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Steve was given this problem:
The radius
r(t) of a circle is increasing at a rate of 3 centimeters per second. At a certain instant
t_(0), the radius is 8 centimeters. What is the rate of change of the area
A(t) of the circle at that instant?
Which equation should Steve use to solve the problem?
Choose 1 answer:
(A)
A(t)=2pi*r(t)
(B)
A(t)=pi[r(t)]^(2)
(C)
A(t)=4pi[r(t)]^(2)
(D)
A(t)=(4)/(3)pi[r(t)]^(3)

Steve was given this problem: $\newline$ The radius $r(t)$ of a circle is increasing at a rate of $3$ centimeters per second. At a certain instant $t_{0}$ , the radius is $8$ centimeters. What is the rate of change of the area $A(t)$ of the circle at that instant? $\newline$ Which equation should Steve use to solve the problem? $\newline$ Choose $1$ answer: $\newline$ (A) $A(t)=2 \pi \cdot r(t)$ $\newline$ (B) $A(t)=\pi[r(t)]^{2}$ $\newline$ (C) $A(t)=4 \pi[r(t)]^{2}$ $\newline$ (D) $A(t)=\frac{4}{3} \pi[r(t)]^{3}$

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Volume of cubes and rectangular prisms: word problems

Full solution

Q. Steve was given this problem:

\newline

The radius

r(t)

of a circle is increasing at a rate of

3

centimeters per second. At a certain instant

t_{0}

, the radius is

8

centimeters. What is the rate of change of the area

A(t)

of the circle at that instant?

\newline

Which equation should Steve use to solve the problem?

\newline

Choose

1

answer:

\newline

(A)

A(t)=2 \pi \cdot r(t)

\newline

(B)

A(t)=\pi[r(t)]^{2}

\newline

(C)

A(t)=4 \pi[r(t)]^{2}

\newline

(D)

A(t)=\frac{4}{3} \pi[r(t)]^{3}

Identify Formula: Identify the correct formula for the area of a circle. $\newline$ The area $A$ of a circle is given by the formula $A = \pi r^2$ , where $r$ is the radius of the circle. This corresponds to option (B) $A(t) = \pi[r(t)]^2$ .
Determine Rate of Change: Determine the rate of change of the area with respect to time. $\newline$ To find the rate of change of the area $A$ with respect to time $t$ , we need to differentiate the area function $A(t)$ with respect to time. This is done using the chain rule of differentiation, since the radius $r$ is a function of time $t$ .
Differentiate Area Function: Differentiate the area function with respect to time. $\newline$ The derivative of $A$ with respect to $t$ is $\frac{dA}{dt}$ , and using the chain rule, we get $\frac{dA}{dt} = \frac{d(\pi r^2)}{dt} = 2\pi r \cdot \frac{dr}{dt}$ . Here, $\frac{dr}{dt}$ is the rate of change of the radius with respect to time, which is given as $3$ centimeters per second.
Substitute Given Values: Substitute the given values into the differentiated formula. $\newline$ At the instant $t_0$ , the radius $r(t)$ is $8$ centimeters, and $\frac{dr}{dt}$ is $3$ centimeters per second. Substituting these values into the formula from Step $3$ , we get $\frac{dA}{dt} = 2\pi \times 8 \, \text{cm} \times 3 \, \text{cm/s}$ .
Calculate Rate of Change: Calculate the rate of change of the area. $\frac{dA}{dt} = 2\pi \times 8\,\text{cm} \times 3\,\text{cm/s} = 48\pi\,\text{cm}^2/\text{s}$ . This is the rate of change of the area of the circle at the instant when the radius is $8$ centimeters.

More problems from Volume of cubes and rectangular prisms: word problems

Question

Chinedu was given this problem:

\newline

A person stands at a distance of

18

meters east of an intersection and watches a car driving away from the intersection to the north at

3

meters per second. The function

y(t)

gives the car's distance, in meters, north of the intersection at time

t

. At a certain instant

t_{0}

\theta\left(t_{0}\right)

between the line of sight from the person to the car and the line of sight from the person to the intersection is

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7

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\theta(t)

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\newline

Which equation should Chinedu use to solve the problem?

