Math 1350 Fall 2012 Abrams / O’Neil
Practice Exam 2 - Solutions
Show all your work. No credit will be given for answers which are not accompanied by supporting
computations. Circle your answer when appropriate. Use the back of the sheet if necessary. As usual, notation
counts! Be clear and precise with your answers, justifying responses appropriately. Make sure to include the
units in your answer when warranted. You may use a standard calculator, one which does not have graphing
capability or a QWERTY keyboard. Good luck!
1. (8 pt total) A particle moves on a vertical line so that its coordinate at time t (in seconds) is y = t3 −12t+3
(in meters) for t ≥ 0.
(a) Find the velocity function of the particle.
v(t) = s0 (t) = 3t2 − 12
(b) Find the acceleration function of the particle.
a(t) = v 0 (t) = (use part a) 6t
(c) At what time or times is the object momentarily not moving?
This is asking for the time(s) at which v(t) = 0. From part (a) we solve and get 3t2 − 12 = 0, so
that 3(t + 2)(t − 2) = 0, which give t = 2 or t = −2. Since we are told t ≥ 0, we get that t = 2 seconds is the
only time at which the object is momentarily not moving.
(d) On what time interval(s) is the particle moving upward (i.e., when is the particle’s position increasing)
and on what time interval(s) is it moving downward?
This asks for times when v(t) > 0 (resp. v(t) < 0). But from part (c) we know v(t) = 0 when t = 2
sec. Plugging in any value between 0 and 2 (e.g. t = 1) shows that v(t) < 0 on this interval. Similarly, we see
that v(t) > 0 when t > 2. So the particle is moving downward initially on the interval 0 ≤ t < 2 sec, and then
upward when t > 2 sec.
(e) Find the distance that the particle travels in the time interval 0 ≤ t ≤ 3.
We need to analyze separately the two intervals indicated in part (d). On the interval 0 ≤ t < 2 the
position of the particle changes from s(0) to s(2), namely position 03 − 12 · 0 + 3 to position 23 − 12 · 2 + 3, or
3 to −13. This is 16 meters. On the interval 2 ≤ t ≤ 3 the position of the particle changes from s (2) to s(3),
namely position 23 − 12 · 2 + 3 to 33 − 12 · 3 + 3, or −13 to −6. This is a change of 7 meters. So the total distance
travelled in the time interval 0 ≤ t ≤ 3 is 16 + 7 = 23 meters.
2. (4 pt total) A student who took Calculus 1 during a recent semester was asked to compute the derivative
of the function g(x) = x2 · e3x . The student’s answer was g 0 (x) = 2x · 3e3x
(a) Describe in words a misconception that this student had about the derivative process.
This student thought that the derivative of a product was just the product of the derivatives. In
other words, they thought that derivatives and products work together in the same way that derivatives and sums
work together. That is NOT true.
(b) Give the correct computation of g 0 (x).
d
d
(e3x ) + dx
(x2 ) · e3x = x2 · e3x · 3 + 2x · e3x (we
We use the product rule for derivatives. g 0 (x) = x2 · dx
d
have used the chain rule to compute dx
(e3x ) ). In factored form this equals e3x (3x2 + 2x).
3. (5 pt total) The cost function for a certain commodity is given as C(x) = 2000 + 3x + 0.01x2 + 0.0002x3
(in thousands of dollars).
(a) Find the marginal cost function.
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Marginal cost is the derivative of the cost function. C 0 (x) = 3 + 0.02x + 0.0006x2 .
(b) Find C 0 (100) and explain its meaning in real-world business terms. What does C 0 (100) predict?
Using what we just did in part (a), C 0 (100) = 3 + 0.02 · 100 + 0.0006(100)2 = 3 + 2 + 6 = 11. Thus the
marginal cost of the 101st unit (i.e., the cost of producing one more unit if we are already producing 100 units)
is 11 , which by the given information indicates an actual cost of $11,000.
(c) Find the exact cost of producing the 101st unit.
We just compute C(101) − C(100). We know that we should get some number close to 11. C(101) −
C(100) = 2000 + 3 · 101 + 0.01(101)2 + 0.0002(101)3 − [2000 + 3 · 100 + 0.01(100)2 + 0.0002(100)3 ]=
(2000 + 303 + 102.01 + 206.0602) − (2000 + 300 + 100 + 200) = 11.07, so that the actual cost is $11,070.
4. (4 pt total) The amount of radioactive waste (in kilograms) in a disposal site at time t (in years) is given
t
by the formula R(t) = t2 +1
.
(a) Find the rate at which R is changing when t = 10 years.
