv = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30

Calculus

Differential Calculus

Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:

Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

Fixed Cost: g(v) = 675 (dollars/hour) is constant w.r.t. speed.

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v it takes 1/vhours to go one mile.

Minimize: C(v) = 675/v + 0.1v2,C′(v) = −675/v2 + 0.2v = (0.2v3 − 675)/v2. Critical Pointv = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30

Calculus

Differential Calculus

Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

Fixed Cost: g(v) = 675 (dollars/hour) is constant w.r.t. speed.

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v it takes 1/vhours to go one mile.

Minimize: C(v) = 675/v + 0.1v2,C′(v) = −675/v2 + 0.2v = (0.2v3 − 675)/v2. Critical Pointv = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30

Calculus

Differential Calculus

Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

Fixed Cost: g(v) = 675 (dollars/hour) is constant w.r.t. speed.

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v it takes 1/vhours to go one mile.

Minimize: C(v) = 675/v + 0.1v2,C′(v) = −675/v2 + 0.2v = (0.2v3 − 675)/v2. Critical Pointv = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30

Calculus

Differential Calculus

Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

Fixed Cost: g(v) = 675 (dollars/hour) is constant w.r.t. speed.

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v it takes 1/vhours to go one mile.

Minimize: C(v) = 675/v + 0.1v2,C′(v) = −675/v2 + 0.2v = (0.2v3 − 675)/v2. Critical Pointv = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30

Calculus

Differential Calculus

Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

Fixed Cost: g(v) = 675 (dollars/hour) is constant w.r.t. speed.

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v it takes 1/vhours to go one mile.

Minimize: C(v) = 675/v + 0.1v2,C′(v) = −675/v2 + 0.2v = (0.2v3 − 675)/v2. Critical Pointv = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30

Calculus

Differential Calculus

Problem 7: The cost of fuel to propel a boat through the water (dollars perhour) is proportional to the cube of the speed. A certain ferry boat uses $100of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the costof running this ferry (labor, maintenance, and so on) is $675 per hour. At whatspeed should it travel so as to minimize the cost per mile traveled?

Solution:Costs must be converted to a per mile basis.

Fuel Cost: f (v) = Cv3 (dollars/hour) where v is speed. f (10) = 100 soC = 0.1.

Fixed Cost: g(v) = 675 (dollars/hour) is constant w.r.t. speed.

Total Cost: h(v) = 675 + 0.1v3 (dollars/hour). At speed v it takes 1/vhours to go one mile.

Minimize: C(v) = 675/v + 0.1v2,C′(v) = −675/v2 + 0.2v = (0.2v3 − 675)/v2. Critical Pointv = 3

√5(675) = 15 miles per hour. (Check that C′′(15) > 0.)

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis9 / 30

Calculus

Integral Calculus

Problem 8: Calculate the definite integral∫ 2

0xex2

dx

Solution:

Use u-substitution: u = x2 so du = 2x dx.

Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!∫ 2

0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30

Calculus

Integral Calculus

Problem 8: Calculate the definite integral∫ 2

0xex2

dx

Solution:

Use u-substitution: u = x2 so du = 2x dx.

Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!∫ 2

0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30

Calculus

Integral Calculus

Problem 8: Calculate the definite integral∫ 2

0xex2

dx

Solution:

Use u-substitution: u = x2 so du = 2x dx.

Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!∫ 2

0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30

Calculus

Integral Calculus

Problem 8: Calculate the definite integral∫ 2

0xex2

dx

Solution:

Use u-substitution: u = x2 so du = 2x dx.

Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!

∫ 2

0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30

Calculus

Integral Calculus

Problem 8: Calculate the definite integral∫ 2

0xex2

dx

Solution:

Use u-substitution: u = x2 so du = 2x dx.

Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!∫ 2

0xex2

dx =12

∫ 4

0eu du.

∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30

Calculus

Integral Calculus

Problem 8: Calculate the definite integral∫ 2

0xex2

dx

Solution:

Use u-substitution: u = x2 so du = 2x dx.

Don’t forget to change the limits of integration: u(0) = 0, u(2) = 4!∫ 2

0xex2

dx =12

∫ 4

0eu du.∫ 2

0xex2

dx =12

eu∣∣∣40

=e4 − 1

2.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis10 / 30

Calculus

Integral Calculus

Problem 9: Determine the indefinite integral∫

x2e−x dx

Solution:

Integration by parts:∫

u dv = u v−∫

v du.

u = x2, du = 2x dx, dv = e−x dx, v = −e−x,∫x2e−x dx = −x2e−x + 2

∫xe−x dx.

u = x, du = dx, dv = e−x dx, v = −e−x dx,∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C

Finally, (incorporating the 2 again)∫x2e−x dx = −x2e−x − 2xe−x − 2e−x + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30

Calculus

Integral Calculus

Problem 9: Determine the indefinite integral∫

x2e−x dx

Solution:

Integration by parts:∫

u dv = u v−∫

v du.

u = x2, du = 2x dx, dv = e−x dx, v = −e−x,∫x2e−x dx = −x2e−x + 2

∫xe−x dx.

u = x, du = dx, dv = e−x dx, v = −e−x dx,∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C

Finally, (incorporating the 2 again)∫x2e−x dx = −x2e−x − 2xe−x − 2e−x + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30

Calculus

Integral Calculus

Problem 9: Determine the indefinite integral∫

x2e−x dx

Solution:

Integration by parts:∫

u dv = u v−∫

v du.

u = x2, du = 2x dx, dv = e−x dx, v = −e−x,∫x2e−x dx = −x2e−x + 2

∫xe−x dx.

u = x, du = dx, dv = e−x dx, v = −e−x dx,∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C

Finally, (incorporating the 2 again)∫x2e−x dx = −x2e−x − 2xe−x − 2e−x + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30

Calculus

Integral Calculus

Problem 9: Determine the indefinite integral∫

x2e−x dx

Solution:

Integration by parts:∫

u dv = u v−∫

v du.