\newline

1

\newline

\tan [\theta(t)]=\frac{y(t)}{18}

\newline

\theta(t)=\frac{18 \cdot y(t)}{2}

\newline

\theta(t)+18+y(t)=180

\newline

[\theta(t)]^{2}=18^{2}+[y(t)]^{2}

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Benjamin invests

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Which expression gives the balance in the account after

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1

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400 \cdot 0.05 \cdot 0.05 \cdot 0.05

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400+(1+0.05)(1+0.05)(1+0.05)

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400+0.05 \cdot 0.05 \cdot 0.05

\newline

400(1+0.05)(1+0.05)(1+0.05)

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Question

f(x)=\frac{x+1}{xe^{x}+e^{x}}

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\newline

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\newline

f(-1)

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1

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1

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e

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Question

v=\sqrt{\frac{3 R T}{M}}

\newline

The root-mean-square speed is the measure of the speed of particles in a gas. Root-mean-square speed,

v

, can be calculated using the equation shown, where

M

is the molar mass of a gas,

R

is the molar gas constant, and

T

is the temperature. Which of the following equations correctly expresses the molar mass of a gas in terms of root-mean-square speed, temperature, and the molar gas constant?

\newline

1

\newline

M=\left(\frac{3 R T}{v}\right)^{2}

\newline

M=\frac{3 R T}{(v)^{2}}

\newline

M=\frac{(v)^{2}}{3 R T}

\newline

M=\frac{\sqrt{3 R T}}{v}

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Question

f=3 \pi\left(\frac{d}{2}\right)^{2} h

\newline

A board foot is a

1

1

1

inch thick volume. The formula estimates the board feet,

f

, available in a tree with a diameter of

d

inches and a height of

h

feet. Which of the following equations correctly gives the diameter in terms of the board feet and the height of the tree?

\newline

1

\newline

d=\sqrt{\frac{2 f}{3 \pi h}}

\newline

d=2 \sqrt{\frac{f}{3 \pi h}}

\newline

d=\frac{2 f^{2}}{3 \pi h}

\newline

d=\frac{f^{2} \sqrt{2}}{3 \pi h}

Posted 7 months ago

Question

The surface area of a cylinder is increasing at a rate of

9 \pi

square meters per hour.

\newline

The height of the cylinder is fixed at

3

\newline

At a certain instant, the surface area is

36 \pi

\newline

What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?

\newline

1

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9 \pi

\newline

27 \pi

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\frac{\pi}{3}

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\frac{1}{2}

\newline

The surface area of a cylinder with base radius

r

h

2 \pi r^{2}+2 \pi r h

\newline

The volume of a cylinder with base radius

r

h

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Question

The radius of the base of a cone is increasing at a rate of

10

meters per second.

\newline

The height of the cone is fixed at

6

\newline

At a certain instant, the radius is

1

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What is the rate of change of the volume of the cone at that instant (in cubic meters per second)?

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1

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20 \pi

\newline

40 \pi

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400 \pi

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200 \pi

\newline

The volume of a cone with radius

r

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Question

The side of the base of a square prism is increasing at a rate of

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2

meters per second.

\newline

At a certain instant, the base's side is

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meters and the height is

7

\newline

What is the rate of change of the volume of the prism at that instant (in cubic meters per second)?

\newline

1

\newline

492

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-492

\newline

-348

\newline

348

\newline

The volume of a square prism with base side

s

h

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Question

The side of the base of a square pyramid is increasing at a rate of

6

meters per minute and the height of the pyramid is decreasing at a rate of

1

meter per minute.

\newline

At a certain instant, the base's side is

3

meters and the height is

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\newline

What is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)?

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1

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-111

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\mathbf{1 1 1}

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-105

\newline

105

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h

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Question

You pick a card at random, put it back, and then pick another card at random.

\newline

4

\newline

5

\newline

6

\newline

7

\newline

What is the probability of picking a number greater than

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and then picking a

4

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Write your answer as a percentage.

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