Calculus I, Math 170 Final Exam Solutions Spring 2008 Directions ...

Calculus I, Math 170
Final Exam Solutions
Spring 2008
Directions:
• Write your name in the top right corner of this page.
• The point total for the exam is 150.
• Show all work leading to your answers.
• Use complete sentences and correct notation wherever appropriate.
• Good luck!
(1) (10 points) First calculate dy
dx if x = y3 − 7y 2 + 2 at (x, y) = (−4, 1). Next compute the
equation of tangent line of the curve x = y 3 − 7y 2 + 2 at (−4, 1). Express your answer in
the form y = mx + b.
• Using implicit differentiation, 1 = 3y 2 y ′ − 14yy ′ . Solve for y ′ to get
y ′ =
• Evaluate at y = 1 to get y ′ = −1/11.
•
y − 1 1
= −
x + 4 11
1
3y 2 − 14y .
⇐⇒ y = −1 7
x +
11 11 .
(2) Differentiate each of the following functions. Simplify your answers.
(a) (4 points) f(x) = e 4x−x2
• Solution:
(b) (6 points) g(t) = (3t 2 − 4t) ln(2t)
• Solution:
(c) (6 points) F (x) = sin −1 (2x 3 − x)
• Solution:
f ′ (x) = (4 − 2x)f(x)
= 2(2 − x)f(x) .
g ′ (t) = (6t − 4) ln(2t) + (3t 2 − 4t) 2
2t
= (6t − 4) ln(2t) + (3t − 4) .
F ′ (x) =
=
1
1 − (2x 3 − x) 2 · (2x3 − x) ′
6x 2 − 1
1 − (2x 3 − x) 2
1

(3) (8 points) Use logarithmic differentiation to calculate the derivative of y = x ln x .
• ln y = ln(x ln x ) = ln x ln x
• So
• Solve for y ′ to get
y ′
y
= 2 ln x
x .
y ′ = 2x −1+ln x ln x .
(4) Find each of the limits
√
given below. If the limit does not exist, then state why. say so.
x + 5 − 3
(a) (4 points) lim
x→4 x − 4
• The limit meets the hypotheses of L’Hospital’s Rule. Thus,
√
x + 5 − 3 1
lim
= lim
x→4 x − 4 x→4 2 √ x + 5
= 1
6 .
• Alternatively, write
√ √
x + 5 − 3 x + 5 + 3
= √
x − 4 x + 5 + 3
x − 4
=
(x − 4)( √ x + 5 + 3)
1
= √
x + 5 + 3
for x = 4.
• Then
(b) (4 points) lim
x→−2− x2 + 2x − 8
x2 − 4
1
lim √ =
x→4 x + 5 + 3 1
6 .
• Write x 2 + 2x − 8 = (x − 2)(x + 4). Then
x 2 + 2x − 8
x 2 − 4
• (4 points) Since lim
x→−2 −
x + 4
=
x + 2
= 1 + 2
x + 2 .
2
= −∞, the original limit does not exist.
x + 2
(c) (4 points) Suppose that lim
x→1 f(x) = 0 and that g is a function satisfying the condition
|g(x)| ≤ 2 for all x = 1. Calculate lim
x→1 f(x)g(x).
• |g(x)| ≤ 2 for x = 1 implies that
−2f(x) ≤ f(x)g(x) ≤ 2f(x) .
• By the Squeeze Theorem, lim
x→1 f(x)g(x) = 0.

(5) (4 points) On the given axes, sketch graphs of the functions described below.
(a) a function that is continuous everywhere except at x = 3 but is continuous from the
left at x = 3
(b) a function that is continuous but not differentiable at x = 1
(6) Find the indefinite or definite integrals of each of the following:
(a) (6 points)
5x
(1 − 2x2 dx
) 2
• Set u = 1 − 2x2 . Then du = −4x dx.
• Then
5x
(1 − 2x2
−5
dx = u
) 2 4
−2 du
= 5
4 u−1 + c
=
y
y
5
4(1 − 2x 2 )
+ c .
x
x

ln 4 e
(b) (8 points)
ln 2
x + e−x ex dx; express your answer as a single logarithmic expression in
− e−x exact form. Do not give a decimal approximation.
• Set u = e x − e −x . Then du = (e x + e −x )dx.
• Then
ln 4
ln 2
ex + e−x ex 15/4 1
dx =
− e−x 3/2 u du
= ln u
15/4
u=3/2
= ln(15/4) − ln(3/2)
= ln(5/2) .
(7) Two particles collide sending one particle north and the other west. The particle moving
north is moving with a position function s(t) = 2t 3 − 14t 2 + 22t − 5 and the one moving
west is given by d(t) = 4t 2 − 16t, t ≥ 0 with t in seconds. At t = 10:
(a) (4 points) What are the velocities of both particles?
• s ′ (t) = 6t 2 − 28t + 22 and d ′ (t) = 8t − 16.
• At t = 10, s ′ = 342 ft/sec and d ′ = 64 ft/sec.
(b) (8 points) At what rate are the particles moving away from each other? Make your
answer accurate to two decimal points. Indicate the units in your answer.
• (The distance z = z(t) between the two particles at time t is determined by the
quadratic equation
z 2 = s 2 + d 2 .
• Then
z ′ z = ss ′ + dd ′ .
• Evaluate at t = 10 to get s = 815 ft and d = 240 ft and z = √ 721825 ≈ 849.60
ft. Then
z ′ =
(342)(815) + (64)(240)
√ 721825
≈ 346.15 ft/sec .
(c) (4 points) What is the acceleration of both particles?
• The acceleration of the particles is s ′′ = 12t − 28 and d ′′ = 8 in units of ft
sec 2 .
• Evaluate at t = 10 to get s ′′ = 92 and d ′′ = 8 ft
sec 2 .
(d) (4 points) When does the velocity of the particle moving north change its sign (+/-)?
• The velocity of the particle moving north changes it sign when s ′ (t) = 0.
• s ′ (t) = 6t 2 − 28t + 22 = (3t − 11)(2t − 2) so that s ′ changes sign at t = 1 and
t = 11/3 seconds.