u = x2, du = 2x dx, dv = e−x dx, v = −e−x,∫x2e−x dx = −x2e−x + 2

∫xe−x dx.

u = x, du = dx, dv = e−x dx, v = −e−x dx,∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C

Finally, (incorporating the 2 again)∫x2e−x dx = −x2e−x − 2xe−x − 2e−x + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30

Calculus

Integral Calculus

Problem 9: Determine the indefinite integral∫

x2e−x dx

Solution:

Integration by parts:∫

u dv = u v−∫

v du.

u = x2, du = 2x dx, dv = e−x dx, v = −e−x,∫x2e−x dx = −x2e−x + 2

∫xe−x dx.

u = x, du = dx, dv = e−x dx, v = −e−x dx,∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C

Finally, (incorporating the 2 again)∫x2e−x dx = −x2e−x − 2xe−x − 2e−x + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30

Calculus

Integral Calculus

Problem 9: Determine the indefinite integral∫

x2e−x dx

Solution:

Integration by parts:∫

u dv = u v−∫

v du.

u = x2, du = 2x dx, dv = e−x dx, v = −e−x,∫x2e−x dx = −x2e−x + 2

∫xe−x dx.

u = x, du = dx, dv = e−x dx, v = −e−x dx,∫xe−x dx = −xe−x +

∫e−x dx = −xe−x − e−x + C

Finally, (incorporating the 2 again)∫x2e−x dx = −x2e−x − 2xe−x − 2e−x + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis11 / 30

Calculus

Integral Calculus

Problem 10: Find the volume of revolution from x = 1 to x = 3 whenf (x) = x− 1 is rotated about the y-axis.

Solution:

Volume =∫ b

a 2πxf (x) dx. (y-axis)

Volume =∫ b

a π(f (x))2 dx. (x-axis)

Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30

Calculus

Integral Calculus

Problem 10: Find the volume of revolution from x = 1 to x = 3 whenf (x) = x− 1 is rotated about the y-axis.

Solution:

Volume =∫ b

a 2πxf (x) dx. (y-axis)

Volume =∫ b

a π(f (x))2 dx. (x-axis)

Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30

Calculus

Integral Calculus

Problem 10: Find the volume of revolution from x = 1 to x = 3 whenf (x) = x− 1 is rotated about the y-axis.

Solution:

Volume =∫ b

a 2πxf (x) dx. (y-axis)

Volume =∫ b

a π(f (x))2 dx. (x-axis)

Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30

Calculus

Integral Calculus

Problem 10: Find the volume of revolution from x = 1 to x = 3 whenf (x) = x− 1 is rotated about the y-axis.

Solution:

Volume =∫ b

a 2πxf (x) dx. (y-axis)

Volume =∫ b

a π(f (x))2 dx. (x-axis)

Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30

Calculus

Integral Calculus

Problem 10: Find the volume of revolution from x = 1 to x = 3 whenf (x) = x− 1 is rotated about the y-axis.

Solution:

Volume =∫ b

a 2πxf (x) dx. (y-axis)

Volume =∫ b

a π(f (x))2 dx. (x-axis)

Here,

Volume =

∫ 3

12πx(x− 1) dx

= 2π[

x3

3− x2

2

] ∣∣∣31

= 2π[

263− 8

2

]=

28π3.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis12 / 30

Calculus

Integral Calculus

Problem 11: Find the area between the curves y = x and y =√

x.

Solution:

The curves bound a region over 0 ≤ x ≤ 1 with√

x ≥ x.

The area is given by the integral of√

x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30

Calculus

Integral Calculus

Problem 11: Find the area between the curves y = x and y =√

x.

Solution:

The curves bound a region over 0 ≤ x ≤ 1 with√

x ≥ x.

The area is given by the integral of√

x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30

Calculus

Integral Calculus

Problem 11: Find the area between the curves y = x and y =√

x.

Solution:

The curves bound a region over 0 ≤ x ≤ 1 with√

x ≥ x.

The area is given by the integral of√

x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30

Calculus

Integral Calculus

Problem 11: Find the area between the curves y = x and y =√

x.

Solution:

The curves bound a region over 0 ≤ x ≤ 1 with√

x ≥ x.

The area is given by the integral of√

x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30

Calculus

Integral Calculus

Problem 11: Find the area between the curves y = x and y =√

x.

Solution:

The curves bound a region over 0 ≤ x ≤ 1 with√

x ≥ x.

The area is given by the integral of√

x− x.

Hence,

Area =

∫ 1

0

√x− x dx

=

[23

x32 − 1

2x2] ∣∣∣1

0

=23− 1

2

=16.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis13 / 30

Calculus

Integral Calculus

Problem 12: Find the indefinite integral∫ 1

x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.

Use cover-up method or common denominator:

1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.

Equate terms: A + B = 0 (x terms) and A = 1 (constant terms) soB = −1.

Finally we integrate:∫1

x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30

Calculus

Integral Calculus

Problem 12: Find the indefinite integral∫ 1

x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.

Use cover-up method or common denominator:

1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.

Equate terms: A + B = 0 (x terms) and A = 1 (constant terms) soB = −1.

Finally we integrate:∫1

x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30

Calculus

Integral Calculus

Problem 12: Find the indefinite integral∫ 1

x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.

Use cover-up method or common denominator:

1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.

Equate terms: A + B = 0 (x terms) and A = 1 (constant terms) soB = −1.

Finally we integrate:∫1

x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30

Calculus

Integral Calculus

Problem 12: Find the indefinite integral∫ 1

x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.

Use cover-up method or common denominator:

1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.

Equate terms: A + B = 0 (x terms) and A = 1 (constant terms) soB = −1.

Finally we integrate:∫1

x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30

Calculus

Integral Calculus

Problem 12: Find the indefinite integral∫ 1

x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.

Use cover-up method or common denominator:

1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.

Equate terms: A + B = 0 (x terms) and A = 1 (constant terms) soB = −1.

