WWSC AC in Runestone Test Problems

slope. The slope of the line between the points $(0.5, s (0.5))$ and $(1, s (1)) .$
curvature. The curvature of the graph of $s$ on the interval $0.5 \leq t \leq 1 .$
area. The area under the graph of $s$ on the interval $0.5 \leq t \leq 1 .$
distance. The distance between the points $(0.5, s (0.5))$ and $(1, s (1)) .$

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What does this value measure physically?

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average velocity. The average velocity of the ball during the interval $0.5 \leq t \leq 1 .$
distance. The distance that the ball travels during the interval $0.5 \leq t \leq 1 .$
acceleration. The acceleration of the ball during the interval $0.5 \leq t \leq 1 .$

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What are the units on

A V_{[0.5, 1]} ?

feet per second
feet
seconds
feet per second squared
foot seconds
none of these

Answer 1.

8

Answer 2.

slope

Answer 3.

average velocity

Answer 4.

feet per second

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Exploration 2.6.2. Lots of dropdowns, using NiceTables, check that answers restore.

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For each of the indefinite integrals below, the main question is to decide whether the integral can be evaluated using a basic antiderivative,

u

-substitution, integration by parts, a combination of

u

-substitution and integration by parts, or none of these other techniques.

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For integrals for which your answer is affirmative (

u

-sub, by parts, combo), you should be able to state what

u

and

d u

are, or what

u

and

d v

are.

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TABLE A

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Method

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\int x^{2} \sin (x^{3}) d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int x^{2} \sin (x) d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \sin (x^{3}) d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int x^{5} \sin (x^{3}) d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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TABLE B

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Method

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\int \frac{1}{1 + x^{2}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \frac{x}{1 + x^{2}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \frac{2 x + 3}{1 + x^{2}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \frac{e^{x}}{1 + (e^{x})^{2}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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TABLE C

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Method

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\int x \ln (x) d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \frac{\ln (x)}{x} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \ln (1 + x^{2}) d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int x \ln (1 + x^{2}) d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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TABLE D

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Method

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\int x \sqrt{1 - x^{2}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \frac{1}{\sqrt{1 - x^{2}}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \frac{x}{\sqrt{1 - x^{2}}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

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\int \frac{1}{x \sqrt{1 - x^{2}}} d x

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Basic Integral
u-Sub
By Parts
Combo
None of These

Answer 1.

u-Sub

Answer 2.

By Parts

Answer 3.

None of These

Answer 4.

Combo

Answer 5.

Basic Integral

Answer 6.

u-Sub

Answer 7.

Combo

Answer 8.

u-Sub

Answer 9.

By Parts

Answer 10.

u-Sub

Answer 11.

None of These

Answer 12.

Combo

Answer 13.

u-Sub

Answer 14.

Basic Integral

Answer 15.

u-Sub

Answer 16.

None of These

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Exploration 2.6.3. Old Checkbox macro.

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The position of an object moving along a line is given by the function

s (t),

where

s

is measured in feet and

t

in seconds. We determine that the velocity is

v (t) = 4 t + 1

feet per second.

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How much does the position change over the time interval

[0, 4] ?

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Change in position = feet

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Suppose you are told that the object’s initial position

s (0) = 7 .

Determine

s (2),

the object’s position

2

seconds later.

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s (2) =

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If you are told instead that the object’s initial position is

s (0) = 3,

what is

s (2) ?

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s (2) =

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If we only know the velocity

v (t) = 4 t + 1

(but not the initial position), which of the following could be the position function?

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Select all that apply.

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$s (t) = 2 t^{2} + t + 37$
$s (t) = 2 t^{2} + t$
$s (t) = 2 t^{2} + t + 5$
$s (t) = 2 t^{2} + t + 11$
$s (t) = 2 t^{2} + t + 20$
None of the above

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If, in addition to knowing the velocity function is

v (t) = 4 t + 1,

we know the initial position

s (0),

how many possibilities are there for

s (t) ?

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0
2
1
5
Infinitely many
none of the above

Answer 1.

36

Answer 2.

10 + 7

Answer 3.

10 + 3

Solution.

\int_{0}^{4} 4 t + 1 d t = 36

feet. Therefore, the position function will be

s_{0} + 2 t^{2} + t,

where

s_{0}

is the initial velocity.

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Exploration 2.6.4. No possible correct answer for dropdown immediately after the first essay box.

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Scenario A

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In the following scenario, we consider the distribution of a quantity along an axis.

