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The radius of the base of a cone is decreasing at a rate of 2 centimeters per minute.
The height of the cone is fixed at 9 centimeters.
At a certain instant, the radius is 13 centimeters.
What is the rate of change of the volume of the cone at that instant (in cubic centimeters per minute)?
Choose 1 answer:
(A)
-78 pi
(B)
-156 pi
(C)
-12 pi
(D)
-507 pi
The volume of a cone with radius
r and height
h is
pir^(2)(h)/(3).

The radius of the base of a cone is decreasing at a rate of $2$ centimeters per minute. $\newline$ The height of the cone is fixed at $9$ centimeters. $\newline$ At a certain instant, the radius is $13$ centimeters. $\newline$ What is the rate of change of the volume of the cone at that instant (in cubic centimeters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $-78 \pi$ $\newline$ (B) $-156 \pi$ $\newline$ (C) $-12 \pi$ $\newline$ (D) $-507 \pi$ $\newline$ The volume of a cone with radius $r$ and height $h$ is $\pi r^{2} \frac{h}{3}$ .

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Find the rate of change of one variable when rate of change of other variable is given

Full solution

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

2

centimeters per minute.

\newline

The height of the cone is fixed at

9

centimeters.

\newline

At a certain instant, the radius is

13

centimeters.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

-78 \pi

\newline

(B)

-156 \pi

\newline

(C)

-12 \pi

\newline

(D)

-507 \pi

\newline

The volume of a cone with radius

r

and height

h

is

\pi r^{2} \frac{h}{3}

.

Given Information: Given: $\newline$ Rate of change of radius $\frac{dr}{dt} = -2$ cm/min (since the radius is decreasing) $\newline$ Height of cone $h = 9$ cm (constant) $\newline$ Radius $r = 13$ cm $\newline$ Volume of a cone $V = \frac{1}{3}\pi r^2 h$ $\newline$ We need to find $\frac{dV}{dt}$ .
Differentiate Volume Formula: Differentiate the volume formula with respect to time $t$ to find $\frac{dV}{dt}$ .
$\frac{dV}{dt} = \frac{1}{3}\pi(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})$
Since the height is constant, $\frac{dh}{dt} = 0$ , so the equation simplifies to:
$\frac{dV}{dt} = \frac{1}{3}\pi(2rh \frac{dr}{dt})$
Substitute Given Values: Substitute the given values into the differentiated equation. $\newline$ $\frac{dV}{dt} = \frac{1}{3}\pi(2 \times 9 \times 13 \times (-2))$ $\newline$ $\frac{dV}{dt} = \frac{1}{3}\pi(2 \times 9 \times 13 \times (-2)) = -234\pi$

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