Finally we integrate:∫1

x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30

Calculus

Integral Calculus

Problem 12: Find the indefinite integral∫ 1

x2+x dx.

Solution:

Partial fractions:1

x2 + x=

1x(x + 1)

=Ax

+B

x + 1.

Use cover-up method or common denominator:

1x(x + 1)

=Ax

+B

x + 1=

A(x + 1) + Bxx(x + 1)

.

Equate terms: A + B = 0 (x terms) and A = 1 (constant terms) soB = −1.

Finally we integrate:∫1

x2 + xdx =

∫1x− 1

x + 1dx = ln x− ln (x + 1) + C.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis14 / 30

Calculus

Centroid and Moment of Inertia

Problem 13: Find the centroid of the region bounded by 0 ≤ x ≤ 2,0 ≤ y ≤ x2.

Solution:

xc = A−1∫

x dA and yc = A−1∫

y dA where A = Area.

A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30

Calculus

Centroid and Moment of Inertia

Problem 13: Find the centroid of the region bounded by 0 ≤ x ≤ 2,0 ≤ y ≤ x2.

Solution:

xc = A−1∫

x dA and yc = A−1∫

y dA where A = Area.

A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30

Calculus

Centroid and Moment of Inertia

Problem 13: Find the centroid of the region bounded by 0 ≤ x ≤ 2,0 ≤ y ≤ x2.

Solution:

xc = A−1∫

x dA and yc = A−1∫

y dA where A = Area.

A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30

Calculus

Centroid and Moment of Inertia

Problem 13: Find the centroid of the region bounded by 0 ≤ x ≤ 2,0 ≤ y ≤ x2.

Solution:

xc = A−1∫

x dA and yc = A−1∫

y dA where A = Area.

A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30

Calculus

Centroid and Moment of Inertia

Problem 13: Find the centroid of the region bounded by 0 ≤ x ≤ 2,0 ≤ y ≤ x2.

Solution:

xc = A−1∫

x dA and yc = A−1∫

y dA where A = Area.

A =

∫ 2

0x2 dx =

83.

x coordinate:

xc =38

∫ 2

0xf (x) dx =

38

14

x4∣∣∣20

=32.

y coordinate:

yc =38

∫ 4

0y(2−√y) dy =

38

(2y− 2

3y

32

) ∣∣∣40

= 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis15 / 30

Calculus

Centroid and Moment of Inertia

Problem 14: Find the moment of inertia of a rectangle with side lengths a andb about its bottom edge (side length a).

Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.

Imagining the bottom edge as the x-axis we want Ix:

Ix =

∫ b

0y2(a− 0) dy =

ab3

3.

Note: We set this up using dy because points on horizontal slices havethe same distance from the axis of rotation. Points on vertical slices (aswith dx) would not.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30

Calculus

Centroid and Moment of Inertia

Problem 14: Find the moment of inertia of a rectangle with side lengths a andb about its bottom edge (side length a).

Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.

Imagining the bottom edge as the x-axis we want Ix:

Ix =

∫ b

0y2(a− 0) dy =

ab3

3.

Note: We set this up using dy because points on horizontal slices havethe same distance from the axis of rotation. Points on vertical slices (aswith dx) would not.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30

Calculus

Centroid and Moment of Inertia

Problem 14: Find the moment of inertia of a rectangle with side lengths a andb about its bottom edge (side length a).

Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.

Imagining the bottom edge as the x-axis we want Ix:

Ix =

∫ b

0y2(a− 0) dy =

ab3

3.

Note: We set this up using dy because points on horizontal slices havethe same distance from the axis of rotation. Points on vertical slices (aswith dx) would not.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30

Calculus

Centroid and Moment of Inertia

Problem 14: Find the moment of inertia of a rectangle with side lengths a andb about its bottom edge (side length a).

Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.

Imagining the bottom edge as the x-axis we want Ix:

Ix =

∫ b

0y2(a− 0) dy =

ab3

3.

Note: We set this up using dy because points on horizontal slices havethe same distance from the axis of rotation. Points on vertical slices (aswith dx) would not.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30

Calculus

Centroid and Moment of Inertia

Problem 14: Find the moment of inertia of a rectangle with side lengths a andb about its bottom edge (side length a).

Solution:

Ix =∫

y2 dA and Iy =∫

x2 dA.

Imagining the bottom edge as the x-axis we want Ix:

Ix =

∫ b

0y2(a− 0) dy =

ab3

3.

Note: We set this up using dy because points on horizontal slices havethe same distance from the axis of rotation. Points on vertical slices (aswith dx) would not.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis16 / 30

Calculus

Vector Calculus

Problem 15: Find the gradient of f (x, y, z) = xey + ln (xz).

Solution:

Gradient is∇f where∇ = ∂∂x~i + ∂

∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.

The gradient is a vector valued function and is the direction of maximumincrease of f .

Directional derivative in direction of unit vector ~u: D~uf = ∇f ·~u.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30

Calculus

Vector Calculus

Problem 15: Find the gradient of f (x, y, z) = xey + ln (xz).

Solution:

Gradient is∇f where∇ = ∂∂x~i + ∂

∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.

The gradient is a vector valued function and is the direction of maximumincrease of f .

Directional derivative in direction of unit vector ~u: D~uf = ∇f ·~u.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30

Calculus

Vector Calculus

Problem 15: Find the gradient of f (x, y, z) = xey + ln (xz).

Solution:

Gradient is∇f where∇ = ∂∂x~i + ∂

∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.

The gradient is a vector valued function and is the direction of maximumincrease of f .

Directional derivative in direction of unit vector ~u: D~uf = ∇f ·~u.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30

Calculus

Vector Calculus

Problem 15: Find the gradient of f (x, y, z) = xey + ln (xz).

Solution:

Gradient is∇f where∇ = ∂∂x~i + ∂

∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.

The gradient is a vector valued function and is the direction of maximumincrease of f .

Directional derivative in direction of unit vector ~u: D~uf = ∇f ·~u.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30

Calculus

Vector Calculus

Problem 15: Find the gradient of f (x, y, z) = xey + ln (xz).

