Practice Problems 8 for 1110 Solutions
October 29, 2016
Exercise 1. A slow freight train chugs along a straight track. The distance it has traveled after x hours is given by a function
f (x). An engineer is walking along the top of the box cars at the rate of 3 mi/hr in the same direction as the train is moving.
The speed of the man relative to the ground is
(a) f (x) + 3
(b) f 0 (x) + 3
(c) f (x) − 3
(d) f 0 (x) − 3
Solution. (b) The speed of the train is the derivative of the trains position, so the train is moving at f 0 (x) mi/hr. The engineer’s
speed relative to the ground is the speed of the train plus his speed relative to the train.
Exercise 2. Suppose you cut a slice of pizza from a circular pizza of radius r, as shown.
As you change the size of the angle θ, you change the area of the slice, A = 21 r2 θ. What is the change in area?
(a) rθ
(b)
1 2
2r
(c) Not enough information
Solution. (b) Since we are changing the angle θ and keeping r constant, make sure to differentiate the area A with respect to
θ.
f (x)
d
Exercise 3. For some function f (x) we know that f (1) = 1 and f 0 (1) = 3. Compute dx
|x=1 . Hint: use the quotient
2
x
rule.
Solution. By the quotient rule we have
2 0
d f (x)
x f (x) − f (x)2x
(1)2 f 0 (1) − f (1)2(1)
|
=
|
=
= 3 − 2 = 1.
x=1
x=1
dx
x2
x4
(1)4
Exercise 4. Write the area of a circle as a function of circumference. What is the instantaneous rate of change of the area of a
circle with respect to circumference?
Solution. The equations for area and circumference for a circle are A = πr2 and C = 2πr. Solving for r in the second
equation gives r = C/2π, and plugging into the first gives A(C) = π(C/2π)2 = C 2 /4π. Then the derivative with respect to C is
A0 (C) = 2C/4π = C/2π.
Exercise 5. Consider a particle moving on a horizontal line. The distance from the start position to the particle at time t is
3
given by f (t) = t3 − 4t2 + 12t + 5.
(a) Compute the velocity and acceleration as a function of time. Graph both of these functions.
(b) When is the particle at rest? How fast is the particle moving at t = 0? What about at t = 3?
(c) When is the particle moving in the positive direction?
(d) When is it moving in the negative direction?
(e) When is the particle’s velocity increasing?
(f) When is the particle’s velocity decreasing?
Solution.
(a) The velocity at time t is v(t) = f 0 (t) = t2 − 8t + 12 = (t − 2)(t − 6). The acceleration at time t is a(t) = v 0 (t) = 2t − 8.
(b) The particle is (instantaneously) at rest at t = 2 and t = 6, the times when v(t) = 0. At t = 0 the particle has velocity
v(0) = 12 and at t = 3 it has velocity v(3) = 32 − 8(3) + 12 = −3.
(c) The particle is moving in the positive direction when v(t) is positive, which is on the intervals [0, 2) and (6, ∞).
(d) The particle is moving in the negative direction when v(t) is negative, which is on the interval (2, 6).
(e) The particle’s velocity is increasing when the acceleration is positive, which is on the interval (4, ∞).
(f) The particle’s velocity is decreasing when the acceleration is negative, which is on the interval [0, 4).
Exercise 6. Compute the derivative of f (x) = tan(x). Use the quotient rule.
Solution. Using the quotient rule we get
d sin(x)
cos(x) cos(x) − sin(x)(− sin(x))
cos2 (x) + sin2 (x)
1
=
=
=
= sec2 (x).
2
dx cos(x)
cos (x)
cos2 (x)
cos2 (x)
We could have also simplified in the following way.
cos2 (x) + sin2 (x)
= 1 + tan2 (x),
cos2 (x)
which also shows that 2 (x) = 1 + tan2 (x).
Exercise 7. Compute the derivative of f (x) = sin(x) cos(x).
Solution. Use the product rule to get f 0 (x) = sin(x)(− sin(x))+cos(x) cos(x) = cos2 (x)−sin2 (x) = cos(2x) where we used a trig
identity in the last equality. This is equal to the following equivalent expressions cos(2x) = cos2 (x) − sin2 (x) = 2 cos2 (x) − 1 =
1 − 2 sin2 (x).