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Suppose that the function

c (x) = 200 + 100 e^{- 0.1 x}

models the density of traffic on a straight road, measured in cars per mile, where

x

is number of miles east of a major interchange, and consider the definite integral

\int_{0}^{2} (200 + 100 e^{- 0.1 x}) d x .

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(i) What are the units on the product

c (x) \cdot △ x ?

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Units:

cars
miles
cars/mile
miles/car
cars*cars
miles*miles

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(ii) What are the units on the definite integral and its Riemann sum approximation given by

\int_{0}^{2} c (x) d x \approx \sum_{i = 1}^{n} c (x_{i}) △ x ?

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Units:

cars
miles
cars/mile
miles/car
cars*cars
miles*miles

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(iii) Evaluate the definite integral

\int_{0}^{2} c (x) d x = \int_{0}^{2} (200 + 100 e^{- 0.1 x}) d x

and write one sentence to explain the meaning of the value you find.

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Scenario B

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In the following scenario, we consider the distribution of a quantity along an axis.

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On a 6 foot long shelf filled with books, the function

B

models the distribution of the weight of the books, measured in pounds per inch, where

x

is the number of inches from the left end of the bookshelf. Let

B (x)

be given by the rule

B (x) = 0.5 + \frac{1}{(x + 1)^{2}} .

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(i) What are the units on the product

B (x) \cdot △ x ?

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Units:

pounds
inches
pound/inch
inch/pound
pound*pound
inch*inch

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(ii) What are the units on the definite integral and its Riemann sum approximation given by

\int_{12}^{36} B (x) d x \approx \sum_{i = 1}^{n} B (x_{i}) △ x ?

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Units:

pounds
inches
pound/inch
inch/pound
pound*pound
inch*inch

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(iii) Evaluate the definite integral

\int_{0}^{72} B (x) d x = \int_{0}^{72} (0.5 + \frac{1}{(x + 1)^{2}}) d x

and write one sentence to explain the meaning of the value you find.

Answer 1.

cars

Answer 2.

cars

Answer 3.

Undefined

Answer 4.

pounds

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Exploration 2.6.5. Hardcoded by Alex, risk of being too long.

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Let

f (x) = \sin (x)

and let

T_{3} (x) = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} .

Our goal is to find the values of

c_{0}, \dots, c_{3}

that make the sine function and its derivative values agree with those of the cubic polynomial

T_{3}

a = 0

and to study the resulting degree

3

approximation of the sine function.

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(a)

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As in previous work, the derivatives of

T_{3} (x)

and their respective values at

a = 0

are those shown in the following table. Compute the various derivatives of

f (x) = \sin (x)

and evaluate them at

a = 0

accordingly, recording your results in the left side of the table.

$f (x) =$	$\sin (x)$	$T_{3} (x) =$	$c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3}$
$f^{'} (x) =$		$T_{3}^{'} (x) =$	$c_{1} + 2 c_{2} x + 3 c_{3} x^{2}$
$f^{″} (x) =$		$T_{3}^{″} (x) =$	$2 c_{2} + 6 c_{3} x$
$f^{‴} (x) =$		$T_{3}^{‴} (x) =$	$6 c_{3}$
$f (0) =$		$T_{3} (0) =$	$c_{0}$
$f^{'} (0) =$		$T_{3}^{'} (0) =$	$c_{1}$
$f^{″} (0) =$		$T_{3}^{″} (0) =$	$2 c_{2}$
$f^{‴} (0) =$		$T_{3}^{‴} (0) =$	$6 c_{3}$

Answer 1.

\cos (x)

Answer 2.

- \sin (x)

Answer 3.

- \cos (x)

Answer 4.

0

Answer 5.

1

Answer 6.

0

Answer 7.

- 1

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(b)

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Now, set

T_{3} (0) = f (0),

T_{3}^{'} (0) = f^{'} (0),

T_{3}^{″} (0) = f^{″} (0),

and

T_{3}^{‴} (0) = f^{‴} (0) .

This implies

c_{0} =

c_{1} =

c_{2} =

, and

c_{3} =

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Putting it all together, what is the resulting formula for

T_{3} (x) ?

Answer 1.

0

Answer 2.

1

Answer 3.

0

Answer 4.

- \frac{1}{6}

Answer 5.

x - \frac{1}{6} x^{3}

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(c)

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Recall that the tangent line approximation

T_{1}

f (x) = \sin (x)

a = 0

T_{1} (x) = x,

as plotted below. On the same axes, we’ve plotted the cubic approximation

T_{3}

you found in part (b).