Solution:

Gradient is∇f where∇ = ∂∂x~i + ∂

∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.

The gradient is a vector valued function and is the direction of maximumincrease of f .

Directional derivative in direction of unit vector ~u: D~uf = ∇f ·~u.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30

Calculus

Vector Calculus

Problem 15: Find the gradient of f (x, y, z) = xey + ln (xz).

Solution:

Gradient is∇f where∇ = ∂∂x~i + ∂

∂y~j + ∂

∂z~k.

So,

∇f (x, y, z) = (ey +1x

)~i + xey~j +1z~k.

The gradient is a vector valued function and is the direction of maximumincrease of f .

Directional derivative in direction of unit vector ~u: D~uf = ∇f ·~u.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis17 / 30

Calculus

Vector Calculus

Problem 16: Let F(x, y, z) = xy~i + y2z~j + xz~k. Find DivF and curlF.

Solution:

If F(x, y, z) = f (x, y, z)~i + g(x, y, z)~j + h(x, y, z)~k then

DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.

The curl of F is given by∇× F,

curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

f (x, y, z) g(x, y, z) h(x, y, z)

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis18 / 30

Calculus

Vector Calculus

Problem 16: Let F(x, y, z) = xy~i + y2z~j + xz~k. Find DivF and curlF.

Solution:

If F(x, y, z) = f (x, y, z)~i + g(x, y, z)~j + h(x, y, z)~k then

DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.

The curl of F is given by∇× F,

curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

f (x, y, z) g(x, y, z) h(x, y, z)

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis18 / 30

Calculus

Vector Calculus

Problem 16: Let F(x, y, z) = xy~i + y2z~j + xz~k. Find DivF and curlF.

Solution:

If F(x, y, z) = f (x, y, z)~i + g(x, y, z)~j + h(x, y, z)~k then

DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.

The curl of F is given by∇× F,

curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

f (x, y, z) g(x, y, z) h(x, y, z)

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis18 / 30

Calculus

Vector Calculus

Problem 16: Let F(x, y, z) = xy~i + y2z~j + xz~k. Find DivF and curlF.

Solution:

If F(x, y, z) = f (x, y, z)~i + g(x, y, z)~j + h(x, y, z)~k then

DivF(x, y, z) = ∇ · F =∂f∂x

+∂g∂y

+∂h∂z

= y + 2yz + x.

The curl of F is given by∇× F,

curlF(x, y, z) =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

f (x, y, z) g(x, y, z) h(x, y, z)

∣∣∣∣∣∣ =

∣∣∣∣∣∣~i ~j ~k∂∂x

∂∂y

∂∂z

xy y2z xz

∣∣∣∣∣∣= (0− y2)~i− (z− 0)~j + (0− x)~k = −y2~i− z~j− x~k.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis18 / 30

Calculus

Vector Calculus

Problem 17: Find the Laplacian of φ = f (x, y, z) = xyz.

Solution:

The Laplacian of φ is ∇2φ = ∂2φ∂x2 + ∂2φ

∂y2 + ∂2φ∂z2 .

φx = yz, φy = xz, and φz = xy.

φxx = 0, φyy = 0, and φzz = 0, so∇2φ(x, y, z) = 0.

Functions for which∇2φ = 0 are called potential functions.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis19 / 30

Calculus

Vector Calculus

Problem 17: Find the Laplacian of φ = f (x, y, z) = xyz.

Solution:

The Laplacian of φ is ∇2φ = ∂2φ∂x2 + ∂2φ

∂y2 + ∂2φ∂z2 .

φx = yz, φy = xz, and φz = xy.

φxx = 0, φyy = 0, and φzz = 0, so∇2φ(x, y, z) = 0.

Functions for which∇2φ = 0 are called potential functions.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis19 / 30

Calculus

Vector Calculus

Problem 17: Find the Laplacian of φ = f (x, y, z) = xyz.

Solution:

The Laplacian of φ is ∇2φ = ∂2φ∂x2 + ∂2φ

∂y2 + ∂2φ∂z2 .

φx = yz, φy = xz, and φz = xy.

φxx = 0, φyy = 0, and φzz = 0, so∇2φ(x, y, z) = 0.

Functions for which∇2φ = 0 are called potential functions.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis19 / 30

Calculus

Vector Calculus

Problem 17: Find the Laplacian of φ = f (x, y, z) = xyz.

Solution:

The Laplacian of φ is ∇2φ = ∂2φ∂x2 + ∂2φ

∂y2 + ∂2φ∂z2 .

φx = yz, φy = xz, and φz = xy.

φxx = 0, φyy = 0, and φzz = 0, so∇2φ(x, y, z) = 0.

Functions for which∇2φ = 0 are called potential functions.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis19 / 30

Calculus

Vector Calculus

Problem 17: Find the Laplacian of φ = f (x, y, z) = xyz.

Solution:

The Laplacian of φ is ∇2φ = ∂2φ∂x2 + ∂2φ

∂y2 + ∂2φ∂z2 .

φx = yz, φy = xz, and φz = xy.

φxx = 0, φyy = 0, and φzz = 0, so∇2φ(x, y, z) = 0.

Functions for which∇2φ = 0 are called potential functions.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis19 / 30

Calculus

Vector Calculus

Problem 17: Find the Laplacian of φ = f (x, y, z) = xyz.

Solution:

The Laplacian of φ is ∇2φ = ∂2φ∂x2 + ∂2φ

∂y2 + ∂2φ∂z2 .

φx = yz, φy = xz, and φz = xy.

φxx = 0, φyy = 0, and φzz = 0, so∇2φ(x, y, z) = 0.

Functions for which∇2φ = 0 are called potential functions.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis19 / 30

Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curvex = t2, y = sin (2πt) at t = 2.

Solution:

Chain Rule: dydt = dy

dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.

The tangent line is thus y = π2 (x− 4).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis20 / 30

Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curvex = t2, y = sin (2πt) at t = 2.

