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The side of a cube is decreasing at a rate of 9 millimeters per minute.
At a certain instant, the side is 19 millimeters.
What is the rate of change of the volume of the cube at that instant (in cubic millimeters per minute)?
Choose 1 answer:
(A) -729
(B) -9747
(C) -6859
(D) -3249

The side of a cube is decreasing at a rate of $9$ millimeters per minute. $\newline$ At a certain instant, the side is $19$ millimeters. $\newline$ What is the rate of change of the volume of the cube at that instant (in cubic millimeters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $-729$ $\newline$ (B) $-9747$ $\newline$ (C) $-6859$ $\newline$ (D) $-3249$

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Calculate unit rates with fractions

Full solution

Q. The side of a cube is decreasing at a rate of

9

millimeters per minute.

\newline

At a certain instant, the side is

19

millimeters.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

-729

\newline

(B)

-9747

\newline

(C)

-6859

\newline

(D)

-3249

Volume formula: Volume of a cube formula is $V = s^3$ , where $s$ is the side length.
Differentiate volume: Differentiate the volume with respect to time to find the rate of change of volume, $\frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt}$ .
Plug in values: Plug in the values: $s = 19 \, \text{mm}$ and $\frac{ds}{dt} = -9 \, \text{mm/min}$ .
Calculate $\frac{dV}{dt}$ : Calculate $\frac{dV}{dt} = 3 \times (19)^2 \times (-9)$ .
Simplify calculation: Simplify the calculation: $\frac{dV}{dt} = 3 \times 361 \times (-9)$ .
Multiply for rate: Multiply to find the rate of change: $\frac{dV}{dt} = 3 \times 361 \times (-9) = 1083 \times (-9)$ .
Final calculation: Final calculation: $\frac{dV}{dt} = -9747 \, \text{mm}^3/\text{min}.$

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