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What do you observe about the approximation

T_{3}

generates compared to the tangent line approximation

T_{1} ?

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The cubic approximation is closer to the function for more values of $x .$
The tangent line approximation is closer to the function for more values of $x .$
Both approximate the function equally well.

Answer.

Choice 1

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(d)

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Compute

f (1) - T_{1} (1)

and

f (1) - T_{3} (1) .

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What do you notice about the size and sign of those differences?

Answer 1.

- 0.158529

Answer 2.

0.00813765

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(e)

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For about what values of

x

does it appear that

| f (x) - T_{1} (x) | < 0.1 ?

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For about what values of

x

does it appear that

| f (x) - T_{3} (x) | < 0.1 ?

Answer 1.

(- 0.9, 0.9)

Answer 2.

(- 1.7, 1.7)

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Exercises Exercises

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1. Runs the risk of being too long, depending on what students enter.

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In this problem, the notation "SIMP(2)" is actually what we have called "SIMP(4)" in our previous work. Different authors use different notation, and the author of this WeBWorK exercise chooses to write "SIMP(n)" where we have written "SIMP(2n)".

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Note: for this problem, because later answers depend on earlier ones, you must enter answers for all answer blanks for the problem to be correctly graded. If you would like to get feedback before you completed all computations, enter a "1" for each answer you did not yet compute and then submit the problem. (But note that this will, obviously, result in a problem submission.)

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(a) What is the exact value of

\int_{0}^{5} e^{x} d x ?

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\int_{0}^{5} e^{x} d x =

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(b)

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Find LEFT(2), RIGHT(2), TRAP(2), MID(2), and SIMP(2); compute the error for each.

	LEFT(2)	RIGHT(2)	TRAP(2)	MID(2)	SIMP(2)
value
error

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(c)

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Repeat part (b) with

n = 4

(instead of

n = 2

	LEFT(4)	RIGHT(4)	TRAP(4)	MID(4)	SIMP(4)
value
error

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(d)

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For each rule in part (b), as

n

goes from

n = 2

n = 4,

does the error go down approximately as you would expect? Explain by calculating the ratios of the errors:

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Error LEFT(2)/Error LEFT(4) =

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Error RIGHT(2)/Error RIGHT(4) =

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Error TRAP(2)/Error TRAP(4) =

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Error MID(2)/Error MID(4) =

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Error SIMP(2)/Error SIMP(4) =

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(Be sure that you can explain in words why these do (or don’t) make sense.)

Answer 1.

e^{5} - 1

Answer 2.

2.5 (1 + e^{2.5})

Answer 3.

2.5 (e^{2.5} + e^{5})

Answer 4.

1.25 (1 + 2 e^{2.5} + e^{5})

Answer 5.

2.5 (e^{1.25} + e^{3 \cdot 1.25})

Answer 6.

\frac{2 \cdot 115.029 + 217.223}{3}

Answer 7.

147.413 - 32.9562

Answer 8.

147.413 - 401.489

Answer 9.

147.413 - 217.223

Answer 10.

147.413 - 115.029

Answer 11.

147.413 - 149.094

Answer 12.

1.25 (1 + e^{1.25} + e^{2.5} + e^{3 \cdot 1.25})

Answer 13.

1.25 (e^{1.25} + e^{2.5} + e^{3 \cdot 1.25} + e^{5})

Answer 14.

0.625 (1 + 2 e^{1.25} + 2 e^{2.5} + 2 e^{3 \cdot 1.25} + e^{5})

Answer 15.

1.25 (e^{0.625} + e^{3 \cdot 0.625} + e^{5 \cdot 0.625} + e^{7 \cdot 0.625})

Answer 16.

\frac{2 \cdot 138.236 + 166.126}{3}

Answer 17.

147.413 - 73.9924

Answer 18.

147.413 - 258.259

Answer 19.

147.413 - 166.126

Answer 20.

147.413 - 138.236

Answer 21.

147.413 - 147.533

Answer 22.

\frac{114.457}{73.4206}

Answer 23.

\frac{- 254.076}{- 110.846}

Answer 24.

\frac{- 69.81}{- 18.713}

Answer 25.

\frac{32.384}{9.177}

Answer 26.

\frac{- 1.681}{- 0.12}

Solution.