Solution:

Chain Rule: dydt = dy

dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.

The tangent line is thus y = π2 (x− 4).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis20 / 30

Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curvex = t2, y = sin (2πt) at t = 2.

Solution:

Chain Rule: dydt = dy

dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.

The tangent line is thus y = π2 (x− 4).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis20 / 30

Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curvex = t2, y = sin (2πt) at t = 2.

Solution:

Chain Rule: dydt = dy

dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.

The tangent line is thus y = π2 (x− 4).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis20 / 30

Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curvex = t2, y = sin (2πt) at t = 2.

Solution:

Chain Rule: dydt = dy

dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.

The tangent line is thus y = π2 (x− 4).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis20 / 30

Calculus

Vector Calculus

Problem 18: Find the equation of the tangent line to the parametric curvex = t2, y = sin (2πt) at t = 2.

Solution:

Chain Rule: dydt = dy

dx ·dxdt

x′(t) = 2t, y′(t) = 2π cos (2πt)

So x(2) = 4, y(2) = 0, and

dydx

∣∣t=2 =

2π cos (4π)

2(2)=π

2.

The tangent line is thus y = π2 (x− 4).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis20 / 30

Differential Equations and Transforms

Differential Equations

Problem 19: Find the particular solution of y′′ − 3y′ + 2y = 0 satisfyingy(0) = 1, y′(0) = 0.

Solution:

Characteristic Equation: r2 − 3r + 2 = 0 or (r − 2)(r − 1) = 0 sor = 1, 2.

The [homogeneous or complementary] solution is y(x) = Aex + Be2x

Then y′(x) = Aex + 2Be2x and

y(0) = 1 = A + B y′(0) = 0 = A + 2B.

B = −1 and A = 2 yield the particular solution: y(x) = 2ex − e2x.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis21 / 30

Differential Equations and Transforms

Differential Equations

Problem 19: Find the particular solution of y′′ − 3y′ + 2y = 0 satisfyingy(0) = 1, y′(0) = 0.

Solution:

Characteristic Equation: r2 − 3r + 2 = 0 or (r − 2)(r − 1) = 0 sor = 1, 2.

The [homogeneous or complementary] solution is y(x) = Aex + Be2x

Then y′(x) = Aex + 2Be2x and

y(0) = 1 = A + B y′(0) = 0 = A + 2B.

B = −1 and A = 2 yield the particular solution: y(x) = 2ex − e2x.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis21 / 30

Differential Equations and Transforms

Differential Equations

Problem 19: Find the particular solution of y′′ − 3y′ + 2y = 0 satisfyingy(0) = 1, y′(0) = 0.

Solution:

Characteristic Equation: r2 − 3r + 2 = 0 or (r − 2)(r − 1) = 0 sor = 1, 2.

The [homogeneous or complementary] solution is y(x) = Aex + Be2x

Then y′(x) = Aex + 2Be2x and

y(0) = 1 = A + B y′(0) = 0 = A + 2B.

B = −1 and A = 2 yield the particular solution: y(x) = 2ex − e2x.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis21 / 30

Differential Equations and Transforms

Differential Equations

Problem 19: Find the particular solution of y′′ − 3y′ + 2y = 0 satisfyingy(0) = 1, y′(0) = 0.

Solution:

Characteristic Equation: r2 − 3r + 2 = 0 or (r − 2)(r − 1) = 0 sor = 1, 2.

The [homogeneous or complementary] solution is y(x) = Aex + Be2x

Then y′(x) = Aex + 2Be2x and

y(0) = 1 = A + B y′(0) = 0 = A + 2B.

B = −1 and A = 2 yield the particular solution: y(x) = 2ex − e2x.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis21 / 30

Differential Equations and Transforms

Differential Equations

Problem 19: Find the particular solution of y′′ − 3y′ + 2y = 0 satisfyingy(0) = 1, y′(0) = 0.

Solution:

Characteristic Equation: r2 − 3r + 2 = 0 or (r − 2)(r − 1) = 0 sor = 1, 2.

The [homogeneous or complementary] solution is y(x) = Aex + Be2x

Then y′(x) = Aex + 2Be2x and

y(0) = 1 = A + B y′(0) = 0 = A + 2B.

B = −1 and A = 2 yield the particular solution: y(x) = 2ex − e2x.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis21 / 30

Differential Equations and Transforms

Differential Equations

Problem 19: Find the particular solution of y′′ − 3y′ + 2y = 0 satisfyingy(0) = 1, y′(0) = 0.

Solution:

Characteristic Equation: r2 − 3r + 2 = 0 or (r − 2)(r − 1) = 0 sor = 1, 2.

The [homogeneous or complementary] solution is y(x) = Aex + Be2x

Then y′(x) = Aex + 2Be2x and

y(0) = 1 = A + B y′(0) = 0 = A + 2B.

B = −1 and A = 2 yield the particular solution: y(x) = 2ex − e2x.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis21 / 30

Differential Equations and Transforms

Differential Equations

Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

Differentiate: y′(x) = 2Ax + B, y′′(x) = 2A.

Substitute:

(2A) + 2(2Ax + B) + (Ax2 + Bx + C) = x2.

Equate like terms: x2 forces A = 1, x forces 4A + B = 0 and 1 leads to2A + 2B + C = 0.

A = 1, B = −4, and C = 6.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis22 / 30

Differential Equations and Transforms

Differential Equations

Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

Differentiate: y′(x) = 2Ax + B, y′′(x) = 2A.

Substitute:

(2A) + 2(2Ax + B) + (Ax2 + Bx + C) = x2.

Equate like terms: x2 forces A = 1, x forces 4A + B = 0 and 1 leads to2A + 2B + C = 0.

A = 1, B = −4, and C = 6.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis22 / 30

Differential Equations and Transforms

Differential Equations

Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

Differentiate: y′(x) = 2Ax + B, y′′(x) = 2A.

Substitute:

(2A) + 2(2Ax + B) + (Ax2 + Bx + C) = x2.

