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) -156 pi (B) -78 pi (C) -507 pi (D) -12 pi The volume of a cone with radius r and height h is pir^(2)(h)/(3).

Volume Formula Derivation: The formula for the volume of a cone is V = (1/3) * pi * r^2 * h. We need to find the derivative of the volume with respect to time, dV/dt.Constant Height Consideration: Given that the height h is constant at 9 cm, we can treat it as a constant when differentiating.Differentiation with Respect to Time: Differentiate the volume formula with respect to time to get dV/dt = (1/3) * pi * h * 2r * dr/dt, since d(r^2)/dt = 2r * dr/dt.Substitution of Values: Plug in the values: h = 9 cm, r = 13 cm, and dr/dt = -2 cm/min (since the radius is decreasing).Calculation of dV/dt: Calculate dV/dt = (1/3) * pi * 9 * 2 * 13 * (-2).Final Simplification: Simplify the expression: dV/dt = (1/3) * pi * 9 * 2 * 13 * (-2) = -156 pi cubic centimeters per minute.

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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)
-156 pi
(B)
-78 pi
(C)
-507 pi
(D)
-12 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) $-156 \pi$ $\newline$ (B) $-78 \pi$ $\newline$ (C) $-507 \pi$ $\newline$ (D) $-12 \pi$ $\newline$ The volume of a cone with radius $r$ and height $h$ is $\pi r^{2} \frac{h}{3}$ .

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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 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)

-156 \pi

\newline

(B)

-78 \pi

\newline

(C)

-507 \pi

\newline

(D)

-12 \pi

\newline

The volume of a cone with radius

r

and height

h

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

Volume Formula Derivation: The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$ . We need to find the derivative of the volume with respect to time, $\frac{dV}{dt}$ .
Constant Height Consideration: Given that the height $h$ is constant at $9$ cm, we can treat it as a constant when differentiating.
Differentiation with Respect to Time: Differentiate the volume formula with respect to time to get $\frac{dV}{dt} = \left(\frac{1}{3}\right) \cdot \pi \cdot h \cdot 2r \cdot \frac{dr}{dt}$ , since $\frac{d(r^2)}{dt} = 2r \cdot \frac{dr}{dt}$ .
Substitution of Values: Plug in the values: $h = 9 \, \text{cm}$ , $r = 13 \, \text{cm}$ , and $\frac{dr}{dt} = -2 \, \text{cm/min}$ (since the radius is decreasing).
Calculation of $\frac{dV}{dt}$ : Calculate $\frac{dV}{dt} = \left(\frac{1}{3}\right) \cdot \pi \cdot 9 \cdot 2 \cdot 13 \cdot (-2)$ .
Final Simplification: Simplify the expression: $\frac{dV}{dt} = \left(\frac{1}{3}\right) * \pi * 9 * 2 * 13 * (-2) = -156 \pi$ cubic centimeters per minute.