Question

the volume of a sphere is increasing at a constant rate of 59 cubic centimeters per minute. at the instant when the radius of the sphere is 2 centimeters, 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: Differentiate volume with respect to time

We know \( V=\frac{4}{3}\pi r^{3} \). Differentiating both sides with respect to time \( t \) (using the chain rule), we get:
\( \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt} \)
We are given \( \frac{dV}{dt} = 59 \) and \( r = 2 \). We can solve for \( \frac{dr}{dt} \).

Substitute the known values into the derivative of volume:
\( 59=4\pi(2)^{2}\frac{dr}{dt} \)
\( 59 = 16\pi\frac{dr}{dt} \)
Then, \( \frac{dr}{dt}=\frac{59}{16\pi} \)

Step2: Differentiate surface area with respect to time

The surface area of a sphere is \( S = 4\pi r^{2} \). Differentiate both sides with respect to time \( t \) (using the chain rule):
\( \frac{dS}{dt}=8\pi r\frac{dr}{dt} \)

Step3: Substitute \( r \) and \( \frac{dr}{dt} \) into the derivative of surface area

We know \( r = 2 \) and \( \frac{dr}{dt}=\frac{59}{16\pi} \). Substitute these values into the formula for \( \frac{dS}{dt} \):
\( \frac{dS}{dt}=8\pi(2)\times\frac{59}{16\pi} \)
Simplify the expression:
The \( \pi \) terms cancel out, and \( 8\times2 = 16 \), so we have \( \frac{dS}{dt}=\frac{16\times59}{16} \)
The 16 in the numerator and denominator cancels, leaving \( \frac{dS}{dt}=59 \)

Answer:

The rate of change of the surface area of the sphere is \( \boldsymbol{59.000} \) square centimeters per minute (rounded to three decimal places).