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).

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

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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

View step-by-step help

Home

Math Problems

Calculus

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

\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. $\newline$ $\frac{dV}{dt} = \frac{1}{3}\pi(2 \times 9 \times 13 \times (-2))$ $\newline$ $\frac{dV}{dt} = \frac{1}{3}\pi(2 \times 9 \times 13 \times (-2)) = -234\pi$