Applications of Derivatives: AP Calculus Practice

Interpreting the Derivative in Context

Question 1

no calculatormedium

The temperature of a cup of coffee is

H (t)

degrees Celsius

t

minutes after it is poured. If

H^{'} (10) = - 1.5

and

H^{''} (10) > 0

, which interpretation is correct at

t = 10

minutes?

Show the answer

The answer is C.

H^{'} (10) = - 1.5 < 0

means the temperature is falling (cooling).

H^{''} (10) > 0

means

H^{'}

is increasing toward

0

, so the cooling rate is slowing down.

Why not A: $H^{'} (10) < 0$ means the temperature is decreasing (cooling), not warming.
Why not B: $H^{''} > 0$ means $H^{'}$ is increasing (becoming less negative), so the cooling rate is decreasing, not increasing.
Why not D: $H^{'} (10) = - 1.5 \neq = 0$ , so the temperature is changing, not constant.

Straight-Line Motion

Question 2

no calculatormedium

A particle moves along a line with velocity

v (t)

given by the graph below, which equals

1

on

[0, 1]

and

- 2

on

[1, 2]

. If the particle starts at the origin, which of the following could be the graph of its position

s (t)

?

Show the answer

The answer is B.

Position is the accumulation of velocity,

s^{'} (t) = v (t)

. With

v = 1 > 0

the particle moves forward at slope

1

, reaching

s (1) = 1

; then

v = - 2 < 0

, so it moves back at slope

- 2

. The position graph is piecewise-linear, turning around at

t = 1

.

Why not A:This is the velocity graph $v (t)$ , not the position graph $s (t)$ .
Why not C:A line that keeps rising means the particle always moves forward, but after $t = 1$ the velocity is negative, so $s$ must decrease.
Why not D:A curved (parabolic) position would mean a changing velocity, but $v$ is constant on each interval, so $s$ is made of straight pieces.

Question 3

no calculatormedium

A particle moves along the

x

-axis with position

x (t) = sin t

for

0 \leq t \leq 2 π

. At which of the following values of

t

does the particle change direction? I.

t = \frac{π}{2}

II.

t = π

III.

t = \frac{3 π}{2}

IV.

t = 2 π

Show the answer

The answer is A.

v (t) = cos t

, which is

0

with a sign change at

t = \frac{π}{2}

(

+ \to -

) and

t = \frac{3 π}{2}

(

- \to +

). At

t = π

and

t = 2 π

,

v = cos t = \mp 1 \neq = 0

, so the particle does not turn there.

Why not B:Included II: at $t = π$ , $v = cos π = - 1 \neq = 0$ , so no turn.
Why not C:Included II ( $v \neq = 0$ at $t = π$ ) and missed III ( $v = cos \frac{3 π}{2} = 0$ with a sign change).
Why not D:Included IV: at $t = 2 π$ , $v = cos 2 π = 1 \neq = 0$ , so no turn.

Related Rates

Question 4

calculatormedium

A boat is pulled toward a dock by a rope that runs over a pulley

9

ft above the point where the rope is tied to the boat. The rope is reeled in at

2

ft/s. How fast (ft/s) is the boat approaching the dock when

15

ft of rope is out? Round to the nearest hundredth.

Show the answer

The answer is C.

Let

x

be the horizontal distance and

L

the rope length:

x^{2} + 9^{2} = L^{2}

. When

L = 15

,

x = 12

. Differentiating,

x \frac{d x}{d t} = L \frac{d L}{d t}

, so

\frac{d x}{d t} = \frac{15}{12} (2) = 2.5

ft/s.

Why not A:That is the rate the rope is reeled in; the boat approaches faster because the rope is the hypotenuse: $\frac{d x}{d t} = \frac{L}{x} \frac{d L}{d t} = \frac{15}{12} (2) = 2.5$ .
Why not B:That is $\frac{x}{L} \frac{d L}{d t} = \frac{12}{15} (2)$ ; you inverted the ratio. Use $\frac{d x}{d t} = \frac{L}{x} \frac{d L}{d t}$ .
Why not D: $x = 1 5^{2} - 9^{2} = 12$ , not $1 5^{2} - 9^{2}$ miscomputed; with $x = 12$ , $\frac{15}{12} (2) = 2.5$ .

Question 5

no calculatorhard

A trough

10

ft long has an isosceles-triangle cross-section that is

4

ft wide at the top and

3

ft deep (vertex down). Water flows in at

5

ft

^{3}

/min. How fast is the water level rising when the water is

2

ft deep?

Show the answer

The answer is A.

