Midterm Exam 2 Review (Solutions)
Math 251, Winter ’16, Gedeon
Midterm Exam 2 is scheduled for Monday, February 15th, at 3:00pm (in class). Remember that
calculators are not allowed for this exam. The following exercises are just a sample of what may be
on the exam. Do not restrict your studying to this review, and it is no guarantee that everything on
this review will be on the midterm. These exercises are all examples of what knowledge/skills you
should have gotten from the material covered so far. Other very important resources for studying
include the WebWork assignments, lecture guides and exercises in the text.
These solutions are not complete solutions. I have provided enough information to hint at how
to do each problem, but I would expect you to show more work on the exam.
1. The Concept Check, True-False Quiz, and Exercises in the Chapter 3 Review of the textbook
(except Exercises 43,44,76).
Answers to odd problems in the textbook.
2. Compute the indicated derivative.
(a) v 0 ( π4 ), where v(t) = tan(t) − 2 csc(t)
√
v 0 (t) = sec2 (t) + 2 csc(t) cot(t), and so v 0 (π/4) = 2 + 2 2
d ex sin(x) (b)
dx x2 + 1
x=0
(x2 + 1)(ex cos x + ex sin x) − ex sin x(2x)
The derivative is
, and so at x = 0 we get 1.
(x2 + 1)2
5
(c) j 00 (x) for j(x) = x4 − √
5
x
0
3
−6/5
j (x) = 4x + x
, and so j 00 (x) = 12x2 − 56 x−11/5
(d)
dy
3x2
for y =
dx
sin(x)
dy
6x sin(x) − 3x2 cos(x)
=
dx
sin2 (x)
√
d2 y
for y = 2x x
2
dx
d2 y
dy
x −3/2 + ln(2)2x x−1/2 + (ln(2))2 2x x1/2
= ln(2)2x x1/2 + 12 2x x−1/2 , and so
= −1
4 2 x
dx
dx2
(f) f 00 (x) with f (x) = sin(cos(x)).
f 0 (x) = − sin(x) cos(cos(x)) and hence f 00 (x) = − sin2 (x) sin(cos(x))−cos(x) cos(cos(x)).
dy
2
(g)
for x2 y 2 + x4 = ey − 1
dx
dy
2xy 2 + 4x3
=
Using implicit differentiation,
dx
2yey2 − 2x2 y
dx
(h)
at the point when x = 2, y = 1, given x arctan(y) = y ln(0.5x)
dy
x
− ln(0.5x)
dx
1+y 2
= y
Using implicit differentation,
. At the point when x = 2 and y = 1,
dy
− arctan(y)
x
dx
4
we get
=
dy
2−π
(e)
1
(t + 1)2 ln(t + 1)
et
t
e (2(t + 1) ln(t + 1) + (t + 1)) − (t + 1)2 ln(t + 1)et
and hence g 0 (0) = 1.
g 0 (t) =
2t
e
√
(j) h0 (x), with h(x) = log2 (csc( x2 − 1))
√
−x cot( x2 − 1)
0
√
h (x) =
ln(2) x2 − 1
(i) g 0 (0), given g(t) =
3. Use implicit differentiation to calculate the derivative of y = tan−1 (x).
dy
1
d
Applying implicit differentiation to tan(y) = x yields
=
and hence
[tan−1 (x)] =
2
dx
sec (y)
dx
1
1 + x2
4. For what values of x is the tangent line to f (x) = x2 ex horizontal?
The tangent line is horizontal when its slope f 0 (x) is equal to zero. Finding f 0 (x) = x2 ex +
2xex = xex (x + 2), we see that f 0 (x) = 0 when x = 0 and x = −2.
5. Write the equation of the tangent line to y = ex − 3x2 + sin(x) at x = 0.
dy dy
Since
= ex − 6x + cos(x), we find that the slope of the line is
= 2. We also know
dx
dx x=0
that when x = 0, y = 1. Thus, the equation of the tangent line (with slope 2 and through
the point (0, 1)) is y = 2x + 1.
6. Suppose that f (2) = −3, g(2) = 2, f 0 (2) = 5, and g 0 (2) = −2. Find h0 (2) if...
(a) h(x) = f (x)g(x)
h0 (x) = f 0 (x)g(x) + f (x)g 0 (x), and so at x = 2, we get h0 (2) = 16.
f (x)
(b) h(x) =
g(x)
g(x)f 0 (x) − f (x)g 0 (x)
h0 (x) =
, and so at x = 2, we get h0 (2) = 1.
g(x)2
(c) h(x) = 3f (x) − 2x + 3 − g(x)
h0 (x) = 3f 0 (x) − 2 − g 0 (x), and so at x = 2, we get h0 (2) = 15.