(a)

\int_{0}^{5} e^{x} d x = e^{x} |_{0}^{5} = e^{5} - e^{0} \approx 147.4132 \dots

(b) Computing the sums directly, since

Δ x = 2.5,

we have LEFT(2)

= 2.5 \cdot e^{0} + 2.5 \cdot e^{2.5} \approx 32.9563

RIGHT(2)

= 2.5 \cdot e^{2.5} + 2.5 \cdot e^{5} \approx 401.489

TRAP(2)

= \frac{1}{2} (32.9563 + 401.489) \approx 217.223

MID(2)

= 2.5 \cdot e^{1.25} + 2.5 \cdot e^{3 \cdot 1.25} \approx 115.029

and SIMP(2) = (2 MID(2) + TRAP(2))/3

\approx 149.094 .

And the errors are just the actual value minus the approximate values, or error(LEFT(2)) = 114.457; error(RIGHT(2)) = -254.076; error(TRAP(2)) = -69.81; error(MID(2)) = 32.3841; and error(SIMP(2)) = -1.681. (c) Similarly, since

Δ x = 0.625,

we have LEFT(4) = 73.9924; error = 73.4207 RIGHT(4) = 258.259; error = -110.846 TRAP(4) = 166.126; error = -18.713 MID(4) = 138.236; error = 9.177 SIMP(4) = 147.533; error = -0.12. (d) For LEFT and RIGHT, we expect the error to go down by 1/2, and this is very roughly what we see. For MID and TRAP, we expect the error to go down by 1/4, and this is approximately what we see. For SIMP, we expect the error to go down by 1/16, again as observed.

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2. Shortened version of question above, but told one student everything was wrong.

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(a) What is the exact value of

\int_{0}^{5} e^{x} d x ?

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\int_{0}^{5} e^{x} d x =

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(b)

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Find LEFT(2), RIGHT(2), TRAP(2), MID(2), and SIMP(4); compute the error for each.

	LEFT(2)	RIGHT(2)	TRAP(2)	MID(2)	SIMP(4)
value
error

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(c)

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Repeat part (b) with

n = 4

(instead of

n = 2

	LEFT(4)	RIGHT(4)	TRAP(4)	MID(4)	SIMP(8)
value
error

Answer 1.

e^{5} - 1

Answer 2.

2.5 (1 + e^{2.5})

Answer 3.

2.5 (e^{2.5} + e^{5})

Answer 4.

1.25 (1 + 2 e^{2.5} + e^{5})

Answer 5.

2.5 (e^{1.25} + e^{3 \cdot 1.25})

Answer 6.

\frac{2 \cdot 115.029 + 217.223}{3}

Answer 7.

147.413 - 32.9562

Answer 8.

147.413 - 401.489

Answer 9.

147.413 - 217.223

Answer 10.

147.413 - 115.029

Answer 11.

147.413 - 149.094

Answer 12.

1.25 (1 + e^{1.25} + e^{2.5} + e^{3 \cdot 1.25})

Answer 13.

1.25 (e^{1.25} + e^{2.5} + e^{3 \cdot 1.25} + e^{5})

Answer 14.

0.625 (1 + 2 e^{1.25} + 2 e^{2.5} + 2 e^{3 \cdot 1.25} + e^{5})

Answer 15.

1.25 (e^{0.625} + e^{3 \cdot 0.625} + e^{5 \cdot 0.625} + e^{7 \cdot 0.625})

Answer 16.

\frac{2 \cdot 138.236 + 166.126}{3}

Answer 17.

147.413 - 73.9924

Answer 18.

147.413 - 258.259

Answer 19.

147.413 - 166.126

Answer 20.

147.413 - 138.236

Answer 21.

147.413 - 147.533

Solution.

(a)

\int_{0}^{5} e^{x} d x = e^{x} |_{0}^{5} = e^{5} - e^{0} \approx 147.4132 \dots

(b) Computing the sums directly, since

Δ x = 2.5,

we have LEFT(2)

= 2.5 \cdot e^{0} + 2.5 \cdot e^{2.5} \approx 32.9563

RIGHT(2)

= 2.5 \cdot e^{2.5} + 2.5 \cdot e^{5} \approx 401.489

TRAP(2)

= \frac{1}{2} (32.9563 + 401.489) \approx 217.223

MID(2)

= 2.5 \cdot e^{1.25} + 2.5 \cdot e^{3 \cdot 1.25} \approx 115.029

and SIMP(4) = (2 MID(2) + TRAP(2))/3

\approx 149.094 .

