Question

the volume of a sphere is increasing at a constant rate of 91 cubic inches per minute. at the instant when the volume of the sphere is 37 cubic inches, what is the rate of change of the surface area of the sphere? the volume of a sphere can be found with the equation $v = \frac{4}{3}\pi r^3$ and the surface area can be found with $s = 4\pi r^2$. round your answer to three decimal places (if necessary).

Explanation:

Step1: Relate \( V \) and \( r \) to find \( r \) when \( V = 37 \)

Given \( V=\frac{4}{3}\pi r^{3} \), solve for \( r \) when \( V = 37 \).
\( 37=\frac{4}{3}\pi r^{3} \)
Multiply both sides by \( \frac{3}{4\pi} \): \( r^{3}=\frac{3\times37}{4\pi}=\frac{111}{4\pi} \)
Take cube root: \( r=\sqrt[3]{\frac{111}{4\pi}} \)
Calculate \( r \): \( r=\sqrt[3]{\frac{111}{4\times3.1416}}\approx\sqrt[3]{\frac{111}{12.5664}}\approx\sqrt[3]{8.833}\approx2.068 \)

Step2: Differentiate \( V \) and \( S \) with respect to time \( t \)

Differentiate \( V=\frac{4}{3}\pi r^{3} \) with respect to \( t \):
\( \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt} \) (using chain rule)
We know \( \frac{dV}{dt} = 91 \), so we can solve for \( \frac{dr}{dt} \):
\( \frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}} \)

Differentiate \( S = 4\pi r^{2} \) with respect to \( t \):
\( \frac{dS}{dt}=8\pi r\frac{dr}{dt} \) (using chain rule)

Step3: Substitute \( \frac{dr}{dt} \) into \( \frac{dS}{dt} \)

Substitute \( \frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}} \) into \( \frac{dS}{dt} \):
\( \frac{dS}{dt}=8\pi r\times\frac{\frac{dV}{dt}}{4\pi r^{2}}=\frac{2\frac{dV}{dt}}{r} \)

Step4: Plug in values of \( \frac{dV}{dt} \) and \( r \)

We have \( \frac{dV}{dt}=91 \) and \( r\approx2.068 \)
\( \frac{dS}{dt}=\frac{2\times91}{2.068}=\frac{182}{2.068}\approx87.998 \)

Answer:

\( \approx 88.000 \) (rounded to three decimal places)