Fall 2014 – MATH 151, Sections 549-551
Quiz #4 Solutions
DIRECTIONS. Read each problem carefully and work it out in the space provided.
Put a box around the answer you believe best answers the question. Calculators are
not allowed.
2
?
Problem 1. What is the derivative of the function f (u) = 1−u
1+u2
0
0
0
with g(u) = 1 − u2 and h(u) = 1 + u2 ,
Solution. Using the quotient rule hg = hg h−gh
2
we have
(1 + u2 )(−2u) − (1 − u2 )(2u)
f (u) =
(1 + u2 )2
−2u − 2u3 − 2u + 2u3
=
(1 + u2 )2
−4u
=
.
(1 + u2 )2
0
So the correct answer is (c).
Problem 2. Find the equation of the tangent line to f (x) = x +
√
x at the point (4, 6).
Solution. The slope of the tangent line to the graph of f (x) at (4, 6) is obtained by
1
evaluating the derivative of f (x) at x = 4. Rewriting f (x) = x + x 2 and using the power
law, we compute:
1 1
1
f 0 (x) = 1 + x− 2 = 1 + √ .
2
2 x
Hence the slope of the tangent line is
1
1
5
f 0 (4) = 1 + √ = 1 + = .
4
4
2 4
Now that we know the slope, and we have a point on the line – namely (4, 6) – we can
use the point-slope form to find the equation:
5
5
y − 6 = (x − 4) = x − 5
4
4
which implies that y = 54 x + 1. Thus (a) is the correct answer.
1
Math 151 Fall 2014
Quiz #4 Solutions
2
Problem 3. The position function of a particle is given by s = t3 − 4.5t2 − 7t, t ≥ 0,
with s(t) given in meters and t in seconds. When does the particle reach a velocity of 5
m/s?
Solution. The velocity function is the derivative of the position function. That is, if v(t)
denotes the velocity of the particle at time t, we have
v(t) = s0 (t) = 3t2 − 9t − 7.
We are asked to find the time t for which v(t) = 5 m/s. That is, we seek the value(s) for
t that satisfy 3t2 − 9t − 7 = 5, or subtracting 5 from both sides,
0 = 3t2 − 9t − 12 = (3t + 3)(t − 4).
We now see that t = −1 and t = 4 satisfy the equation. Since in the statment of the
problem, we are told t ≥ 0, the correct answer is t = 4, which is choice (b).
Problem 4. What is
dy
dx
of y = cos x − 2 tan x?
Solution. By the sum rule, we have
d
d
dy
=
cos x − 2 tan x.
dx
dx
dx
Now,
d
dx
cos x = − sin x and
d
dx
tan x = sec2 x. Thus,
dy
= − sin x − sec2 x.
dx
So the correct answer is (a).