√
√
7. Use linear approximation of f (x) = x at x = 100 to estimate 99.8. Is this an overestimate
or underestimate?
For values of x near 100, f (x) ≈ f 0 (100)(x − 100) + f (100). Computing f 0 (100) and f (100),
√
1
we have that for values of x near 100, x ≈ 20
(x − 100) + 10. In particular, when x = 99.8,
√
we have 99.8 ≈ 9.99. This is an overestimate because f 00 (x) < 0 near x = 100.
8. For each of the following functions, on what interval(s) is the function increasing? Decreasing?
Concave up? Concave down?
π (a) f (x) = ln cos
x
with −1 < x < 1
2
f 0 (x) = − π2 tan( π2 x), which is positive for −1 < x < 0 and negative for 0 < x < 1.
Hence, f is increasing on the interval (−1, 0) and decreasing on the interval (0, 1).
2
f 00 (x) = − π2 sec2 ( π2 x), which is always negative. Hence, f is concave down on the
interval (−1, 1).
2
(b) g(x) = ln(x2 + 1)
2x
g 0 (x) = 2
, which is positive for x > 0 and negative for x < 0. Hence, g is increasing
x +1
on the interval (0, ∞) and decreasing on the interval (−∞, 0).
−2(x − 1)(x + 1)
g 00 (x) =
, which is positive for −1 < x < 1 and negative for x < −1 or
(x2 + 1)2
x > 1. Hence, g is concave up on the interval (−1, 1) and concave down on the intervals
(−∞, −1) and (1, ∞).
9. When the brightness x of a light source is increased, the eye reacts by decreasing the area R
of the pupil. The experimental formula
R=
40 + 24x0.4
1 + 4x0.4
has been used to model the dependence of R on x when R is measured in square millimeters
and x is measured in appropriate units of brightness. Find the rate of change of the reaction
with respect to x, called the sensitivity.
The sensitivity to brightness is given by
−54.4x−0.6
dR
=
square millimeters per unit
dx
(1 + 4x0.4 )2
brightness.
10. Decide whether each of the following statements is true or false.
(a) It is impossible to find y 0 from an equation in which it is impossible to isolate y.
(b) The acceleration of an object is the instantaneous rate of change in its velocity.
p(x)
p0 (x)
(c) If r(x) =
, then r0 (x) = 0
.
q(x)
q (x)
(a) is False (use implicit differentiation); (b) is True (by definition); (c) is False (use quotient
rule).
11. Suppose that an ant is walking around on the edge of a yardstick for eight seconds. Its
position (in inches) after t seconds (0 ≤ t ≤ 8) is given by s = 31 t3 − 3t2 + 5t + 10.
(a) Find the ant’s velocity (including units) after t seconds.
s0 (t) = t2 − 6t + 5 inches per second.
(b) Find the ant’s acceleration (including units) after t seconds.
s00 (t) = 2t − 6 in/s2 .
(c) When is the ant moving in the positive direction?
The ant is moving in the positive direction when s0 (t) > 0, and so this happens on the
intervals (0, 1) and (5, 8).
(d) When is the ant speeding up?
The ant is speeding up when both velocity and acceleration are positive, on (5, 8), and
when both velocity and acceleration are negative, on (1, 3).
(e) What is the total distance traveled by the ant during the first two seconds?
Note that the ant changes direction at t = 1. In the first second, he travels |s(1)−s(0)| =
7/3 inches. In the second second, he travels |s(2)−s(1)| = 5/3 inches. In total, he travels
4 inches during the first two seconds.
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12. An oscillating lemming population is modeled by the function
π P (t) = 200 cos
t + 60t
10
where t is measured in hours from now. Use linear approximation to estimate the population
5 minutes from now.
π
P 0 (t) = −20π sin( 10
t) + 60. We will use linear approximation at t = 0 since 5 minutes
(t = 5/60 = 1/12 hours) is close to 0. We have P (0) = 200 and P 0 (0) = 60, and so for values
of t near 0, P (t) ≈ 60t + 200. In particular, P (1/12) ≈ 205.
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