2
(by the quotient rule), which simplifies to
Rate of change is the derivative. So R0 (t) = (t +1)·1−t(2t)
(t2 +1)2
1−t2
.
(t2 +1)2
So R0 (10) =
1−102
(102 +1)2
=
−99
10201
= −.0097 kilograms per year.
(b) Is R(t) growing or shrinking after 10 years? Justify.
Since the derivative R0 (10) is negative, R(t) is shrinking after ten years.
5. (2 pt total) Complete these two versions of the Chain Rule:
(i) If y is a differentiable function of u, and in turn u is a differentiable function of x, then
dy du
dy
= du
· dx
dx
(ii)
d
(f (g(x))
dx
= f 0 (g(x)) · g 0 (x).
(3 pt) Find the slope of the tangent line to the curve y = cos(3x) when x = π2 .
y 0 = − sin(3x) · 3 = −3 sin(3x). So the slope of the tangent line when x =
(−3)(−1) = 3.
6.
π
2
is −3 sin(3 π2 ) =
7. (6 pt total) In this problem f (x) denotes the function (2x + 1)3 .
(a) Using the chain rule, compute f 0 (x).
f 0 (x) = 3(2x + 1)2 · 2 = 6(2x + 1)2 .
(b) Do some algebra and ’multiply out’ the expression for f (x). Then, compute f 0 (x) from the resulting
expression.
f (x) = (2x + 1)3 = 8x3 + 3 · 4x2 + 3 · 2x + 1 = 8x3 + 12x2 + 6x + 1.
From this, we see f 0 (x) = 24x2 + 24x + 6.
(c)
TRUE, we have just seen an example of this in parts (a) and (b).
If you are given
a function f (x) and asked to compute its derivative, there might be more than one formula from the table of
derivatives which can be used to complete the task.
(d) Find the equation of the tangent line to the curve y = (2x + 1)3 when x = 0.
Using EITHER form of f 0 (x) from parts (a) or (b), we see that when x = 0, then f 0 (0) = 6. Since
f (0) = (2 · 0 + 1)3 = 1, the equation of the tangent line is given by y − 1 = 6(x − 0), which simplifies to y = 6x + 1.
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8. (2 pt) Derivatives and inverse functions.
Suppose that by actually computing the appropriate limit you are able to compute f 0 (x) for some
function f . Suppose also that f has an inverse function, call it g. Describe in words how you might go about
finding g 0 (x) without computing a limit.
dy
. Trade in the equation y = g(x) for its inverse
Write the equation y = g(x), We want to find dx
d
−1
equation x = g (y) = f (y). Now compute dx of both sides of this equation using implicit differentiation.
dy
Finally, solve for dx
.
9.
(1 pt each) Derivatives of specific functions.
(a) If y = arcsin(x), then y 0 =
(b) If y = ln(x), then y 0 =
√ 1
1−x2
1
x
10. (6 pt) Find the equation of the tangent line to the curve 4x3 + xy − y 3 = −6 at the point (−1, 1).
dy
dy
d
of both sides. We get 12x2 + x · dx
+ 1 · y − 3y 2 dx
= 0. Now
Use implicit differentiation. Take dx
−12(−1)2 −1
dy
dy
dy
−12x2 −y
dy
2 dy
2
solve for dx : x · dx + −3y dx = −12x − y, so dx = x−3y2 . So at the point (−1, 1) we get dx = (−1)−3(1)2 = 13
.
4
13
13
17
So the equation of the tangent line is y − 1 = 4 (x − (−1)), which yields y = 4 x + 4 .
11. (5 pt) A boat is pulled into a dock by a rope attached to the bow of the boat and passing through
a pulley on the dock that is 1 meter higher than the bow of the boat. If the rope is pulled in at a rate of 1
meter/second, how fast is the boat approaching the dock when it is 5 meters from the dock?
Let x denote the amount of rope between the bow of the boat and the pulley, and let y denote the
distance from the boat to the dock. Then y 2 + 12 = x2 . We are told that dx/dt = −1m/ sec . We are asked to
= 2x dx
, so
find dy/dt when y = 5.
Take dtd of both sides of the equation, and use the chain rule to get 2y dy
dt
dt
that solving for
dy
dt
gives
dy
dt
=
x dx
dt
y
. The only unknown piece of information is the value of x when y = 5. But
√
√
26(−1)
plugging into the original equation gives 52 + 11 = x2 , so that x = 26. Thus dy
=
= −1.0198m/ sec.
dt
5
So the boat is approaching the dock at 1.0198 meters/second.
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