The radius of the base of a cone is increasing at a rate of 10 meters per second. The height of the cone is fixed at 6 meters. At a certain instant, the radius is 1 meter. What is the rate of change of the volume of the cone at that instant (in cubic meters per second)? Choose 1 answer: (A) 40 pi (B) 200 pi (C) 400 pi (D) 20 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 dV/dt, the rate of change of volume with respect to time.Given Parameters: Given: dr/dt = 10 m/s (rate at which radius increases), h = 6 m (height of the cone is constant), and r = 1 m (radius at the instant we are considering).Differentiation with Respect to Time: Differentiate the volume formula with respect to time t to find dV/dt: dV/dt = (1/3) * pi * d(r^2)/dt * h.Substitution and Calculation: Since r^2 differentiates to 2r * dr/dt, we substitute dr/dt = 10 m/s and r = 1 m into the equation: dV/dt = (1/3) * pi * (2 * 1 m * 10 m/s) * 6 m.Simplification: Simplify the expression: dV/dt = (1/3) * pi * 20 m^2/s * 6 m.Final Result: dV/dt = (1/3) * pi * 120 m^3/s.Multiply out the constants: dV/dt = 40 pi m^3/s.

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The radius of the base of a cone is increasing at a rate of 10 meters per second.
The height of the cone is fixed at 6 meters.
At a certain instant, the radius is 1 meter.
What is the rate of change of the volume of the cone at that instant (in cubic meters per second)?
Choose 1 answer:
(A)
40 pi
(B)
200 pi
(C)
400 pi
(D)
20 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 increasing at a rate of $10$ meters per second. $\newline$ The height of the cone is fixed at $6$ meters. $\newline$ At a certain instant, the radius is $1$ meter. $\newline$ What is the rate of change of the volume of the cone at that instant (in cubic meters per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $40 \pi$ $\newline$ (B) $200 \pi$ $\newline$ (C) $400 \pi$ $\newline$ (D) $20 \pi$ $\newline$ 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 6

Volume of cubes and rectangular prisms: word problems

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

10

meters per second.

\newline

The height of the cone is fixed at

6

meters.

\newline

At a certain instant, the radius is

1

meter.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

40 \pi

\newline

(B)

200 \pi

\newline

(C)

400 \pi

\newline

(D)

20 \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 $\frac{dV}{dt}$ , the rate of change of volume with respect to time.
Given Parameters: Given: $\frac{dr}{dt} = 10 \, \text{m/s}$ (rate at which radius increases), $h = 6 \, \text{m}$ (height of the cone is constant), and $r = 1 \, \text{m}$ (radius at the instant we are considering).
Differentiation with Respect to Time: Differentiate the volume formula with respect to time $t$ to find $\frac{dV}{dt}$ : $\frac{dV}{dt} = \frac{1}{3} \cdot \pi \cdot \frac{d(r^2)}{dt} \cdot h$ .
Substitution and Calculation: Since $r^2$ differentiates to $2r \cdot \frac{dr}{dt}$ , we substitute $\frac{dr}{dt} = 10 \, \text{m/s}$ and $r = 1 \, \text{m}$ into the equation: $\frac{dV}{dt} = \left(\frac{1}{3}\right) \cdot \pi \cdot (2 \cdot 1 \, \text{m} \cdot 10 \, \text{m/s}) \cdot 6 \, \text{m}$ .
Simplification: Simplify the expression: $\frac{dV}{dt} = \left(\frac{1}{3}\right) \cdot \pi \cdot 20 \, \text{m}^2/\text{s} \cdot 6 \, \text{m}$ .
Final Result: $\frac{dV}{dt} = \frac{1}{3} \times \pi \times 120\, \text{m}^3/\text{s}$ .
Final Result: $\frac{dV}{dt} = \frac{1}{3} \cdot \pi \cdot 120 \, \text{m}^3/s$ .Multiply out the constants: $\frac{dV}{dt} = 40 \pi \, \text{m}^3/s$ .