(8) (16 points) A cylindrical can with no top is to be made to hold 12 cm 3 of liquid. Find the
dimensions of the can that will minimize the cost of the metal needed to make the can.
• Let r denote the radius and h the height of the can. Then
• In particular, h = 12/πr 2 .
• The surface area is
•
• f ′ (r) = 0 when r = (12/π) 1/3 .
85
80
75
70
65
60
V = πr 2 h .
f(r) = 2πrh + πr 2
= 24
r + πr2 .
f ′ (r) = −24
+ 2πr .
r2 • It can be checked by either the first or second derivative test that f has a local min at
(12/π) 1/3 and hence the absolute minimum of f on (0, ∞) occurs at this value as well.
• Solving for h when r = (12/π) 1/3 55
50
, we get h = r.
45
40
35
30
25
20
15
10
5
-1 1 2 3 4 5 6 7 8
V = 1.66
-5
-10
-15
-20
-25
-30
-35
-40
-45
-50
-55
g( x)
= 2⋅V
x +π⋅x2
xB = 0.78
g(x)) V B
(x,
1
V 3
= 0.81
π
g( xB)
= 6.17

(9) A function F is defined on the interval [0, 4]; its derivative is F ′ (x) = e sin x − 2 cos(3x).
With the use of a graphing calculator, answer the following questions.
(a) (4 points) Sketch F ′ in the window [0, 4] × [−2, 5].
(b) (2 points) On what interval is F increasing?
4
2
-2
-4
(c) (4 points) At what value(s) x, does F have a local maximum? Justify your answer.
-6
2 4
• F is increasing on the interval [0.27, 3.75] where the the graph of F ′ lies in the
first quadrant.
• F has a local extreme value at each x-intercept.
• F has a local maximum at the greater intercept c ≈ 3.75 since near c, F
Show Text Objects
′ goes
from being positive to negative.
(d) (4 points) How many inflection points does F have? Justify your answer.
• Inflections points are points of the form (a, F (a)) where the concavity changes.
• Because F ′′ is continuous, the inflection points correspond to the solutions of the
equation F ′′ (x) = 0.
• The graph of F ′ indicates that there are three such points, namely where F ′ has
either a local max or min.
(e) (4 points) Given that F (0) = 0, find F (4) accurate to three decimal places.
• Use the FTC
F (x) =
x
0
F ′ (t) dt
(f) (2 points) How many points on the graph of F ′ in the interval (0, 4) satisfy the Mean
Old Value Theorem?
• The slope of line thru (0, −1) and (4, −1.22) is ≈ −0.055.
• The graph indicates that there are 3 points x ∈ (0, 4) where the slope of the
tangent is −0.0055.

(10) Let R be the region given below in the first quadrant that is bounded by the graphs of
y = cos x and y = sin x. Give exact answers, not decimal approximations, to each of the
following questions.
(a) (4 points) Find the area of R.
• The two graphs intersect at (π/4, π/4).
R =
π/4
0
4
3
2
(cos x − sin x) dx = √ 2 − 1 .
1
R
-2 2 4 6
-1
-2
π/4
V = 2π (cos
-3 0
2 (x) − sin 2 x) dx .
• Use the trig identity cos(2x) = cos2 x − sin2 x.
-4
-5
-6
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(b) (6 points) Find the volume of the solid created by rotating R around the x-axis.
• The volume is
• Then
V = 2π
π/4
0
= π sin(2x)
= π .
cos(2x) dx
π/4
0

(11) Consider the differential equation y ′ = x(y − 1) 2 .
(a) (4 points) On the axes provided, sketch a slope field for the differential equation at the
2
indicated points.
3
1
-4 -2 2 4
(b) (6 points) Find the particular solution y = -2f(x)
to the differential equation with the
initial condition f(0) = −1.
• The equation is separable. Rewrite as y ′ (y − 1) −2 = x. Integrating leads to
-1
-3
−(y − 1) −1 = x2
• y(0) = −1 implies that c = 1/2.
• The solution is y = 1 − 2
x 2 +1 .
(c) (4 points) What is the range of the solution found in part b)?
2
-4
-5
-6
+ c
• The range of y is [−1, 1).
• The minimum −1 occurs at x = 0: since 2
x 2 +1
1 − 2
x2 ≤ 1 − 2 = −1 .
+ 1
• The horizontal asymptote of y is 1 since 2
x 2 +1
≤ 2,
→ 0 as x → ±∞.
(12) (6 points) Let f be a function that is defined on some interval containing a number a. Let
L ∈ R. The truth of the sentence “lim f(x) = L” is defined as the truth of the following
x→a
implication: given any ɛ > 0, there exists a δ > 0 such that for any number x,
0 < |x − a| < δ =⇒ |f(x) − L| < ɛ .
Prove that lim
x→2 (3x − 5) = 1 by showing directly that the above definition holds. In other
words, using nothing more than elementary algebra, verify the definition.
Proof. Observe that |f(x) − L| = |3x − 6| = 3|x − 2|. It follows that
|f(x) − L|
|x − 2|
= 3
for x = 2. Given ɛ > 0, choose 0 < δ ≤ ɛ/3. Then
0 < |x − 2| < δ =⇒ |3x − 6| < ɛ .