Question

The radius r of a sphere is increasing at a rate of 8 inches per minute. (a) Find the rate of change of the volume when r = 8 inches. in.3/min (b) Find the rate of change of the volume when r = 38 inches. in.3/min

Answer 1

Answer: a) 2421.743 in³/ min , b) 54437.52 in³/ min

Explanation: The formulae for the volume of a sphere is given below as

V = (4π/3) ×r³ where V =volume of sphere and r = radius of sphere.

By taking the time derivative of the formulae, we have the rate of change of volume with time (we do so by using implicit diffrenciation) , hence we have that

dV/dt = (4π/3) × 3r²×dr/dt

Where dV/dt = rate of change of volume and dr/dt= rate of change of radius.

A)

r = 8 inches, dr/dt = 3 in/min

By substituting this into the formulae, we have that

dV/dt = (4π/3) × 3r²×dr/dt

dV/dt = (4π/3) × 3(8)² × 3

dV/dt = (4π/3) × 576

dV/dt = 2421.743 in³/ min

B)

r = 38 inches, dr/dt = 3 in/min

By substituting this into the formulae, we have that

dV/dt = (4π/3) × 3r²×dr/dt

dV/dt = (4π/3) × 3(38)² × 3

dV/dt = (4π/3) × 12996

dV/dt = 54437.52 in³/ min

Answer 2

a) The rate of change of the volume when r = 8 inches is $2411.52cm^3/s$

b) The rate of change of the volume when r = 38 inches is $54,409.92cm^3/s$

The formula for calculating the volume of a sphere is expressed as:

$V=\frac{4}{3}\pi r^3$

The rate of change of volume is expressed as:

$\frac{dV}{dt}= \frac{dV}{dr} \cdot \frac{dr}{dt}\\ \frac{dV}{dt}=3(\frac{4}{3} )\pi r^2\cdot \frac{dr}{dt}\\ \frac{dV}{dt}=4 \pi r^2\cdot \frac{dr}{dt}$

$\frac{dr}{dt}$ is the rate at which the radius is increasing

If r = 8in and $\frac{dr}{dt}=3in^2/mi n$

Substitute the given values into the formula:

$\frac{dV}{dt}=4 (3.14)(8)^2 \times 3\\\frac{dV}{dt} = 2,411.52cm^3/s$

Similarly if If r = 38in and $\frac{dr}{dt}=3in^2/mi n$

$\frac{dV}{dt}=4 (3.14)(38)^2 \times 3\\\frac{dV}{dt} = 54,409.92m^3/s$

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