Δ x = 0.625,

we have LEFT(4) = 73.9924; error = 73.4207 RIGHT(4) = 258.259; error = -110.846 TRAP(4) = 166.126; error = -18.713 MID(4) = 138.236; error = 9.177 SIMP(8) = 147.533; error = -0.12. (d) For LEFT and RIGHT, we expect the error to go down by 1/2, and this is very roughly what we see. For MID and TRAP, we expect the error to go down by 1/4, and this is approximately what we see. For SIMP, we expect the error to go down by 1/16, again as observed.

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3. Uses old table macros, but parserCheckboxList.

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The graph below shows the distance traveled,

D

(in miles) as a function of time,

t

(in hours).

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(Click on the graph to get a larger version.)

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a) For each of the intervals, find the values of

Δ D

and

Δ t

between the indicated start and end times. Enter your answers in their respective columns in the table below.

Time Interval	$Δ D$	$Δ t$
t = 1 to t = 4
t = 1.5 to t = 3
t = 2 to t = 4

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b) Based on your results from (a) it follows that the average rate of change of

D

is constant, it does not depend over which interval of time you choose. What is the constant rate of change of

D ?

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\frac{Δ D}{Δ t} =

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c) Which of the statements below CORRECTLY explains the significance of your answer to part (b)? Select ALL that apply (more than one may apply).

It is the total distance the car travels in five hours.
It represents the car’s velocity.
It is the acceleration of the car over the five hour time interval.
It is how far the car will travel in a half-hour.
It is the slope of the line.
It is the average velocity of the car over the first two hours.
None of the above.

Answer 1.

135

Answer 2.

3

Answer 3.

67.5

Answer 4.

1.5

Answer 5.

90

Answer 6.

2

Answer 7.

45

Answer 8.

B, E, F

Solution.

a) Over the first interval

t = 1

t = 4,

Δ D = 180 - 45 = 135

and

Δ t = 4 - 1 = 3

Similarly, over the second interval

t = 1.5

t = 3,

Δ D = 135 - 67.5 = 67.5

and

Δ t = 3 - 1.5 = 1.5

Finally, over the last interval

t = 2

t = 4,

Δ D = 180 - 90 = 90

and

Δ t = 4 - 2 = 2

b) Notice that for each of the three intervals in (a) (as well as any other interval),

\frac{Δ D}{Δ t} = 45 .

c) There are three correct statements: It represents the car’s velocity, it is the average velocity of the car over the first two hours, and it is the slope of the line.

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4. Uses NiceTables.

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The figure below shows

f .

🔗

(Click on the graph for a larger version.)

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F^{'} = f

and

F (0) = 0,

find

F (b)

for

b =

1, 2, 3, 4, 5, 6, and fill these values in the following table.

$b$	1	2	3	4	5	6
$F (b)$

Answer 1.

1

Answer 2.

2

Answer 3.

2.5

Answer 4.

2

Answer 5.

1

Answer 6.

0.5

Solution.

Since

F (0) = 0,

F (b) = \int_{0}^{b} f (t) d t .

For each

b

we determine

F (b)

graphically: for regions where

f

is constant, the integral over that region is just (length)(value of f); for regions that are triangular or trapezoidal, we calculate the area using either of

\frac{1}{2}

(base)(height) (where the height is given by the value of

f

and may be negative), or (length)(average height). Thus,

F (0) = 0

F (1) = F (0) + 1 = 0 + 1 = 1

F (2) = F (1) - 0.5 = 1 - 0.5 = 2

F (3) = F (2) + 0.5 = 2 + 0.5 = 2.5

F (4) = F (3) - 0.5 = 2.5 - 0.5 = 2

F (5) = F (4) - 1 = 2 - 1 = 1

and, finally,

F (6) = F (5) - 0.5 = 1 - 0.5 = 0.5 .

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5. Units Help, once activated and clicked on.

🔗

A rod has length 5 meters. At a distance

x

meters from its left end, the density of the rod is given by

δ (x) = 1 + 2 x

g/m.

🔗

(a) Complete the Riemann sum for the total mass of the rod (use

D x

in place of

Δ x

🔗

mass =

Σ

🔗

(b) Convert the Riemann sum to an integral and find the exact mass.

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mass =

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(include units ¹)

Answer 1.

(1 + 2 x) D x

Answer 2.

30 g

Solution.

(a) Suppose we choose an

x,

0 \leq x \leq 5 .

Δ x

is a small fraction of a meter, then the density of the rod is approximately

δ (x)

anywhere from

x

x + Δ x

meters from the left end of the rod. The mass of the rod from

x

x + Δ x

meters is therefore approximately

δ (x) Δ x = (1 + 2 x) Δ x .

