CALCU-1. A 1000-gallon tank is used to collect rai

Solution by Steps

step 1

To find

r'(4)

, we need to differentiate the piecewise function

r(t)

for the interval 2 < t \leq 5

step 2

The derivative of

r(t)

for 2 < t \leq 5 is

r'(t) = 50t \cos(0.2 t^2)

step 3

Substitute

t = 4

into

r'(t)

to find

r'(4)

step 4

Calculate

r'(4) = 50 \cdot 4 \cdot \cos(0.2 \cdot 4^2)

[question a] Answer

r'(4) = 200 \cos(3.2)

gallons per hour

Key Concept

Differentiation of piecewise functions

Explanation

To find the rate of change at a specific point within a given interval of a piecewise function, we differentiate the relevant piece of the function and evaluate at that point.

step 1

To explain why

\lim_{t \to 2} r(t)

does not exist, we need to consider the left-hand limit and the right-hand limit as

t

approaches 2

step 2

Calculate the left-hand limit using the first part of the piecewise function:

\lim_{t \to 2^-} 200e^{-0.21t}

step 3

Calculate the right-hand limit using the second part of the piecewise function:

\lim_{t \to 2^+} 115 + 25\sin(0.2t^2)

step 4

Compare the two limits to determine if they are equal

[question b] Answer

The limits as

t

approaches 2 from the left and right are not equal, hence

\lim_{t \to 2} r(t)

does not exist.

Key Concept

Limits of piecewise functions

Explanation

For a limit to exist at a point where a function is piecewise-defined, the left-hand limit and the right-hand limit must be equal. If they are not, the limit does not exist.

step 1

To find the amount of rainwater from

0 \leq t \leq 2

, we integrate the rate function

r(t)

over this interval

step 2

Set up the integral

\int_0^2 200e^{-0.21t} dt

for the given interval

step 3

Calculate the integral to find the total amount of rainwater

[question c] Answer

The total amount of rainwater is the result of the integral from 0 to 2.

Key Concept

Integration of rate functions

Explanation

The integral of a rate function over an interval gives the total amount of change over that interval.

step 1

To find the total amount of rainwater at

t=5

, we need to integrate the rate function

r(t)

from

0

to

2

and from

2

to

5

, and subtract the constant rate of the pump from

2

to

5

step 2

Set up the integral for the first part

\int_0^2 200e^{-0.21t} dt

and calculate it

step 3

Set up the integral for the second part

\int_2^5 (115 + 25\sin(0.2t^2)) dt

and calculate it

step 4

Subtract the constant rate of the pump over the interval from

2

to

5

, which is

100 \times (5-2)

step 5

Add the results of the two integrals and subtract the pump's contribution to find the total amount of rainwater at

t=5

[question d] Answer

The total amount of rainwater in the tank at

t=5

is the sum of the two integrals minus the pump's contribution.

Key Concept

Integration and constant rates

Explanation

To find the total amount of a quantity that changes at a variable rate and is affected by a constant rate, we integrate the variable rate and account for the constant rate separately.

Generated Graph

Solution by Steps

step 1

To find the acceleration vector at time

t=5

, we need to differentiate the velocity components

x'(t)

and

y'(t)

with respect to time

step 2

The acceleration in the

x

-direction is

x''(t) = \frac{d}{dt}(3\cos(\sqrt{1+t^2}))

step 3

The acceleration in the

y

-direction is

y''(t) = \frac{d}{dt}(-5+4\sqrt{2+\sin t})

step 4

Evaluate

x''(5)

and

y''(5)

to find the acceleration components at

t=5

[question a] Answer

The acceleration vector at

t=5

is

(x''(5), y''(5))

.

Key Concept

Acceleration vector components

Explanation

The acceleration vector is found by differentiating the velocity components with respect to time.

step 1

To find the total distance traveled, we need to integrate the speed function over the given time interval

step 2

The speed function is the magnitude of the velocity vector:

v(t) = \sqrt{(x'(t))^2 + (y'(t))^2}

step 3

Integrate

v(t)

from

t=0

to

t=4

to find the total distance traveled

[question b] Answer

The total distance traveled from

t=0

to

t=4

is the integral of

v(t)

over that interval.

