Final exam practice UCLA: Math 3B, Fall 2016math.ucla.edu/~noah/math3b-f16/final-practice-s.pdf · Final exam practice UCLA: Math 3B, Fall 2016 Instructor: Noah White Date: Monday,

Final exam practice

UCLA: Math 3B, Fall 2016

Instructor: Noah WhiteDate: Monday, November 28, 2016Version: practice.

• This exam has 7 questions, for a total of 84 points.

• Please print your working and answers neatly.

• Write your solutions in the space provided showing working.

• Indicate your final answer clearly.

• You may write on the reverse of a page or on the blank pages found at the back of the booklet howeverthese will not be graded unless very clearly indicated.

• Non programmable and non graphing calculators are allowed.

Name:

ID number:

Discussion section:

Question Points Score

1 20

2 15

3 11

4 12

5 10

6 8

7 8

Total: 84

UCLA: Math 3B Final exam practice (solutions) Fall, 2016

1. Each of the following questions has exactly one correct answer. Choose from the four options presentedin each case. No partial points will be given.

(a) (2 points) The function f(x) = e−x − ex is

A. Always increasing.

B. Always decreasing.

C. Always concave up.

D. Always concave down.

(b) (2 points) The function f(x) = x2

x−1 has a

A. Horizontal asymptote at y = 1.

B. Vertical asymptote at x = 2.

C. Slanted asymptote with gradient −1.

D. Slanted asymptote with gradient 1.

(c) (2 points) The function f(x) = 4x− 11−x has a critical point at

A. x = 0

B. x = 0.5

C. x = −1

D. x = −0.5

(d) (2 points) The function f(x) = 1− e−(2x−4)2 has a

A. minimum at x = 2.

B. maximum at x = 2.

C. minimum x = 0.

D. maximum x = 0.

(e) (2 points) The definite integral∫ 1

−2 3x2 + 1 dx has a value of

A. 12.

B. 4.

C. 0.

D. −2.


(f) (2 points) The definite integral∫ π/20

sin3 x cosx dx has a value of

A. 0.

B. 14 .

C. 0.5.

D. 2π.

(g) (2 points) The solution of the differential equation dydt = 2y when y(0) = 3 has

A. y(1) = 2e2

B. y(0.5) = 3e.

C. y(1) = 3e.

D. y(0.5) = 3e0.5.

(h) (2 points) The solution of the differential equation dydt = 2te−y when y(0) = 0 has

A. y(1) = ln 4

B. y(0.5) = ln 2.

C. y(1) = ln 2.

D. y(0.5) = ln 4.

(i) (2 points) The differential equation dydt = y(1− y)(y − 3) has a

A. unstable equilibrium at y = 2.

B. unstable equilibrium at y = 0.

C. stable equilibrium at y = 3.

D. stable equilibrium at y = 1.

(j) (2 points) The differential equation dydt = ln(y) has a

A. stable equilibrium at y = 0.

B. unstable equilibrium at y = e−1.

C. stable equilibrium at y = e.

D. unstable equilibrium at y = 1.


2. Let f(x) = 11+e−2x . Note that f ′(x) = 2e−2x

(1+e−2x)2 and f ′′(x) = 4e−2x(e−2x−1)(1+e−2x)3 .

(a) (2 points) Find the x and y intercepts of f(x).

Solution: No x-intercepts, y-intercept at y = 0.5.

(b) (1 point) Does f(x) have any horizontal asymptotes? If so what are they?

Solution: We need to evaluate

limn→∞

f(x) = limn→∞

1

1 + e−2x= 1.

and

limn→−∞

f(x) = limn→−∞

1

1 + e−2x= 0.

So we have horizontal asymptote in the positive direction at y = 1 and in the negative directionat y = 0.

(c) (1 point) Does f(x) have any vertical asymptotes? If so what are they?

Solution: The denominator of f(x) can never be zero so the function is always well defined,thus no vertical asymptotes.

(d) (2 points) For what x is the first derivative f ′(x) positive?

Solution: The denominator is always positive, as is the numerator. Thus f ′(x) is positive forall x.


(e) (2 points) For what x is the second derivative f ′′(x) positive?

