The radius of a sphere is decreasing at a rate of 1 meter per hour. At a certain instant, the radius is 4 meters. What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?

Given Information: Given: Rate of change of radius (dr/dt) = -1 meter per hour (negative because the radius is decreasing) Radius (r) at a certain instant = 4 meters We need to find the rate of change of the volume (dv/dt) at that instant. The volume of a sphere is given by the formula V = (4/3)πr^3.Differentiate Volume Formula: Differentiate the volume formula with respect to time (t) to find the rate of change of the volume. (dV/dt) = (d/dt)((4/3)πr^3) Using the chain rule, we get: (dV/dt) = (4/3)π * 3r^2 * (dr/dt) (dV/dt) = 4πr^2 * (dr/dt)Substitute Values: Substitute the given values of r and dr/dt into the differentiated volume formula. r = 4 meters dr/dt = -1 meter per hour (dV/dt) = 4π(4 meters)^2 * (-1 meter per hour) (dV/dt) = 4π * 16 * (-1) (dV/dt) = -64π cubic meters per hourFinal Result: The rate of change of the volume of the sphere at the instant when the radius is 4 meters is -64π cubic meters per hour.

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The radius of a sphere is decreasing at a rate of 1 meter per hour.
At a certain instant, the radius is 4 meters.
What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?

The radius of a sphere is decreasing at a rate of $1$ meter per hour. $\newline$ At a certain instant, the radius is $4$ meters. $\newline$ What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?

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Calculus

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

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Q. The radius of a sphere is decreasing at a rate of

1

meter per hour.

\newline

At a certain instant, the radius is

4

meters.

\newline

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

Given Information: Given: $\newline$ Rate of change of radius $\frac{dr}{dt} = -1$ meter per hour (negative because the radius is decreasing) $\newline$ Radius $r$ at a certain instant $= 4$ meters $\newline$ We need to find the rate of change of the volume $\frac{dv}{dt}$ at that instant. $\newline$ The volume of a sphere is given by the formula $V = \frac{4}{3}\pi r^3$ .
Differentiate Volume Formula: Differentiate the volume formula with respect to time $t$ to find the rate of change of the volume. $\newline$ $(\frac{dV}{dt}) = (\frac{d}{dt})((\frac{4}{3})\pi r^3)$ $\newline$ Using the chain rule, we get: $\newline$ $(\frac{dV}{dt}) = (\frac{4}{3})\pi \cdot 3r^2 \cdot (\frac{dr}{dt})$ $\newline$ $(\frac{dV}{dt}) = 4\pi r^2 \cdot (\frac{dr}{dt})$
Substitute Values: Substitute the given values of $r$ and $\frac{dr}{dt}$ into the differentiated volume formula. $\newline$ $r = 4$ meters $\newline$ $\frac{dr}{dt} = -1$ meter per hour $\newline$ $\frac{dV}{dt} = 4\pi(4 \text{ meters})^2 * (-1 \text{ meter per hour})$ $\newline$ $\frac{dV}{dt} = 4\pi * 16 * (-1)$ $\newline$ $\frac{dV}{dt} = -64\pi$ cubic meters per hour
Final Result: The rate of change of the volume of the sphere at the instant when the radius is $4$ meters is $-64\pi$ cubic meters per hour.