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Coursera Calculus I Fake and Real Finals (April 2013) Jabari Zakiya Fake Question 1 Evaluate the limit a) 1 b) 17 c) 0 d) 10 Because use L' Hopital's Rule: = 3(3) 2 – 12(3) + 10 = 27 – 36 + 10 = 1 Real Question 1 Evaluate the limit a) 8 b) 0 c) 128 d) 17 Because use L' Hopital's Rule: = 8(4) – 15 = 32 – 15 = 17 Fake Question 2 Evaluate the limit a) b) c) d) = = = = Real Question 2 Evaluate the limit a) 1 b) 0 c) d) –1 e) = = = = = –1 Fake Question 3 Let f(x) = –x 4 + x 3 – 3x 2 – 2x + 4. What is ? a) 16x 3 – 15x 2 – 4x – 2 b) 20x 3 + 9x 2 + 4x – 2 c) –4 x 3 – 6x 2 – 8x – 2 d) –4 x 3 + 3 x 2 – 6 x 2 Real Question 3 Let f(x) = 2x 5 + 3x 3 + 4x 2 + 2x + 1. What is ? a) 10x 4 + 3x 2 + 4x + 2 b) 10 x 4 + 9 x 2 + 8 x + 2 c) 10x 4 + 9x 2 + 4x + 2 d) 10x 4 + 3x 2 + 8x + 2

Coursera Calculus I Fake and Real Finals (April 2013)

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Page 1: Coursera Calculus I Fake and Real Finals (April 2013)

Coursera Calculus I Fake and Real Finals (April 2013)

Jabari Zakiya

Fake Question 1

Evaluate the limit a) 1 b) 17 c) 0 d) 10

Because use L' Hopital's Rule: = 3(3)2 – 12(3) + 10 = 27 – 36 + 10 = 1

Real Question 1

Evaluate the limit a) 8 b) 0 c) 128 d) 17

Because use L' Hopital's Rule: = 8(4) – 15 = 32 – 15 = 17

Fake Question 2

Evaluate the limit a) b) c) d)

= = = =

Real Question 2

Evaluate the limit a) 1 b) 0 c) d) –1 e)

= = = = = –1

Fake Question 3

Let f(x) = –x4 + x3 – 3x2 – 2x + 4. What is ?

a) 16x3 – 15x2 – 4x – 2 b) 20x3 + 9x2 + 4x – 2 c) –4 x3 – 6x2 – 8x – 2 d) –4 x 3 + 3 x 2 – 6 x – 2

Real Question 3

Let f(x) = 2x5 + 3x3 + 4x2 + 2x + 1. What is ?

a) 10x4 + 3x2 + 4x + 2 b) 10 x 4 + 9 x 2 + 8 x + 2 c) 10x4 + 9x2 + 4x + 2 d) 10x4 + 3x2 + 8x + 2

Page 2: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 4

To the right is the graph of y = f(x).

Which graph below is most likely the graph of y = f '(x)?

a) b)

c) d)

The answer is a) because the graph of f(x) shows it has 7 local minima/maxima, which means f '(x) must have 7 places where its graph crosses the x-axis (where f '(x) = 0)

Page 3: Coursera Calculus I Fake and Real Finals (April 2013)

Real Question 4

To the right is the graph of y = f(x).

Which graph below is most likely the graph of y = f '(x)?

a) b)

c) d)

The answer is d) because the graph of f(x) shows it has 3 local minima/maxima, which means f '(x) must have 3 places where its graph crosses the x-axis (where f '(x) = 0).

Page 4: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 5

Let f(x) = . What is the derivative of f with respect to x?

a) b)

c) d)

f '(x) =

Real Question 5

Let f(x) = . What is the derivative of f with respect to x?

a) b) c) d)

f '(x) =

Fake Question 6

Consider the function f(x) = sin(cos(log x)). Here, log refers to the natural logarithm. Use the chain ruleto compute f '(x).

a) b)

c) d)

f '(x) = cos(cos(log(x))) (–sin(log(x))) (1/x)

Real Question 6

Consider the function f(x) = log(log(x2). Here, log refers to the natural logarithm. Use the chain rule to compute f '(x).

a) b) c) d)

f '(x) = = = also f(x) = log(2log(x)) and then f '(x) =

Page 5: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 7

Let f(x) = arctan(esin(x)). What is f '(x)?