Equate like terms: x2 forces A = 1, x forces 4A + B = 0 and 1 leads to2A + 2B + C = 0.

A = 1, B = −4, and C = 6.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis22 / 30

Differential Equations and Transforms

Differential Equations

Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

Differentiate: y′(x) = 2Ax + B, y′′(x) = 2A.

Substitute:

(2A) + 2(2Ax + B) + (Ax2 + Bx + C) = x2.

Equate like terms: x2 forces A = 1, x forces 4A + B = 0 and 1 leads to2A + 2B + C = 0.

A = 1, B = −4, and C = 6.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis22 / 30

Differential Equations and Transforms

Differential Equations

Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

Differentiate: y′(x) = 2Ax + B, y′′(x) = 2A.

Substitute:

(2A) + 2(2Ax + B) + (Ax2 + Bx + C) = x2.

Equate like terms: x2 forces A = 1, x forces 4A + B = 0 and 1 leads to2A + 2B + C = 0.

A = 1, B = −4, and C = 6.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis22 / 30

Differential Equations and Transforms

Differential Equations

Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

Differentiate: y′(x) = 2Ax + B, y′′(x) = 2A.

Substitute:

(2A) + 2(2Ax + B) + (Ax2 + Bx + C) = x2.

Equate like terms: x2 forces A = 1, x forces 4A + B = 0 and 1 leads to2A + 2B + C = 0.

A = 1, B = −4, and C = 6.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis22 / 30

Differential Equations and Transforms

Differential Equations

Problem 20: Find a particular solution of y′′ + 2y′ + y = x2.

Solution:

Method of undetermined coefficients. Assume yp(x) = Ax2 + Bx + C.

Differentiate: y′(x) = 2Ax + B, y′′(x) = 2A.

Substitute:

(2A) + 2(2Ax + B) + (Ax2 + Bx + C) = x2.

Equate like terms: x2 forces A = 1, x forces 4A + B = 0 and 1 leads to2A + 2B + C = 0.

A = 1, B = −4, and C = 6.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis22 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:

Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:

Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Differential Equations

Problem 21: Find the general solution of y′ − 3y = sin (2x).

Solution:Method of undetermined coefficients. Assumeyp(x) = A cos (2x) + B sin (2x).

Differentiate: y′(x) = −2A sin (2x) + 2B cos (2x).

Substitute:

(−2A sin (2x) + 2B cos (2x))− 3(A cos (2x) + B sin (2x)) = sin (2x).

Equate like terms, sin (2x): −2A− 3B = 1 and cos (2x): 2B− 3A = 0.

A = 2/13, B = 3/13 so yp(x) = 2/13 cos (2x) + 3/13 sin (2x).

Char. polynomial is r− 3 = 0, so yh(x) = Ae3x and y(x) = yh(x) + yp(x).

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis23 / 30

Differential Equations and Transforms

Fourier Series

Problem 22: Find the Fourier Series of f (x) = x, 0 ≤ x ≤ 1, as a 1-periodicfunction.

Solution:

Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.

In this case, a0 = 1, an = 0, n ≥ 1, and

bn = 2∫ 1

0x sin 2πnx dx = (Integrate by parts) = − 1

πn.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis24 / 30

Differential Equations and Transforms

Fourier Series

Problem 22: Find the Fourier Series of f (x) = x, 0 ≤ x ≤ 1, as a 1-periodicfunction.

Solution:

Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.

In this case, a0 = 1, an = 0, n ≥ 1, and

bn = 2∫ 1

0x sin 2πnx dx = (Integrate by parts) = − 1

πn.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis24 / 30

Differential Equations and Transforms

Fourier Series

Problem 22: Find the Fourier Series of f (x) = x, 0 ≤ x ≤ 1, as a 1-periodicfunction.

Solution:Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.

In this case, a0 = 1, an = 0, n ≥ 1, and

bn = 2∫ 1

0x sin 2πnx dx = (Integrate by parts) = − 1

πn.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis24 / 30

Differential Equations and Transforms

Fourier Series

Problem 22: Find the Fourier Series of f (x) = x, 0 ≤ x ≤ 1, as a 1-periodicfunction.

Solution:Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.

In this case, a0 = 1, an = 0, n ≥ 1, and

bn = 2∫ 1

0x sin 2πnx dx = (Integrate by parts) = − 1

πn.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis24 / 30

Differential Equations and Transforms

Fourier Series

Problem 22: Find the Fourier Series of f (x) = x, 0 ≤ x ≤ 1, as a 1-periodicfunction.

Solution:Fourier Series:

f (x) ≈ a0

2+∞∑

n=1

[an cos

2πnT

x + bn sin2πnT

x].

Coefficients:

an =2T

∫ T

0f (x) cos

2πnT

x dx bn =2T

∫ T

0f (x) sin

2πnT

x dx.

In this case, a0 = 1, an = 0, n ≥ 1, and

bn = 2∫ 1

0x sin 2πnx dx = (Integrate by parts) = − 1

πn.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis24 / 30

Differential Equations and Transforms

Laplace Transform

Problem 23: Find the Laplace transform of f (t) = eat.

Solution:

Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)

Laplace transforms behave nicely with derivatives:

L{f ′(t)} = sF(s)− y(0).

We can use this to solve differential equations.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis25 / 30

Differential Equations and Transforms

Laplace Transform

Problem 23: Find the Laplace transform of f (t) = eat.

Solution:

Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)

Laplace transforms behave nicely with derivatives:

L{f ′(t)} = sF(s)− y(0).

We can use this to solve differential equations.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis25 / 30

Differential Equations and Transforms

Laplace Transform

Problem 23: Find the Laplace transform of f (t) = eat.

Solution:Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)

Laplace transforms behave nicely with derivatives:

L{f ′(t)} = sF(s)− y(0).

We can use this to solve differential equations.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis25 / 30

Differential Equations and Transforms

Laplace Transform

Problem 23: Find the Laplace transform of f (t) = eat.