Key Concept

Total distance traveled

Explanation

The total distance is the integral of the speed function over the time interval.

step 1

To find when the vertical distance above the line

y=0

is greatest, we need to find the maximum value of

y(t)

for

0 \leq t \leq 4

step 2

Differentiate

y(t)

to find

y'(t)

and set it equal to zero to find critical points

step 3

Evaluate

y(t)

at the critical points and endpoints to find the maximum value

[question c] Answer

The time

t

when the particle's vertical distance above the line

y=0

is greatest is found by evaluating

y(t)

at critical points and endpoints.

Key Concept

Finding maximum value of a function

Explanation

The maximum value of

y(t)

is found by evaluating the function at critical points and endpoints within the interval.

step 1

To determine when the tangent line is vertical, we need to find when

x'(t) = 0

because the slope of the tangent line is given by

\frac{dy}{dx}

step 2

Solve

x'(t) = 0

for

t

in the interval

(0, 4)

[question d] Answer

The time

t

when the tangent line to the particle's path is vertical is found by solving

x'(t) = 0

within the interval

(0, 4)

.

Key Concept

Vertical tangent line

Explanation

A vertical tangent line occurs when the horizontal component of velocity,

x'(t)

, is zero.

Solution by Steps

step 1

To find the slope of the tangent line at

t=1

, we need to evaluate

\frac{dW}{dt}

at

t=1

step 2

Given

\frac{dW}{dt} = W \cdot e^{-t}

and

W(1) = 2000

, we substitute

t=1

and

W=2000

into the differential equation

step 3

Calculating the derivative at

t=1

gives

\frac{dW}{dt}\bigg|_{t=1} = 2000 \cdot e^{-1} = 2000 \cdot \frac{1}{e}

step 4

Simplifying, we find the slope of the tangent line at

t=1

is

\frac{2000}{e}

step 5

To approximate

W(2)

, we use the tangent line at

t=1

:

W(t) \approx W(1) + \frac{dW}{dt}\bigg|_{t=1} \cdot (t-1)

step 6

Substituting

t=2

into the tangent line equation gives

W(2) \approx 2000 + \frac{2000}{e} \cdot (2-1)

step 7

Simplifying, we find

W(2) \approx 2000 + \frac{2000}{e}

[question a] Answer

The slope of the tangent line at

t=1

is

\frac{2000}{e}

, and the approximate value of

W(2)

using the tangent line is

2000 + \frac{2000}{e}

.

Key Concept

Finding the slope of a tangent line to a curve at a given point using a derivative

Explanation

The slope of the tangent line to the graph of a function at a particular point is given by the value of the derivative of the function at that point.

step 8

Since \frac{d^{2}W}{dt^{2}} < 0 for t > 0, the function

W(t)

is concave down

step 9

A tangent line to a concave down function lies above the function

step 10

Therefore, the approximation for

W(2)

using the tangent line at

t=1

is an overestimate

[question b] Answer

The approximation found in part (a) is an overestimate for

W(2)

.

Key Concept

Concavity and estimation

Explanation

When a function is concave down and we use a tangent line to estimate values, the tangent line will give an overestimate because it lies above the function.

step 11

To solve the differential equation using separation of variables, we start with

\frac{dW}{dt} = W \cdot e^{-t}

step 12

Separating variables, we get

\frac{1}{W} dW = e^{-t} dt

step 13

Integrating both sides, we have

\ln|W| = -e^{-t} + C

step 14

Exponentiating both sides to eliminate the natural logarithm, we get

W = e^{-e^{-t} + C}

step 15

Using the initial condition

W(1) = 2000

, we solve for

C

step 16

Substituting

t=1

into

W = e^{-e^{-t} + C}

, we get

2000 = e^{-e^{-1} + C}

step 17

Solving for

C

, we find

C = \ln(2000) + e^{-1}

step 18

The particular solution is

W(t) = e^{-e^{-t} + \ln(2000) + e^{-1}}

[question c] Answer

The particular solution to the differential equation with initial condition

W(1)=2000

is

W(t) = e^{-e^{-t} + \ln(2000) + e^{-1}}

.