a) b) c) d)

f '(x) =

Real Question 7

Let f(x) = arctan(x2 + 1). What is f '(x)?

a) b) c) d)

f '(x) =

Fake Question 8

Compute

a) - cos ( cos ( x )) sin ( x ) b) cos(cos(x)) sin(x) c) -cos(cos(x)) cos(x) d) cos(cos(x)) cos(x)

f '(x) = cos(cos(x)) (–sin(x))

Real Question 8

Compute

a) sin(x) cos(cos(x)) b) sin(x) sin(cos(x)) c) –sin(x) sin(cos(x)) d) – sin ( x )cos ( cos ( x ))

f '(x) = cos(cos(x)) (–sin(x))

Page 6: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 9

A ladder is 5 m long, and rests against the side of a building. The bottom of the ladder is presently 4 m from the bottom of the wall, and is being pushed toward the wall at a rate of 2 m/s. Let y be the distance between the top of the ladder and the ground; how quickly is y changing? That is, what is dy/dt ?

a) m/s b) m/s c) m/s d) m/s

Real Question 9

A certain ladder is 4 m long, and rests against the side of a building. The bottom of the ladder is presently 3 m from the bottom of the wall, and is being pulled away from the wall at a rate of 1m/s. How quickly is the top of the ladder moving down the wall?

a) m/s b) m/s c) m/s d) m/s

For both problems, the ladder length L2 = x2 + y2

implicit differentiation gives: = 0 = so =

For Fake Question 9: L = 5, x = 4, y = = 3, and = -2 gives = = m/sec

For Real Question 9: L = 4, x = 3, y = = , and = 1 gives = – = m/sec

Fake Question 10

The function y = x3 – 9x2 + 24x + 2 has a single local maximum. At what point does that local maximum occur?a) –4 b) 2 c) 4 d) –2

y' = 3x2 – 18x + 24 = 0 => x2 – 6x + 8 = 0 gives (x – 4)(x – 2) = 0 so x = 2 or 4 for local min/max.

y'' = 6x – 18 and y''(4) = 6(4) – 18= 6 and y''(2) = 6(2) – 18= -6 is negative so local maximum at x=2.

Real Question 10

The function y = 2x3 + 3x2 – 36x – 14 has a single local minimum. At what point does that local minimum occur?a) -2 b) 3 c) –3 d) 2

y' = 6x2 + 6x – 36 = 0 => x2 + x – 6 = 0 gives (x + 3)(x – 2) = 0 so x = 2 or -3 for local min/max.

y'' = 12x + 6 and y''(-3) = 12(-3)+6= -30 and y''(2) = 12(2)+6= 30 is positive so local minimum at x=2.

Page 7: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 11

A soup can is to be produced to hold exactly 162п cubic centimeters of soup. How tall should the can be to minimize the surface area of the can?

a) cm b) cm c) cm d) cm

Real Question 11A soup can is to be produced to hold exactly 128п cubic centimeters of soup. How tall should the can be to minimize the surface area of the can? a) 10 cm b) 4 cm c) 8 cm d) 6 cm

The geometry and process to solve both problems are the same, with just different can volumes.To solve, write the can's surface area in terms of h, then differentiate, set to 0, and solve for h.

Volume of cylinder: V = and thus

Can Surface Area (SA) = Area of top + Area of bottom + Area of side SA = +

SA = +

SA = +

Now that we have SA in term of just h we can differentiate it with respect to h and set it to zero.

= + = 0

= =

h =

h =

h =

For Fake Problem 11: V = 162п cm so h = = = = cm

For Real Problem 11: V = 128п cm so h = = = = = 8 cm

[You can check this is a minima (not maxima) problem be plotting SA (set V = 1 for easier math) against h to see the curve has just one local minimum.]

Page 8: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 12It is perhaps a bit funny to observe that

sin 29634 ≈ 0.6000001629810547737228990440413

cos 29634 ≈ -0.799999877764182979445839271152

In light of this, which value listed below do you believe to be the best approximation to sin 29634.001?