Solution:Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)

Laplace transforms behave nicely with derivatives:

L{f ′(t)} = sF(s)− y(0).

We can use this to solve differential equations.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis25 / 30

Differential Equations and Transforms

Laplace Transform

Problem 23: Find the Laplace transform of f (t) = eat.

Solution:Laplace transform:

F(s) =

∫ ∞0

f (t)e−st dt.

In this case,

F(s) =

∫ ∞0

e−(s−a)t dt = − 1(s− a)

e−(s−a)t∣∣∣∞0

=1

s− a. (s > a)

Laplace transforms behave nicely with derivatives:

L{f ′(t)} = sF(s)− y(0).

We can use this to solve differential equations.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis25 / 30

Differential Equations and Transforms

Laplace Transform

Problem 24: Use Laplace transforms to solve y′ + 2y = 0, y(0) = 2.

Solution:

Apply the Laplace transform to the equation: sY(s)− y(0) + 2Y(s) = 0.

So Y(s) = y(0)/(s + 2) = 2/(s + 2).

We find y(t) by recognizing the form of Y(s) using known Laplacetransforms:

y(t) = 2L−1{

1s + 2

}= 2e−2t.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis26 / 30

Differential Equations and Transforms

Laplace Transform

Problem 24: Use Laplace transforms to solve y′ + 2y = 0, y(0) = 2.

Solution:

Apply the Laplace transform to the equation: sY(s)− y(0) + 2Y(s) = 0.

So Y(s) = y(0)/(s + 2) = 2/(s + 2).

We find y(t) by recognizing the form of Y(s) using known Laplacetransforms:

y(t) = 2L−1{

1s + 2

}= 2e−2t.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis26 / 30

Differential Equations and Transforms

Laplace Transform

Problem 24: Use Laplace transforms to solve y′ + 2y = 0, y(0) = 2.

Solution:Apply the Laplace transform to the equation: sY(s)− y(0) + 2Y(s) = 0.

So Y(s) = y(0)/(s + 2) = 2/(s + 2).

We find y(t) by recognizing the form of Y(s) using known Laplacetransforms:

y(t) = 2L−1{

1s + 2

}= 2e−2t.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis26 / 30

Differential Equations and Transforms

Laplace Transform

Problem 24: Use Laplace transforms to solve y′ + 2y = 0, y(0) = 2.

Solution:Apply the Laplace transform to the equation: sY(s)− y(0) + 2Y(s) = 0.

So Y(s) = y(0)/(s + 2) = 2/(s + 2).

We find y(t) by recognizing the form of Y(s) using known Laplacetransforms:

y(t) = 2L−1{

1s + 2

}= 2e−2t.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis26 / 30

Differential Equations and Transforms

Laplace Transform

Problem 24: Use Laplace transforms to solve y′ + 2y = 0, y(0) = 2.

Solution:Apply the Laplace transform to the equation: sY(s)− y(0) + 2Y(s) = 0.

So Y(s) = y(0)/(s + 2) = 2/(s + 2).

We find y(t) by recognizing the form of Y(s) using known Laplacetransforms:

y(t) = 2L−1{

1s + 2

}= 2e−2t.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis26 / 30

Differential Equations and Transforms

Laplace Transform

Problem 25: Find the inverse Laplace transform of Y(s) = s+4s2+4 .

Solution:

Consulting a table, the relevant transforms are:

L{cosωt} =s

s2 + ω2 L{sinωt} =ω

s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).

Often we must use partial fractions to first factor an expression intorecognizable pieces.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis27 / 30

Differential Equations and Transforms

Laplace Transform

Problem 25: Find the inverse Laplace transform of Y(s) = s+4s2+4 .

Solution:

Consulting a table, the relevant transforms are:

L{cosωt} =s

s2 + ω2 L{sinωt} =ω

s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).

Often we must use partial fractions to first factor an expression intorecognizable pieces.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis27 / 30

Differential Equations and Transforms

Laplace Transform

Problem 25: Find the inverse Laplace transform of Y(s) = s+4s2+4 .

Solution:Consulting a table, the relevant transforms are:

L{cosωt} =s

s2 + ω2 L{sinωt} =ω

s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).

Often we must use partial fractions to first factor an expression intorecognizable pieces.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis27 / 30

Differential Equations and Transforms

Laplace Transform

Problem 25: Find the inverse Laplace transform of Y(s) = s+4s2+4 .

Solution:Consulting a table, the relevant transforms are:

L{cosωt} =s

s2 + ω2 L{sinωt} =ω

s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).

Often we must use partial fractions to first factor an expression intorecognizable pieces.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis27 / 30

Differential Equations and Transforms

Laplace Transform

Problem 25: Find the inverse Laplace transform of Y(s) = s+4s2+4 .

Solution:Consulting a table, the relevant transforms are:

L{cosωt} =s

s2 + ω2 L{sinωt} =ω

s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).

Often we must use partial fractions to first factor an expression intorecognizable pieces.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis27 / 30

Differential Equations and Transforms

Laplace Transform

Problem 25: Find the inverse Laplace transform of Y(s) = s+4s2+4 .

Solution:Consulting a table, the relevant transforms are:

L{cosωt} =s

s2 + ω2 L{sinωt} =ω

s2 + ω2 .

We haveY(s) =

s + 4s2 + 4

=s

s2 + 4+ 2

2s2 + 4

.

So y(t) = cos (2t) + 2 sin (2t).

Often we must use partial fractions to first factor an expression intorecognizable pieces.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis27 / 30

Differential Equations and Transforms

Separable ODEs

Problem 26: Solve the initial value problem: (1 + x2) dydx = 3xy, y(0) = 1.

Solution:

This differential equation is separable: (integrate RHS using u = 1 + x2)∫1y

dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.

Exponentiation leads to y = A(1 + x2)32 and since y(0) = 1, we have

A = 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis28 / 30

Differential Equations and Transforms

Separable ODEs

Problem 26: Solve the initial value problem: (1 + x2) dydx = 3xy, y(0) = 1.