Solution: Again, the denominator is always positive. The numerator is positive when

e−2x > 1

i.e. when −2x > 0 thus when x < 0.

(f) (3 points) On the graph provided, sketch f(x)

y

x


(g) (4 points) List the local maximums and minimums of f ′(x) (note: this question if asking about theextrema of the derivative of f !)

Solution: To find the local minimums and maximums, we first find the critical points. Thisis were the derivative of f ′(x) equal zero, i.e. where f ′′(x) = 0 or where f ′′(x) is undefined.

0 = f ′′(x)

=4e−2x(e−2x − 1)

(1 + e−2x)3

0 = 4e−2x(e−2x − 1)

0 = e−2x − 1

e−2x = 1

−2x = 0

x = 0

Since f ′′(x) is defined everywhere, the only critical point occurs at x = 0. Using the firstderivative test, we see that this is a local maximum.


3. A heavy chain hangs over a ledge. The chain is 4 m in length and weighs a total of 10 kg. Attached tothe end of the chain is a full bucket weighing 20 kg. In this question we will calculate the work neededto pull the chain and bucket up to the top of the ledge. Recall that

W = Fd

that is, the work done, is the force needed multiplied by the distance moved. You may assume theacceleration due to gravity is 10 m/s2.

(a) (1 point) If we ignore the weight of the chain, how much work needs to be done to pull the bucketup?

Solution:10 · 20 · 4 = 800 J

(b) (1 point) Suppose we divide the chain into n intervals, of constant length ∆h. What is ∆h in termsof n?

Solution:

∆h =4

n

(c) (1 point) Let h = 0 be the top of the chain and h = 4 be the bottom. Let hk be the depth of thetop of the kth interval of chain below the ledge (where the first interval is the 0th), thus h0 = 0.What is hk in terms of ∆h and k?

Solution:hk = k∆h

]


(d) (2 points) How much does the kth interval, of chain weigh? Your answer should be in terms of ∆h.

Solution:2.5∆h

(e) (2 points) If n is large enough, we can say approximately that the kth interval of chain needs to bepulled up hk m. Using this approximation, how much work is needed to be done to pull up the kth

interval of chain?

Solution: We need to pull the interval of chain a distance of hk so

2.5∆h · 10 · hk = 25hk∆h


(f) (2 points) Write a Riemann sum which represents the total amount of work needed to pull thechain and bucket up.

Solution:

W = 800 + limn→∞

n−1∑k=0

25hk∆h

(g) (2 points) Use an integral to evaluate the Riemann sum above.

Solution:

W = 800 +

∫ 4

0

25h dh

= 800 +

[25

2h2]40

= 800 +

(25

216

)= 1000 J


4. Solve the following differential equations. If no initial condition is given, find the general solution.

(a) (4 points) dydt = 2− y.

Solution: We separate variables and integrate∫1

2− ydy =

∫dt

So we get that− ln(2− y) = t+ C

Thusy(t) = 2− Ce−t

is the general solution.

(b) (4 points) dydt = y2(1− 2t) where y(0) = 1.

Solution: We separate variables and integrate∫1

y2dy =

∫1− 2t dt

So we get that

−1

y= t− t2 + C

So we get that

y(t) =1

t2 − t+ C.

Using the fact that y(0) = 1 we have that

1 =1

C

so C = 1 and the solution is

y(t) =1

t2 − t+ 1.

(c) (4 points) dydt = e−y sin t where y(0) = 0.

Solution: We separate variables and integrate∫ey dy =

∫sin t dt

So we get thatey = − cos t+ C

Thus the general solution isy(t) = ln(C − cos t).

Using the fact that y(0) = 0 we get that 0 = ln(C − 1), so C = 2. Thus the solution is

y(t) = ln(2− cos t).


5. In this question we will investigate the behaviour of the solutions of

dy

dt= a− (ln y)2 − 2

(a) (4 points) Draw a bifurcation diagram for this equation with parameter a. Make sure to label theregions of your diagram with up/down arrows according to the direction of the derivative.

y

x

(Thecomputer has added some weird bumps to the function above. Please ignore them!)