Key Concept

Separation of variables in differential equations

Explanation

Separation of variables is a method to solve a differential equation by separating the variables on different sides of the equation and then integrating.

step 19

To approximate the number of walleye as time approaches infinity, we evaluate the limit of

N(t) = \frac{11,000}{2+3e^{-t}}

as

t \to \infty

step 20

As

t \to \infty

,

e^{-t} \to 0

step 21

Substituting

e^{-t} = 0

into

N(t)

, we get

N(\infty) = \frac{11,000}{2+3(0)}

step 22

Simplifying, we find

N(\infty) = \frac{11,000}{2}

[question d] Answer

As time approaches infinity, the number of walleye in the lake is approximately

\frac{11,000}{2}

.

Key Concept

Limits at infinity for rational functions

Explanation

As

t

approaches infinity, terms involving

e^{-t}

approach zero, simplifying the expression for the limit.

Solution by Steps

step 1

To find the average rate of change of

f'

over the interval

3 \leq x \leq 12

, we use the values of

f'

at

x=3

and

x=12

step 2

The average rate of change is given by the formula

\frac{f'(x_2) - f'(x_1)}{x_2 - x_1}

step 3

Substituting the given values, we get

\frac{9 - 5}{12 - 3}

step 4

Simplifying, we find the average rate of change to be

\frac{4}{9}

[question 4a] Answer

The average rate of change of

f'

over the interval

3 \leq x \leq 12

is

\frac{4}{9}

.

Key Concept

Average Rate of Change

Explanation

The average rate of change of a function over an interval is the change in the function values divided by the change in the input values.

step 1

To explain why there must be at least one value of

x

for

3 \leq x \leq 12

such that

m(x) = 0

, we use the Intermediate Value Theorem

step 2

Since

f

is twice-differentiable, it is also continuous. Therefore,

m(x) = f(f(x))

is continuous on the interval

[3, 12]

step 3

We look for sign changes in

f(x)

within the interval

[3, 12]

step 4

From the table,

f(3) = -2

and

f(12) = 15

. Since there is a sign change from negative to positive, by the Intermediate Value Theorem, there exists a

c

in

(3, 12)

such that

f(c) = 0

step 5

If

f(c) = 0

, then

m(c) = f(f(c)) = f(0)

. Since

f

is continuous, there must be at least one value of

x

in

[3, 12]

where

m(x) = 0

[question 4b] Answer

There must be at least one value of

x

for

3 \leq x \leq 12

such that

m(x) = 0

due to the Intermediate Value Theorem.

Key Concept

Intermediate Value Theorem

Explanation

The Intermediate Value Theorem states that if a function is continuous on a closed interval and takes on different signs at the endpoints, then it must have a zero within that interval.

step 1

To approximate the value of

\int_{3}^{15}(f''(x))^2 dx

using a right Riemann sum, we use the given subintervals and the values of

f''(x)

step 2

The right Riemann sum is given by the sum of the products of the function values at the right endpoints and the widths of the subintervals

step 3

For the interval

[3,5]

, the width is

5-3=2

and the right endpoint value is

(f''(5))^2=(-1)^2=1

step 4

For the interval

[5,12]

, the width is

12-5=7

and the right endpoint value is

(f''(12))^2=(2)^2=4

step 5

For the interval

[12,15]

, the width is

15-12=3

and the right endpoint value is

(f''(15))^2=(-6)^2=36

step 6

Adding the products, we get

2 \cdot 1 + 7 \cdot 4 + 3 \cdot 36

step 7

Simplifying, we find the approximate value to be

2 + 28 + 108 = 138

[question 4c] Answer

The approximate value of

\int_{3}^{15}(f''(x))^2 dx

using a right Riemann sum is 138.