At depth

h

the surface width is

w = \frac{4}{3} h

, so the triangular area is

\frac{1}{2} w h = \frac{2}{3} h^{2}

and

V = 10 \cdot \frac{2}{3} h^{2} = \frac{20}{3} h^{2}

. Then

\frac{d V}{d t} = \frac{40}{3} h \frac{d h}{d t}

; at

h = 2

,

\frac{80}{3} \frac{d h}{d t} = 5 \Rightarrow \frac{d h}{d t} = \frac{3}{16}

.

Why not B:Include the $10$ -ft length: $V = area \times 10$ . Dropping it gives $\frac{15}{8}$ instead of $\frac{3}{16}$ .
Why not C:The surface width grows with depth, $w = \frac{4}{3} h$ , not a constant $4$ ft; the answer is $\frac{3}{16}$ .
Why not D:The triangular cross-section is $\frac{1}{2} w h$ ; dropping the $\frac{1}{2}$ gives $\frac{3}{32}$ instead of $\frac{3}{16}$ .

Extreme Values & Critical Points

Question 6

no calculatormedium

Suppose

f

is continuous on the closed interval

[1, 5]

. Which statement must be true?

Show the answer

The answer is C.

By the Extreme Value Theorem, a function continuous on a closed interval attains both an absolute maximum and an absolute minimum on that interval.

Why not A:Continuity does not imply differentiability; $f$ could have a corner inside the interval.
Why not B:A continuous function need not have an interior critical point; its extremes could occur at the endpoints.
Why not D:This would require differentiability and is the conclusion of Rolle's/MVT under extra hypotheses; mere continuity does not guarantee it.

First Derivative Test

Question 7

no calculatorhard

Let

g

be a differentiable function such that

g (x) < 0

for all real numbers

x

, and let

f^{'} (x) = (x^{2} - 4) g (x)

. Which of the following is true?

Show the answer

The answer is B.

f^{'} (x) = (x - 2) (x + 2) g (x)

with

g < 0

. So

f^{'}

has sign:

x < - 2

:

(+) (-) = -

;

- 2 < x < 2

:

(-) (-) = +

;

x > 2

:

(+) (-) = -

. Thus

f^{'}

goes

-, +, -

: a relative minimum at

x = - 2

and a relative maximum at

x = 2

.

Why not A:This ignores that $g (x) < 0$ flips the sign of $f^{'}$ : with the negative factor, $f^{'}$ goes $-, +, -$ , giving min at $- 2$ and max at $2$ .
Why not C: $f^{'}$ changes sign differently at $- 2$ ( $- \to +$ , a min) and at $2$ ( $+ \to -$ , a max).
Why not D: $f^{'}$ does not have the same sign change at both; at $- 2$ it is $- \to +$ (a minimum).

Question 8

no calculatorhard

The graph of

f^{'}

(the derivative of

f

) is shown. At what

x

does

f

have a local maximum?

Show the answer

The answer is A.

On the

f^{'}

graph,

f

has a local max where

f^{'}

crosses zero

+ \to -

: that's

x = 1

.

Why not B:At $x = 3$ , $f^{'}$ crosses $- \to +$ , so that's a local minimum of $f$ .
Why not C: $x = 2$ is where $f^{'}$ bottoms out (a min of $f^{'}$ ), i.e. an inflection point of $f$ , not an extremum.
Why not D: $f^{'} (0) = 2 > 0$ ; $f$ is increasing at $x = 0$ , not at a max.

Question 9

no calculatorhard

A differentiable function

f

has

f^{'} (x) = x^{2} (x - 2)

. Using the First Derivative Test,

f

has

Show the answer

The answer is D.

f^{'} = x^{2} (x - 2)

is negative for

x < 0

and

0 < x < 2

, and positive for

x > 2

. So

f^{'}

changes sign only at

x = 2

(negative to positive: a minimum); at

x = 0

the sign does not change, so no extremum.

Why not A:At $x = 0$ the factor $x^{2}$ keeps $f^{'}$ negative on both sides, so there is no extremum; and $x = 2$ is a minimum, not a maximum.
Why not B: $f^{'}$ does not change sign at $x = 0$ (stays negative), so $x = 0$ is not a maximum. Only $x = 2$ is an extremum (a minimum).
Why not C: $x = 0$ is not an extremum because $f^{'}$ keeps the same sign there; only $x = 2$ is a relative minimum.

Concavity & the Second Derivative Test

Question 10

no calculatormedium

If

f^{''} (x) = x (x + 1) (x - 2)^{2}

, then the graph of

f

has inflection points when

x =

Show the answer

The answer is C.

Inflection points occur where

f^{''}

changes sign. The simple roots

x = - 1

and

x = 0

give sign changes; the double root

x = 2

(even multiplicity) does not, so

f^{''}

stays the same sign there. Inflection points:

x = - 1

and

0

only.