Solution:

This differential equation is separable: (integrate RHS using u = 1 + x2)∫1y

dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.

Exponentiation leads to y = A(1 + x2)32 and since y(0) = 1, we have

A = 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis28 / 30

Differential Equations and Transforms

Separable ODEs

Problem 26: Solve the initial value problem: (1 + x2) dydx = 3xy, y(0) = 1.

Solution:

This differential equation is separable: (integrate RHS using u = 1 + x2)∫1y

dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.

Exponentiation leads to y = A(1 + x2)32 and since y(0) = 1, we have

A = 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis28 / 30

Differential Equations and Transforms

Separable ODEs

Problem 26: Solve the initial value problem: (1 + x2) dydx = 3xy, y(0) = 1.

Solution:

This differential equation is separable: (integrate RHS using u = 1 + x2)∫1y

dy =

∫3x

1 + x2 dx ln y =32

ln (1 + x2) + C.

Exponentiation leads to y = A(1 + x2)32 and since y(0) = 1, we have

A = 1.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis28 / 30

Differential Equations and Transforms

Euler’s Method

Problem 27: Use Euler’s Method to predict y(0.5) if dydx = 1 + xy, y(0) = 0,

and ∆x = 0.25.

Solution:

Euler’s Method: yk+1 = yk + dydx

∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.

Interpret y2 ≈ y(x0 + 2∆x) = y(0.5) so y(0.5) ≈ 0.516.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis29 / 30

Differential Equations and Transforms

Euler’s Method

Problem 27: Use Euler’s Method to predict y(0.5) if dydx = 1 + xy, y(0) = 0,

and ∆x = 0.25.

Solution:

Euler’s Method: yk+1 = yk + dydx

∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.

Interpret y2 ≈ y(x0 + 2∆x) = y(0.5) so y(0.5) ≈ 0.516.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis29 / 30

Differential Equations and Transforms

Euler’s Method

Problem 27: Use Euler’s Method to predict y(0.5) if dydx = 1 + xy, y(0) = 0,

and ∆x = 0.25.

Solution:

Euler’s Method: yk+1 = yk + dydx

∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.

Interpret y2 ≈ y(x0 + 2∆x) = y(0.5) so y(0.5) ≈ 0.516.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis29 / 30

Differential Equations and Transforms

Euler’s Method

Problem 27: Use Euler’s Method to predict y(0.5) if dydx = 1 + xy, y(0) = 0,

and ∆x = 0.25.

Solution:

Euler’s Method: yk+1 = yk + dydx

∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.

Interpret y2 ≈ y(x0 + 2∆x) = y(0.5) so y(0.5) ≈ 0.516.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis29 / 30

Differential Equations and Transforms

Euler’s Method

Problem 27: Use Euler’s Method to predict y(0.5) if dydx = 1 + xy, y(0) = 0,

and ∆x = 0.25.

Solution:

Euler’s Method: yk+1 = yk + dydx

∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.

Interpret y2 ≈ y(x0 + 2∆x) = y(0.5) so y(0.5) ≈ 0.516.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis29 / 30

Differential Equations and Transforms

Euler’s Method

Problem 27: Use Euler’s Method to predict y(0.5) if dydx = 1 + xy, y(0) = 0,

and ∆x = 0.25.

Solution:

Euler’s Method: yk+1 = yk + dydx

∣∣∣(xk,yk)

·∆x.

y0 = 0 is given, so y1 = 0 + (1 + 0(0)) · 0.25 = 0.25

Now, y2 = 0.25 + (1 + (0.25)(0.25)) · 0.25 = 0.515625.

Interpret y2 ≈ y(x0 + 2∆x) = y(0.5) so y(0.5) ≈ 0.516.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis29 / 30

Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:

The bisection method starts with points, L0 and R0 for which one off (L0), f (R0) is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 3,R2 = 4.

At this point our estimate of the root would be x = 3.5. Noticef (3.5) = −0.25 which is much closer.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30

Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:

The bisection method starts with points, L0 and R0 for which one off (L0), f (R0) is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 3,R2 = 4.

At this point our estimate of the root would be x = 3.5. Noticef (3.5) = −0.25 which is much closer.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30

Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:The bisection method starts with points, L0 and R0 for which one off (L0), f (R0) is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 3,R2 = 4.

At this point our estimate of the root would be x = 3.5. Noticef (3.5) = −0.25 which is much closer.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30

Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:The bisection method starts with points, L0 and R0 for which one off (L0), f (R0) is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 3,R2 = 4.

At this point our estimate of the root would be x = 3.5. Noticef (3.5) = −0.25 which is much closer.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30

Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:The bisection method starts with points, L0 and R0 for which one off (L0), f (R0) is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 3,R2 = 4.

At this point our estimate of the root would be x = 3.5. Noticef (3.5) = −0.25 which is much closer.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30

Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:The bisection method starts with points, L0 and R0 for which one off (L0), f (R0) is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 3,R2 = 4.

At this point our estimate of the root would be x = 3.5. Noticef (3.5) = −0.25 which is much closer.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30

Numerical Analysis

Bisection Method

Problem 28: Use two iterations of the bisection method to approximate a rootof f (x) = x2 − 3x− 2.

Solution:The bisection method starts with points, L0 and R0 for which one off (L0), f (R0) is negative and the other is positive. The intermediate valuetheorem says a root lies in between (assuming f is continuous).

Notice f (0) = −2 < 0 and f (4) = 2 > 0, so we can use L0 = 0 andR0 = 4.

First iteration: Check f (0.5(L0 + R0)) = f (2) = −10, so set L1 = 2,R1 = 4.

Second iteration: Check f (0.5(L1 + R1)) = f (3) = −2, so set L2 = 3,R2 = 4.

At this point our estimate of the root would be x = 3.5. Noticef (3.5) = −0.25 which is much closer.

Brody Dylan Johnson (St. Louis University) Fundamentals of Engineering Calculus, Differential Equations & Transforms, and Numerical Analysis30 / 30