Key Concept

Right Riemann Sum

Explanation

A right Riemann sum approximates the value of an integral by summing the products of the function values at the right endpoints of subintervals and the widths of those subintervals.

step 1

To find

g(5)

, we evaluate the integral from

-2

to

3 \cdot 5 = 15

of

f'(t) dt

and add 7

step 2

We use the values of

f'(x)

from the table to approximate the integral

step 3

The integral from

-2

to

15

can be approximated by adding the areas under the curve represented by

f'(x)

, which are the products of the function values and the widths of the intervals

step 4

We have intervals

[-2,3]

,

[3,5]

,

[5,12]

, and

[12,15]

step 5

The approximate integral is

(3+2) \cdot 12 + (5-3) \cdot 5 + (12-5) \cdot 0 + (15-12) \cdot 9

step 6

Simplifying, we get

5 \cdot 12 + 2 \cdot 5 + 0 + 3 \cdot 9 = 60 + 10 + 0 + 27 = 97

step 7

Adding 7 to the integral, we find

g(5) = 97 + 7 = 104

step 8

To find

g'(5)

, we use the Fundamental Theorem of Calculus which states that

g'(x) = f'(3x) \cdot 3

step 9

Substituting

x=5

, we get

g'(5) = f'(15) \cdot 3

step 10

From the table,

f'(15) = 2

, so

g'(5) = 2 \cdot 3 = 6

[question 4d] Answer

g(5) = 104

and

g'(5) = 6

.

Key Concept

Fundamental Theorem of Calculus

Explanation

The Fundamental Theorem of Calculus relates the derivative of an integral function to the original function, allowing us to find the derivative at a specific point.

Solution by Steps

step 1

To evaluate the limit

\lim _{x \rightarrow 4}\left(\frac{6-2 f(x)}{4-x}\right)

, we recognize this as the definition of a derivative at

x=4

step 2

The derivative of

f

at

x=4

is given by

f'(4)

, which can be found in the table

step 3

Substituting the value from the table, we have

f'(4) = 14

step 4

Therefore, the limit is equal to

-2 \cdot f'(4) = -2 \cdot 14 = -28

[question a] Answer

The limit is

-28

.

Key Concept

The limit as

x

approaches a value that results in the form

\frac{0}{0}

can often be evaluated using the definition of the derivative.

Explanation

The limit given is in the form of the definition of a derivative, which is the rate of change of the function at a particular point. By using the derivative value from the table, we can find the limit.

step 1

To evaluate the integral

\int_{0}^{4} x f''(x) dx

, we use the Fundamental Theorem of Calculus and integration by parts

step 2

Let

u = x

and

dv = f''(x)dx

. Then

du = dx

and we need to find

v

such that

dv = f''(x)dx

step 3

Since

v

is the antiderivative of

f''(x)

,

v = f'(x)

step 4

Applying integration by parts, we get

\int_{0}^{4} x f''(x) dx = [x f'(x)]_0^4 - \int_{0}^{4} f'(x) dx

step 5

We evaluate the first term

[x f'(x)]_0^4

using the values from the table:

4 \cdot f'(4) - 0 \cdot f'(0) = 4 \cdot 14 - 0 = 56

step 6

The second term

\int_{0}^{4} f'(x) dx

is the change in

f(x)

from

0

to

4

, which is

f(4) - f(0) = 3 - 11 = -8

step 7

Combining the two terms, we get

56 - (-8) = 64

[question b] Answer

The value of the integral is

64

.

Key Concept

Integration by parts allows us to transform the integral of a product of functions into a simpler form.

Explanation

By using integration by parts and the values from the table, we can evaluate the integral of

x f''(x)

over the interval

[0, 4]

.

step 1

To evaluate the integral

\int_{1}^{\infty} \frac{f''(1/x)}{x^2} dx

, we use the given substitution

u = \frac{1}{x}

,

du = -\frac{1}{x^2} dx

step 2

Changing the limits of integration, when

x = 1

,

u = 1

and as

x \rightarrow \infty

,

u \rightarrow 0

step 3

The integral becomes

\int_{1}^{0} -f''(u) du

step 4

Since the limits are in the reverse order, we can change them and multiply by

-1

to get

\int_{0}^{1} f''(u) du

step 5

This integral represents the change in

f'(u)

from

0

to

1

step 6

Using the table, we find

f'(1) - f'(0) = -7 - (-2) = -5

[question c] Answer

The value of the integral is

-5

.