Warning: The approximation below is better than what Calculus One might give you, so choose the value closest to your best guess.

a) 0.599199863236404284272396125743 b) 0.599920162993278355424954856486c) 0.600800162858818956718998622509 d) 0.600080162968831192024677762636

Euler's Method from problem gives, with h = 0.001:

sin 29634.001 = sin 29634 + h*cos 29634 = 0.6000001629810547737228990440413 – (0.001)(0.799999877764182979445839271152) = 0.6000001629810547737228990440413 – 0.000799999877764182979445839271152 = 0.5992001631032906

Real Question 12Let f(x) = , the tenth root function. Note that since f(1024) = 2. Calculate f '(1024) and use that value to approximate the tenth root of f(1040), a significant number in the United States on April 15.

a) 2 + 1/322 b) 2 + 1/5120 c) 2 + 1/320 d) 2 + 4/1289 e) 2 + 821/264562

f '(x) = x-9/10/10 and h = 1040 – 1024 = 16 so Euler's Method gives:

f(1040) = f(1024) + h*f '(1024) = 2 + 16(1/10*10249/10)= 2 + 24/10*29 = 2 + 1/10*25 = 2 + 1/320

Fake Question 13

Find .

a) - 4 x 4 – 3 x 3 – 3 x 2 – 3 x + C b) 4x4 + 3x3 + 3x2 + 3x + C c) -48x4 – 18x – 6 + C d) 48x2 + 18x + 6 + C

Real Question 13

Find .

a) -3x3 + 4x2 – x + C b) -18x + 4 + C c) -3x3 + 2x2 – x + C d) -3x 3 + 2x 2 – x + C e) -3x3 + 4x2 + x + C

Page 9: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 14

Suppose I wish to approximate

using a Riemann sum; I want to divide [-3,3] into 100 equal-sized subintervals, and I want to use the left hand endpoint to sample the function in each subinterval. What sum should I evaluate to achieve my goals?

a) b) c)

d) e)

The length of the interval [-3,3] is 3 – (-3) = 6 units, and each subinterval length is dx = 6/100.Left hand rule, f(x) starts at left side of rectangles, at x = -3, (-3+6/100), (-3+12/100), (-3+18/100)...

Real Question 14

Suppose I wish to approximate

using a Riemann sum; I want to divide [-1,2] into 6 equal-sized subintervals, and I want to use the righthand endpoint to sample the function in each subinterval. What sum should I evaluate to achieve my goals?

a) b) c) d)

The length of the interval [-1,2] is 2 – (-1) = 3 units, and each subinterval length is dx = 3/6 = 1/2.Right hand rule, f(x) starts at right side of rectangles, at x = (-1+1/2), (-1+2/2)....(-1+5/2), (-1+6/2).

Page 10: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 15

Evaluate the definite integral

Hint: You can make this problem much easier if you think about the graph of y = x8sin x

a) b) c) d) e)

Since y = x8sin x is the multiplication of an even function x8 by an odd function sin x its integral over asymmetric region [-1,1] is zero, so the original integral is just for x4. Therefore:

= + = 0 + = = =

Real Question 15

Evaluate the definite integral .

a) b) c) 4 d) e) 2

= = = = =

Fake Question 16

Evaluate the definite integral .

a) b) c) d)

Using substitution: u = x4, du = 4x3 dx

= = = = – 0 =

Real Question 16

Evaluate the definite integral .

a) b) c) d)

Using substitution: u = x2, du = 2x dx

= = = =

Page 11: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 17

Evaluate the definite integral .

a) b) c) d)

Use integration by parts: let u = x, dv = cos(x) dx, then du = dx, v = sin(x)

= xsin(x) – = xsin(x) + = =

Real Question 17

Evaluate the definite integral .

a) -cos(1) – sin(1) b) cos(1) – sin(1) c) cos(1) + sin(1) d) sin (1) – cos (1)

Use integration by parts: let u = x, dv = sin(x) dx, then du = dx, v = -cos(x)

= -xcos(x) + = -xcos(x) + = -cos(1) + sin(1)

Fake Question 18Consider the region above the graph of y = x4 and below the graph of y = 5x + 6. What is the area of this region?