Why not A:At $x = 2$ the factor $(x - 2)^{2}$ has even multiplicity, so $f^{''}$ does not change sign there, so it is not an inflection point.
Why not B:Counts every zero of $f^{''}$ ; but $x = 2$ is a double root, where $f^{''}$ keeps the same sign, so it is not an inflection point.
Why not D: $x = 2$ is exactly where $f^{''}$ does NOT change sign (squared factor); the sign changes occur at the simple roots $- 1$ and $0$ .

Connecting f, f', and f''

Question 11

no calculatormedium

The graph of

f

is shown below. Which of the following is the graph of

f^{'}

?

Show the answer

The answer is A.

f

has a local minimum near

x = - 1

and a local maximum near

x = + 1

, so

f^{'} = 0

at both and

f^{'} > 0

only on the interval

(- 1, 1)

where

f

is increasing. That is the downward parabola

f^{'} (x) = 3 - 3 x^{2}

, which crosses zero at

x = \pm 1

.

Why not B:This opens the wrong way: it is positive outside $[- 1, 1]$ and negative between, but $f$ is increasing (so $f^{'} > 0$ ) on $(- 1, 1)$ .
Why not C:This is the graph of $f$ itself, not its derivative.
Why not D:A line crosses zero only once, but $f$ has two turning points, so $f^{'}$ must have two zeros.

Question 12

no calculatorhard

The graph of

f^{'}

, the derivative of

f

, is shown below. Which of the following statements must be true?

I. f

has a relative minimum at

x = 3

.

II. f

has a point of inflection at

x = 1

.

III. f

is concave up for

x < 1

.

Show the answer

The answer is D.

From the graph,

f^{'} < 0

on

(- 1, 3)

and

f^{'} > 0

outside, so

f

has a relative MIN at

x = 3

(I true).

f^{'}

has its minimum at

x = 1

, where

f^{''}

changes sign, giving an inflection point of

f

at

x = 1

(II true). For

x < 1

,

f^{'}

is decreasing so

f^{''} < 0

:

f

is concave DOWN, so III is false. Hence I and II only.

Why not A:II is also true: $f^{'}$ has a minimum at $x = 1$ , so $f^{''}$ changes sign there, which makes it a point of inflection of $f$ .
Why not B:I is also true: $f^{'}$ changes from negative to positive at $x = 3$ , so $f$ has a relative minimum there.
Why not C:III is false: for $x < 1$ , $f^{'}$ is decreasing, so $f^{''} < 0$ and $f$ is concave DOWN, not up.

Optimization

Question 13

no calculatormedium

The cost of producing

x

items is

C (x) = x^{2} + 400

dollars. The factory can produce at most

15

items. For

1 \leq x \leq 15

, how many items should be produced to minimize the average cost per item?

Show the answer

The answer is B.

The average cost is

\frac{C ( x )}{x} = x + \frac{400}{x}

, with derivative

1 - \frac{400}{x ^{2}}

. This is

0

at

x = 20

, outside the allowed interval, and it is negative on

[1, 15]

. So the average cost decreases the whole way, and the minimum on

[1, 15]

is at the endpoint

x = 15

.

Why not A:At $x = 10$ the average cost is $50$ dollars, but it keeps falling up to $x = 15$ , where it is about $41.67$ dollars.
Why not C: $x = 20$ is where the average cost would be smallest with no limit, but the factory cannot make more than $15$ items.
Why not D: $40$ is the minimum average cost (in dollars) without the limit, $20 + \frac{400}{20}$ , not a number of items.

Question 14

no calculatorhard

What is the smallest possible distance from the point

(0, 3)

to a point on the parabola

y = x^{2}

?

Show the answer

The answer is A.

Minimize

D^{2} = x^{2} + (x^{2} - 3)^{2}

. Then

\frac{d}{d x} D^{2} = 2 x + 2 (x^{2} - 3) (2 x) = 2 x (2 x^{2} - 5) = 0

, giving

x^{2} = \frac{5}{2}

. Then

D^{2} = \frac{5}{2} + (\frac{5}{2} - 3)^{2} = \frac{11}{4}

, so

D = \frac{11}{2}

.

Why not B: $3$ is the distance to the vertex $(0, 0)$ , but a point with $x^{2} = \frac{5}{2}$ is closer, giving distance $\frac{11}{2} \approx 1.66$ .
Why not C:Minimizing $D^{2} = x^{2} + (x^{2} - 3)^{2}$ gives $x^{2} = \frac{5}{2}$ and $D^{2} = \frac{11}{4}$ , so $D = \frac{11}{2}$ , not $3$ .
Why not D: $\frac{5}{2}$ is the $x$ -coordinate $5/2$ confusion; the minimum distance is $11 /2$ .