Key Concept

Substitution in integrals can simplify the integral and change the limits of integration.

Explanation

By substituting

u = \frac{1}{x}

, we can transform the integral into a definite integral with known values from the table, allowing us to evaluate it.

Generated Graph

Solution by Steps

step 1

To find

f'(0)

, we differentiate the given Maclaurin series term by term

step 2

The first derivative of the function

f(x) = 3 - 4x^2 - \frac{x^4}{4} + \frac{x^6}{192}

is

f'(x) = -8x - x^3 + \frac{x^5}{32}

step 3

Evaluating at

x=0

, we get

f'(0) = 0

step 4

To find

f''(0)

, we differentiate

f'(x)

to get the second derivative

step 5

The second derivative of

f(x)

is

f''(x) = -8 - 3x^2 + \frac{5x^4}{32}

step 6

Evaluating at

x=0

, we get

f''(0) = -8

step 7

Since

f'(0) = 0

and f''(0) < 0,

f

has a relative maximum at

x=0

[1] Answer

f'(0) = 0

,

f''(0) = -8

, and

f

has a relative maximum at

x=0

.

Key Concept

Finding the first and second derivatives at a point to determine the nature of extrema.

Explanation

The first derivative test is used to find where the function has a slope of zero, indicating potential extrema. The second derivative test then determines whether it is a maximum or minimum.

step 8

To use the Lagrange error bound, we need the maximum value of the fifth derivative of

f

on the interval

[-\frac{1}{2}, \frac{1}{2}]

step 9

From the graph, we see that the maximum value of

|f^{(5)}(x)|

is 2

step 10

The Lagrange error bound formula is

E \leq \frac{M|x-a|^n}{n!}

, where

M

is the maximum value of the

n

th derivative on the interval,

x

is the value we are approximating,

a

is the center of the Taylor series, and

n

is the degree of the next term not included in the polynomial

step 11

Plugging in the values, we get

E \leq \frac{2|\frac{1}{2}-0|^5}{5!}

step 12

Simplifying, we find

E \leq \frac{2(\frac{1}{2})^5}{120} = \frac{1}{240}

step 13

Since \frac{1}{240} < \frac{1}{1000}, the error bound is satisfied

[2] Answer

The Lagrange error bound shows that |f(\frac{1}{2}) - P_4(\frac{1}{2})| < \frac{1}{1000}.

Key Concept

Using the Lagrange error bound to estimate the error of a Taylor polynomial approximation.

Explanation

The Lagrange error bound provides an upper limit on the error between the actual function value and the Taylor polynomial approximation, ensuring the approximation is within a desired accuracy.

step 14

The Maclaurin series for

\sin x

is

x - \frac{x^3}{3!} + \cdots

. The first two nonzero terms are

x

and

-\frac{x^3}{6}

step 15

The Maclaurin series for

f'(x)

is

-8x - x^3 + \frac{x^5}{32}

. The first two nonzero terms are

-8x

and

-x^3

step 16

To find the coefficient of

x^4

in the Maclaurin series for

f'(x) \sin x

, we multiply the series for

f'(x)

and

\sin x

and look for terms that result in

x^4

step 17

The

x^4

term comes from multiplying the

-8x

term from

f'(x)

with the

-\frac{x^3}{6}

term from

\sin x

step 18

Multiplying these terms, we get

(-8x)(-\frac{x^3}{6}) = \frac{8x^4}{6} = \frac{4x^4}{3}

step 19

The coefficient of

x^4

is

\frac{4}{3}

[3] Answer

The coefficient of the fourth-degree term in the Maclaurin series for

f'(x) \sin x

is

\frac{4}{3}

.

Key Concept

Finding the coefficient of a specific term in the product of two series.

Explanation

By multiplying corresponding terms from the series of

f'(x)

and

\sin x

, we can find the coefficient of any term in the resulting series.

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