Vertical rectangle slices have height: (5x + 6) – (x4) and width dx, and curves meet when x4 = 5x + 6 gives x4 – 5x – 6 = 0 so x = -1 and x = 2 means x integrated from [-1,2], therefore integrate:

= = = =

Real Question 18Consider the region above the graph of y = x2 and below the graph of y = x + 2. What is the area of this region?

a) b) c) 3 d)

Vertical rectangle slices have height: (x + 2) – (x2) and width dx, and curves meet when x2 = x + 2 gives x2 – x – 2 = 0 and (x – 2)( x + 1) = 0 means x integrated from [-1,2], therefore integrate:

= = = =

Page 12: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 19Let A be the region in the plane consisting of those points (x,y) such that 0 ≤ y ≤ cos x, -П/2 ≤ x ≤ П/2. Rotate A around the x-axis in space. What is the volume of the resulting solid?

a) b c) d)

Real Question 19Let A be the region in the plane consisting of those points (x,y) such that 0 ≤ y ≤ 1 – x2. Rotate A aroundthe x-axis in space. What is the volume of the resulting solid?

a) b) c) d)

For each problem, the distance from the x-axis to each curve is the radius of circular cross-sectional area with r = y which form slices of cylinders with height dx, so that the solid's volume is the sum of infinitesimally thin cylinders with volume V = = , which span from y(-x) = 0 to y(x) =0.

For Fake Question 19: y = cos(x)

V = = = = =

For Real Question 19: y = 1 – x2

V = = = = =

Page 13: Coursera Calculus I Fake and Real Finals (April 2013)

Fake Question 20A paper cone is to be formed by starting with a disk of radius 12 cm, cutting out a circular sector, and gluing the new edges together. What is the maximum volume of such a cone?

a) cm3 b) cm3 c) cm3 d) 6 cm3 e) cm3

Real Question 20A paper cone is to be formed by starting with a disk of radius 9 cm, cutting out a circular sector, and gluing the new edges together. The size of the circular sector is chosen to maximize the volume of the resulting cone. How tall is the cone?

a) cm b) cm c) cm d) cm

[Circular Sector: https://en.wikipedia.org/wiki/Circular_sector]The geometry of both problems is the same, so the process ofsolving both is similar.

Let the disk radius be R, and the radius of the base of the resulting cone be r, and its height h.

Thus R2 = h2 + r2, and h = and r = with volume of cone V =

To find the maximum cone volume, write V in terms of just r or h and differentiate accordingly,however, the math comes out easier by replacing r in terms of h.

volume of cone: V = and writing r in term of h gives:

V = =

= = 0

= 0 and so h =

For Fake Question 20: h = = = cm and r2 = = 96 (so r = cm)

so then the maximum volume is: VMax = = = cm3

For Real Question 20: h = = = cm and r2 = = 54 (so r = cm)

so then the maximum volume is: VMax = = = cm3

However, in order to make the cone, we need to know the value of θ that maximizes the cone's volume.

From geometry, the arc length (perimeter) of the arcsection subtended by θ is: s = RθWhen rolled into a cone the arc length s becomes the circumference of the cone's circular base, so

s = Rθ = and thus θ = /R radians and using radians = 180 degrees

For Fake Question 20: θ = /R = /12 = /3 radians or deg = 293.94 degFor Real Question 20: θ = /R = /9 = /3 radians or deg = 293.94 deg

Page 14: Coursera Calculus I Fake and Real Finals (April 2013)

From above it seems the maximum θ for any R is /3 radians or ≈ 294 degrees. We worked backwards to find it. How do we directly prove this is true (or false) with calculus?

First, write V in terms of θ.

From above: s = Rθ = gives r = Rθ/

and then: h = =

and so: V = = =

and thus: V =

Now that we have V in term of just θ we can differentiate it with respect to θ and set it to zero.

= = 0

0 = = = = = = Max = radians

Thus, we proved that the circular sector θ that maximizes V for any R is radians (293.94 deg).

Then using Max in V = , the maximum cone volume becomes just: VMax =

This gives for R = 9 cm, Vmax = cm3, and for R = 12 cm, VMax = cm3, which match the values previously obtained.

So to make a cone of maximum volume cut from a circular sector of a disk, cut out an angular sectionof 360 – 293.94 = 66.06 (~66) degrees and use the bigger piece to form the maximum volume cone.

[Experiment: make cones from 3 same sized circles with 3 different θ s; θ = 294 deg, θ > 294 deg (say 310 deg), and θ < 294 deg (say 280 deg), and see which one holds the most salt.]