The radius of the base of a cylinder is decreasing at a rate of 12 kilometers per second. The height of the cylinder is fixed at 2.5 kilometers. At a certain instant, the radius is 40 kilometers. What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)? Choose 1 answer: (A) -1600 pi (B) -2400 pi (C) -4000 pi (D) -360 pi The volume of a cylinder with radius r and height h is pir^(2)h.

Write Known Values: The formula for the volume of a cylinder is V = πr^2h. We need to find the derivative of the volume with respect to time, dV/dt.Find Derivative: First, let's write down what we know: The radius is decreasing, so dr/dt = -12 km/s (negative because it's decreasing). The height is constant, so dh/dt = 0 km/s. The radius at the instant we're considering is r = 40 km. The height is h = 2.5 km.Apply Chain Rule: Now, let's find the derivative of the volume with respect to time using the chain rule: dV/dt = d(πr^2h)/dt = πh * 2r * dr/dt.Plug in Values: Plug in the values we know: dV/dt = π * 2.5 * 2 * 40 * (-12).Calculate Rate of Change: Calculate the rate of change: dV/dt = π * 2.5 * 2 * 40 * (-12) = -2400π cubic kilometers per second.

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The radius of the base of a cylinder is decreasing at a rate of 12 kilometers per second.
The height of the cylinder is fixed at 2.5 kilometers.
At a certain instant, the radius is 40 kilometers.
What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)?
Choose 1 answer:
(A)
-1600 pi
(B)
-2400 pi
(C)
-4000 pi
(D)
-360 pi
The volume of a cylinder with radius
r and height
h is
pir^(2)h.

The radius of the base of a cylinder is decreasing at a rate of $12$ kilometers per second. $\newline$ The height of the cylinder is fixed at $2$ . $5$ kilometers. $\newline$ At a certain instant, the radius is $40$ kilometers. $\newline$ What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $-1600 \pi$ $\newline$ (B) $-2400 \pi$ $\newline$ (C) $-4000 \pi$ $\newline$ (D) $-360 \pi$ $\newline$ The volume of a cylinder with radius $r$ and height $h$ is $\pi r^{2} h$ .

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Math Problems

Grade 7

Volume of cubes and rectangular prisms: word problems

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Q. The radius of the base of a cylinder is decreasing at a rate of

12

kilometers per second.

\newline

The height of the cylinder is fixed at

2

5

kilometers.

\newline

At a certain instant, the radius is

40

kilometers.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

-1600 \pi

\newline

(B)

-2400 \pi

\newline

(C)

-4000 \pi

\newline

(D)

-360 \pi

\newline

The volume of a cylinder with radius

r

and height

h

\pi r^{2} h

Write Known Values: The formula for the volume of a cylinder is $V = \pi r^2 h$ . We need to find the derivative of the volume with respect to time, $\frac{dV}{dt}$ .
Find Derivative: First, let's write down what we know: $\newline$ The radius is decreasing, so $\frac{dr}{dt} = -12$ km/s (negative because it's decreasing). $\newline$ The height is constant, so $\frac{dh}{dt} = 0$ km/s. $\newline$ The radius at the instant we're considering is $r = 40$ km. $\newline$ The height is $h = 2.5$ km.
Apply Chain Rule: Now, let's find the derivative of the volume with respect to time using the chain rule: $\frac{dV}{dt} = \frac{d(\pi r^2 h)}{dt} = \pi h \cdot 2r \cdot \frac{dr}{dt}$ .
Plug in Values: Plug in the values we know: $\frac{dV}{dt} = \pi \times 2.5 \times 2 \times 40 \times (-12)$ .
Calculate Rate of Change: Calculate the rate of change: $\frac{dV}{dt} = \pi \times 2.5 \times 2 \times 40 \times (-12) = -2400\pi$ cubic kilometers per second.