the radius of a sphere is decreasing at a constant rate of 7 centimeters per minute. at the instant when the…

the radius of a sphere is decreasing at a constant rate of 7 centimeters per minute. at the instant when the radius of the sphere is 8 centimeters, what is the rate of change in the volume? the volume of a sphere can be found with the equation v = 4/3πr³. round your answer to three decimal places.

the radius of a sphere is decreasing at a constant rate of 7 centimeters per minute. at the instant when the radius of the sphere is 8 centimeters, what is the rate of change in the volume? the volume of a sphere can be found with the equation v = 4/3πr³. round your answer to three decimal places.

Answer

Explanation:

Step1: Differentiate volume formula

We have $V=\frac{4}{3}\pi r^{3}$. Differentiating with respect to time $t$ using the chain - rule, $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$.

Step2: Substitute given values

We know that $\frac{dr}{dt}=- 7$ cm/min (negative because the radius is decreasing) and $r = 8$ cm. Substitute these values into the derivative formula: $\frac{dV}{dt}=4\pi(8)^{2}(-7)$.

Step3: Calculate the result

First, $(8)^{2}=64$. Then $4\pi\times64\times(-7)=4\times\pi\times64\times(-7)=-1792\pi\approx - 5632.920$ $cm^{3}/min$.

Answer:

$-5632.920$ $cm^{3}/min$