MAT01B1: Calculus with parametric curves

Dr Craig

16 October 2018

My details:

I acraig@uj.ac.za

I Consulting hours:

Any time Wed, Thu or Fri except

08h30 – 11h30 Thu, 13h00 – 15h30 Thu,

10h30 – 12h30 Friday.

I Office C-Ring 508

https://andrewcraigmaths.wordpress.com/

(Or, just google ‘Andrew Craig maths’.)

Semester Test 2

I Saturday 20 October

I 09h00 B-LES 100/B-LES 101

I Scope:

4.5, 4.7, 3.9

6.1, 6.2, 6.3, 6.5

(including proof of MVT for Integrals)

8.1, 8.2

9.3, 9.5

10.1, 10.2 (up to Example 2)

Recap: last week we learned to work with

curves defined in the following way:

x = f (t) y = g(t)

These curves contain more information than

just the set of points (x, y) = (f (t), g(t)).

The additional information is the fact that a

curve is at a particular point (x′, y′) at a

particular value of t. Also, parametric curves

have direction.

Recap: sometimes we can rewrite the

equations x = f (t) and y = g(t) without

using t.

(For example: x = r cos t, y = r sin t gives

us r2 = x2 + y2.) This can sometimes make

it easier to sketch the curve.

There will also sometimes be limits on the

values of t (and hence x and y). These will

either be implicitly imposed by the domain of

f (t) and/or g(t), or explicitly stated (e.g.

0 6 t 6 2π).

Derivatives of parametric curves

The derivative of a curve

dy

dx=

dy

dtdx

dt

is defined wheneverdx

dt6= 0

Horizontal tangent whendy

dt= 0.

Vertical tangent whendx

dt= 0 and

dy

dt6= 0

Second derivatives

To get the second derivative of a

parametric curve, we differentiate the first

derivative with respect to x:

d2y

dx2=

d

dx

(dy

dx

)=

d

dt

(dy

dx

)dx

dtWe can use the second derivative to explore

the concavity of a parametric curve.

Example: consider the curve defined by

x = f (t) = t2 y = g(t) = t3 − 3t

(a) Show that the curve has two tangents at

the point (3, 0) and find their equations.

(b) Find the points (x, y) on the curve where

the tangent is horizontal or vertical.

(c) Determine where the curve is concave

upward or downward.

(d) Sketch the curve.

x = f (t) = t2 y = g(t) = t3 − 3t

Example: consider the cycloid

x = r(θ − sin θ) y = r(1− cos θ)

(a) Find the tangent at the point where

θ = π/3.

(b) At what points is the tangent horizontal?

(Give (x, y) coordinates of the points.)

(c) At what points is the tangent vertical?

Vertical tangents of the cycloid

We get vertical tangents at t = k.2π (or

x = k.2πr) for all k ∈ Z. We can see this by

calculating

limθ→k.2π+

dy

dx=∞ and lim

θ→k.2π−dy

dx= −∞

Area under a parametric curve

We are used to the following situation:

A =

∫ b

a

y dx

However, now y is no longer a function of x (it is afunction of t) so we aren’t able to find theantiderivative.

We use the fact that dxdt = f ′(t) and hence

dx = f ′(t) dt. We change the x-bounds to t-boundsby solving a = f(α) and b = f(β). Finally we have

A =

∫ β

α

g(t)f ′(t) dt

Example

Find the area under one arch of the cycloid

x = r(θ − sin θ) y = r(1− cos θ)

A = 3πr2

Example:

Find the area between the x-axis and the

parametric curve

x = 1 + et y = t− t2

Solution:

A = 3− e

Arc length from Chapter 8

If y = h(x) and a 6 x 6 b and h′(x) is

continuous, then

L =

∫ b

a

√1 +

(dy

dx

)2

dx

Now when f (α) = a and f (β) = b we have

L =

∫ β

α

√1 +

(dy

dx

)2dx

dtdt

and therefore

L =

∫ β

α

√(dx

dt

)2

+

(dy

dt

)2

dt

Theorem: If C is described by x = f (t),

y = g(t), α 6 t 6 β and f ′(t) and g′(t)

are continuous on [α, β], and C is

traversed exactly once as t increases from

α to β, then the length of C is

L =

∫ β

α

√(dx

dt

)2

+

(dy

dt

)2

dt

Arc length examples:

Find the arc length of the two curves from

last time.

(a) x = cos t, y = sin t, 0 6 t 6 2π

(b) x = sin 2t, y = cos 2t, 0 6 t 6 2π

Arc length example:

Find the length of one arch of the cycloid

x = r(θ − sin θ) y = r(1− cos θ)

Surface area

Previously, if we have y = f (x) and we

rotated a piece of that function (a 6 x 6 b)

around the x-axis, we would calculate the

surface area as follows:∫ b

a

2πf (x)√

1 + [f ′(x)]2 dx

Now suppose that we rotate a parametric

curve about the x-axis.

Surface area: for x = f (t), y = g(t),

α 6 t 6 β rotated about the x-axis (with f ′

and g′ continuous and g(t) > 0) we have

S =

∫ β

α

2πy

√(dx

dt

)2

+

(dy

dt

)2

dt

Note that we will use y = g(t) so that we

can integrate with respect to t:

S = 2π

∫ β

α

g(t)

√(dx

dt

)2

+

(dy

dt

)2

dt

Surface area

If we rotate x = f (t), y = g(t), α 6 t 6 β

about the y-axis (with f ′ and g′ continuous

and f (t) > 0) we have

S =

∫ β

α

2πf (t)

√(dx

dt

)2

+

(dy

dt

)2

dt

Example: show that the surface area of a

sphere of radius r is 4πr2. (This can be done

by rotating a semicircle either around the

x-axis or around the y-axis.)