(b) (2 points) Draw a phase diagram when a = 4 and sketch the solution if y(0) = 1.

Solution:


t

y

0

e−√2

e√2


(c) (2 points) Draw a phase diagram when a = 2 and sketch the solution if y(0) = 0.5.

Solution:

t

y

0

1

(d) (2 points) Draw a phase diagram when a = 1 and sketch the solution if y(0) = 1.

Solution:

t

y

0


6. (8 points) A new internet service provider (ISP) opens. They have horrible customer service so theycalculate that the probability a given customer is still a customer in t months time is

e−0.2t.

In order to remain profitable, the ISP must have approximately 3000 customers at any given time (inthe long term). How many customers must the ISP gain per month in order to remain profitable?

Solution: Let a fix a number x and find the number of customers the ISP has x months into thefuture. If we use t to denote time then now will be t = 0 and t = x will be x months into the future.

Suppose the ISP is gaining customers at a rate of a customers per month. The if we split the periodfrom t = 0 to t = x into n subintervals, we have a∆t customers arriving in each time period (where∆t = x/n).

Let tk be the time at the start of the kth subinterval. Now, let us concentrate on the kth subinterval.Each of the customers that arrive at the ISP during this time will have been with the ISP for x− tkmonths. So when t = x

a∆te−0.2(x−tk)

of these customers will still be with the ISP. Adding all of these contributions up (and taking thelimit as n→∞) we see that the total number of customers with the ISP at t = x is

limn→∞

n−1∑k=0

a∆te−0.2(x−tk).

This is a Riemann sum so we can turn it into an integral and evaluate:∫ x

0

ae−0.2(x−t) dt =[ a

0.2e−0.2(x−t)

]x0

=a

0.2

(1− e−0.2x

).

Thus in the long term (when x → ∞) the ISP will have a/0.2 = 5a customers. Since we want theISP to have 3000 customers, we see that we need a to be at least 600.


7. (8 points) A straight, and very narrow steel bar is to be carried along a 1 meter wide corridor. Thecorridor makes a 90 degree turn. What is the longest length of steel pipe that can be carried down thecorridor and around the corner?

Solution: This is a tricky question! Consider the diagram below.

x

1 m

1 m

Here is what this picture represents. Imagine if one end of the pipe is x m from the corner. Then thegreen line shows the maximum length that the pipe could possible extend down the hallway. Callthis length L(x). Now L(x) depends on x and we want to work out what is the maximum lengthof pipe we can carry down the hallway. In other words, we want to know, what is the minimum ofL(x)? First let us find an expression for L(x). We split this up into two pieces.

On the diagram we see two triangles. L(x) is the sum of the lengths of the hypotenuses. The lowertriangle obviously has a hypotenuse of length√

1 + x2.

Now the upper triangle is similar to the lower triangle. If the hypotenuse has length a then bycomparing the ratios of the side lengths we see that

a√1 + x2

=1

x

so we see that

L(x) =√

1 + x2 +

√1 + x2

x=

(1 +

1

x

)√1 + x2.

To find the minimum of this function, first find its critical points. Using the product rule:

L′(x) =

(− 1

x2

)√1 + x2 +

(1 +

1

x

)1

2

2x√1 + x2

=x3 − 1

x2√

1 + x2.


(after a little bit of algebra!)

Now we see that the function L has critical points at x = 1 and x = 0. However when x → 0 wehave L(x) → ∞ so this cannot be a minimum. To check that x = 1 is a minimum we can just usethe first derivative test.

x < 1 1 > 1L′(x) − 0 +

so it is indeed a minimum. Thus L(1) = 2√

2 is the maximum length of pipe we can carry down thehallway.


This page has been left intentionally blank. You may use it as scratch paper. It will not be graded unlessindicated very clearly here and next to the relevant question.


This page has been left intentionally blank. You may use it as scratch paper. It will not be graded unlessindicated very clearly here and next to the relevant question.

Documents

Final exam practice UCLA: Math 3B, Fall 2016math.ucla.edu/~noah/math3b-f16/final-practice-s.pdf · Final exam practice UCLA: Math 3B, Fall 2016 Instructor: Noah White Date: Monday,