If we slice the rod into

N

pieces, then a Riemann sum is

Σ_{i = 1}^{N} (1 + 2 x_{i}) Δ x,

or, more loosely,

Σ (1 + 2 x) Δ x .

(b) The definite integral is

M = \int_{0}^{5} δ (x) d x = \int_{0}^{5} (1 + 2 x) d x = (1 x + 1 x^{2}) |_{0}^{5} = 30 grams .

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6. Check Units being Accepted when MathQuill is being used.

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The temperature,

H,

in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by

H = f (t),

where

t

is in minutes since the coffee was put on the counter.

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Is $f^{'} (t)$ positive or negative?
- positive
- negative
(Be sure that you are able to give a reason for your answer.)
What are the units of $f^{'} (10) ?$
Suppose that $| f^{'} (10) | = 0.8$ and $f (10) = 72 .$ Fill in the blanks (including units where needed) and select the appropriate terms to complete the following statement about the temperature of the coffee in this case.
At minutes after the coffee was put on the counter, its
- derivative
- temperature
- change in temperature
is and will
- increase
- decrease
by about in the next 30 seconds.

Answer 1.

negative

Answer 2.

degC/min

Answer 3.

10

Answer 4.

temperature

Answer 5.

72 d e g C

Answer 6.

decrease

Answer 7.

0.4 d e g C

Solution.

As the cup of coffee cools, the temperature decreases, so $f^{'} (t)$ is negative.
Since $f^{'} (t) = d H / d t,$ the units are degrees Celsius per minute (degC/min).
The quantity $f^{'} (10)$ represents the rate at which the coffee is cooling, in degrees per minute, 10 minutes after the cup is put on the counter. Thus we can complete the statement:

At 10 minutes after the coffee was put on the counter, its temperature is 72 degC and will decrease by about 0.4 degC in the next 30 seconds.

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7. Click on the HelpLink when activated.

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For the following indefinite integral, find the full power series centered at

t = 0

and then give the first 5 nonzero terms of the power series and the open interval of convergence.

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f (t) = \int \frac{t}{1 + t^{5}} d t

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f (t) = C + \sum_{n = 0}^{\infty}

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f (t) = C +

+

+

+

+

+ \dots

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The open interval of convergence is: (Give your answer in interval notation ².)

Answer 1.

\frac{{(- 1)}^{n} t^{5 n + 2}}{5 n + 2}

Answer 2.

\frac{t^{5 \cdot 0 + 2}}{5 \cdot 0 + 2}

Answer 3.

\frac{- 1 t^{5 \cdot 1 + 2}}{5 \cdot 1 + 2}

Answer 4.

\frac{t^{5 \cdot 2 + 2}}{5 \cdot 2 + 2}

Answer 5.

\frac{- 1 t^{5 \cdot 3 + 2}}{5 \cdot 3 + 2}

Answer 6.

\frac{t^{5 \cdot 4 + 2}}{5 \cdot 4 + 2}

Answer 7.

(- 1, 1)

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8. Weird Error in the Static Rendering.

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Match the following equations with their direction field. Clicking on each picture will give you an enlarged view. While you can probably solve this problem by guessing, it is useful to try to predict characteristics of the direction field and then match them to the picture.

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Here are some handy characteristics to start with -- you will develop more as you practice.

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Set $y$ equal to zero and look at how the derivative behaves along the $x$ -axis.
Do the same for the $y$ -axis by setting $x$ equal to $0$
Consider the curve in the plane defined by setting $y^{'} = 0$ -- this should correspond to the points in the picture where the slope is zero.
Setting $y^{'}$ equal to a constant other than zero gives the curve of points where the slope is that constant. These are called isoclines, and can be used to construct the direction field picture by hand.

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$y^{'} = - \frac{(2 x + y)}{(2 y)}$
$y^{'} = \frac{y}{x} + 3 \cos (2 x)$
$y^{'} = y + x e^{- x} + 1$
$y^{'} = 2 \sin (3 x) + 1 + y$

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Error: PGbasicmacros: imageRow: Unknown displayMode: PTX.

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Error: PGbasicmacros: imageRow: Unknown displayMode: PTX.

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9. My note only said weird... not sure why.

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Given the differential equation

x^{'} (t) = - x^{4} + 1 x^{3} + 13 x^{2} - 12.25 x - 9.1875 .

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List the constant (or equilibrium) solutions to this differential equation in increasing order and indicate whether or not these equilibria are stable, semi-stable, or unstable. (It helps to sketch the graph.³ )

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