Erika was given this problem: $\newline$ The side $s(t)$ of a cube is increasing at a rate of $2$ kilometers per hour. At a certain instant $t_{0}$ , the side is $1$ . $5$ kilometers. What is the rate of change of the volume $V(t)$ of the cube at that instant? $\newline$ Which equation should Erika use to solve the problem? $\newline$ Choose $1$ answer: $\newline$ (A) $V(t)=\sqrt{3[s(t)]^{2}}$ $\newline$ (B) $V(t)=[s(t)]^{2}$ $\newline$ (C) $V(t)=6[s(t)]^{2}$ $\newline$ (D) $V(t)=[s(t)]^{3}$

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Identify equivalent linear expressions: word problems

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Q. Erika was given this problem:

\newline

The side

s(t)

of a cube is increasing at a rate of

2

kilometers per hour. At a certain instant

t_{0}

, the side is

1

5

kilometers. What is the rate of change of the volume

V(t)

of the cube at that instant?

\newline

Which equation should Erika use to solve the problem?

\newline

Choose

1

answer:

\newline

(A)

V(t)=\sqrt{3[s(t)]^{2}}

\newline

(B)

V(t)=[s(t)]^{2}

\newline

(C)

V(t)=6[s(t)]^{2}

\newline

(D)

V(t)=[s(t)]^{3}

Volume Formula: To find the rate of change of the volume of the cube, we need to use the formula for the volume of a cube, which is the side length cubed, $V(t) = [s(t)]^3$ . This is because the volume of a cube is calculated by multiplying the length of the side by itself three times.
Differentiation with Chain Rule: Since the side $s(t)$ is changing with time, we need to differentiate the volume function $V(t)$ with respect to time $t$ to find the rate of change of the volume. This will give us $\frac{dV}{dt}$ , the rate of change of the volume with respect to time.
Given Rates and Values: Differentiating $V(t) = [s(t)]^3$ with respect to time $t$ , we use the chain rule which states that $\frac{dV}{dt} = 3[s(t)]^2 \cdot \frac{ds}{dt}$ , where $\frac{ds}{dt}$ is the rate of change of the side length with respect to time.
Substitution and Simplification: We are given that the rate of change of the side length, $\frac{ds}{dt}$ , is $2$ kilometers per hour. We also know that at the instant $t_0$ , the side length $s(t)$ is $1.5$ kilometers.
Calculation of Rate: Substituting the given values into the differentiated volume formula, we get $\frac{dV}{dt} = 3[1.5]^2 \times 2$ . This simplifies to $\frac{dV}{dt} = 3 \times 2.25 \times 2$ .
Calculation of Rate: Substituting the given values into the differentiated volume formula, we get $\frac{dV}{dt} = 3[1.5]^2 \times 2$ . This simplifies to $\frac{dV}{dt} = 3 \times 2.25 \times 2$ . Calculating the value, we get $\frac{dV}{dt} = 3 \times 2.25 \times 2 = 13.5 \text{ km}^3/\text{hour}$ . This is the rate of change of the volume of the cube at the instant $t_0$ .