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ABCalculusMultipleChoiceTest

FromCalculusCourseDescription–EffectiveFall2010(pp.17‐27)

Solution: This is the definition of the derivative of the function cos , evaluated at

.

cos

sin

sin 1 AnswerA

lim→

lim→

Two equivalent definitions of a derivative:

ABCalculusMultipleChoiceTest Fall2012

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being negative means the curve is decreasing.

being negative means the curve is concave down.

Let’s consider each of the five points:

Point : Curve is increasing. Curve is concave down.

Point : Curve is decreasing. Curve is concave down. Bingo!

Point : Curve is decreasing. Curve is concave up.

Point : Curve is flat. Curve is concave up.

Point : Curve is increasing. Curve is concave up.

The only point that meets the conditions specified is Point . AnswerB

6 0

∙ 1 ∙ 3 ∙ 2 ∙ 2 0

3 2 2 0

2 3 2 0

23 2

1 2 ∙ 2 ∙ 13 ∙ 2 ∙ 1 2 ∙ 2 ∙ 1

Use implicit differentiation.

Evaluated at the point 2, 1 , we get:


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First, find the points of intersection of the two curves. Set equal and .

has the solutions , .

Now, let’s define the integration we need.

We must use the washer method because there is a gap between the region we

are revolving and its reflection across the axis of revolution, .

Revolving about a horizontal line means our disks are vertical, as shown, and

we should integrate with respect to .

The height of the larger disk is the distance between the axis of revolution,

and the curve , so the height is .

The height of the smaller disk is the distance between the axis of revolution,

and the curve , so the height is .

We move the disks from left to right in the region, i.e., from to .

The volume is determined from the formula: . Therefore,


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To answer this question, we must find the critical values and possible inflection points.

2

4 6 2 2 3 Critical values exist at 0,

′ 12 12 12 1 Possible inflection points: 0,1

Then, build an Ault Table with intervals separated by the critical values and the ‐values of any possible inflection points:

From above, the values of that define the intervals in the table are 0, 1, .

Note: In developing the table below, identify the signs (i.e., “ , “ “) first. The word descriptors are based on the signs.

∞, , , . . , ∞

decreasing decreasing decreasing increasing

and is: increasing decreasing increasing increasing

: concave up concave down concave up concave up

What can we say about extrema and inflection points based on this?

There is a minimum at 1.5 because the curve changes from decreasing to increasing.

There are two inflections points, at 0, 1 because concavity changes at those points.

So, this curve has one relative extreme and two inflection points. AnswerC


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We will use the power rule and the chain rule here:

sin 3

2 sin 3 ∙ sin 3

2 sin 3 ∙ cos 3 ∙ 3

2 sin 3 ∙ cos 3 ∙ 1 2 sin 3 ∙ cos 3

0 2 sin 3 0 ∙ cos 3 0

∙ AnswerB

This narrows our solutions down to Answer C and Answer D. Again, we need to consider the

point on the curve: 2, 2 . Let’s try it in each of these answers.

Answer C: √4 12 ⇒ 2 √4 ∙ 2 12

Answer D: √4 12 ⇒ 2 √4 ∙ 2 12 X

Therefore, our solution is AnswerC

2 4 2

12

Since 2, 2 is a point on

the curve, we get:

4

4

4

12

2

4

4 12

√4 12


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6 48 90 6 8 15 ⇒ critical values: 3, 5

′ 12 48 ⇒ Possible inflection points at: 4

It’s time for an Ault Table:

, , , , ∞

increasing decreasing decreasing increasing

and is: decreasing decreasing increasing increasing

: concave down concave down concave up concave up

The signs are the same on the intervals: , and , ∞ . AnswerE

Speed is increasing on the interval(s) where the first and second

derivatives have the same sign.

2 24 90 7 for 0

66 03 0

The average rate of change over an interval is the slope of the line connecting

its endpoints.

The endpoints are determined as follows:

5

For 0, 0 0 5 ∙ 0 0 So a point is 0, 0 .

For 3, 3 3 5 ∙ 3 66 So a point is 3, 66 .

The slope, then, is:


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is the curve on top, and is the curve on the bottom. They intersect where

⇒ 3 0, i.e., at 0, 3 .

The area of the shaded region, then, is:

3 13

32

30

13∙ 3

32∙ 3

13∙ 0

32∙ 0 9

272

1 √

AnswerD

lim→

42 4

lim→

21 8

lim→

21 8

lim→

28

The limit is inderminate as →∞. Use L’Hospital’s Rule:

The limit is still inderminate as → ∞. Use L’Hospital’s Rule:


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5

∙ 2 0

0 5 5

1 5

You can solve for the

equation implied by the

differential equation if

you like. However, it is

easier to find a couple of

slopes in the field.

I like to find the slopes at

points that are not near

the origin.

Try:

Point 5, 5 , 1

Point 5, 5 , 1

Point 5, 5 , 1

Point 5, 5 , 1

Answer E is the only one

with these slopes.


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We might think this a straightforward integration, but we must be careful. If velocity is both

positive and negative over the interval, we must break the interval into multiple sections and

integrate each.

3 ∙ sin 2

Since 0 everywhere, 3 ∙ sin 2 0 when sin 2 0.

sin 2 0 when 0, in the domain 0 2.

So, our solution is:

3 ∙ sin 2 3 ∙ sin 2

2.055 0.206 .


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A visualization of this situation is provided above. The people who live within one mile of the lake

live between 1 mile and 2 miles from the center of the lake.

We can think of this as a volume problem with a circular cross section of radius and height

. The volume of solid of revolution is given by the cylindrical shell formula:

2

Then,

In Problem 17, we are

looking for the ‐value of

a point where the curve

is not continuous but the

limit exists. This implies

a hole in the graph in a

location where the

function would be

continuous if the hole

were filled in.

This occurs only at 0.


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Let’s look at average slopes in the applicable intervals shown:

Average slope from 1.1 to 1.2: . .

. .2.0.

Average slope from 1.2 to 1.3: . .

. .1.8.

Since 0 over the entire interval, the curve is concave down over the interval, and so the slope is decreasing throughout the interval. Therefore, the slope of the curve at 1.2 must be

between . and . . AnswerD

Position Functions: sin 1

Velocity Functions: ′ cos ′ 2

Then we set the velocity functions equal and see

how many solutions exist on the interval 0. 10 .

cos 2

cos 2 0

A look at the graph of this function tells us that

there are three points where the velocity functions

are equal on the interval 0. 10 . AnswerD


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1

1

is the area under the curve from 1

to 0.

Based on the lines in the graph, the area is bounded by

a triangle with base 1 and height 2. So, the

area of the triangle is ∙ 1 ∙ 2 1 Then, .

1 ∙

The average value of a function

over an interval is given by:

Note that this is also the average

area under the graph over the

interval.

In Problem 21, we are looking for a

function with non‐zero area under

the graph. That is, we are looking

for a function that does not have

equal positive and negative areas

over the interval , .

All of the graphs shown have equal

positive and negative areas over

this interval except Graph E.


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2,000 . 10.2

2,000 . 100 10,000 . 10

0

10,000 1 , gallons AnswerD

2 0 sin2

2 0 sin2

.157 . 1

Clearly, this is a problem involving differentials, so we need the differential

formula:

∙

Let’s find the pieces to go into this formula.

4.8. So, let our reference point be 5

We are given: 5 3 and 5 4

Then: 4.8 5 5 ∙ 4.8 5 3 4 ∙ 0.2 .

AnswerA

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