Question

V=4/3? r^2 Suppose that, for the sphere in the video, instead of being told how fast the radius is changing, we're told that the volume is increasing at a constant rate of dV/dt=4 cubic centimeters per second. How fast is the radius increasing at the instant when the radius is r=10 centimeters? dr/dt= centimeters per second.
Instead of thinking about the volume, suppose that we are interested in how the surface area of the sphere is changing. Use the surface area formula S=4? r^2 to determine how fast the surface area is changing at the instant when the radius is r=20 cm and the radius is increasing at dr/dt=2 centimeters per second. dS/dt=

How to get this answer?

Answer 1

Answer:

$\frac{dr}{dt}=0.01$cm per second

$\frac{dS}{dt}=320 square centimeter per secondStep-by-step explanation:We are given that volume of sphere [tex]V=\frac{4}{3}\pi r^3$

Volume of sphere is increasing at a constant rate

$\frac{dV}{dt}=4$ cubic centimeters per second

We have to find the rate of  radius at which increasing

when r= 10 cm

Differentiating w.r.t time

$\frac{dV}{dt}=\frac{4}{3}\pi\cdot3r^2\frac{dr}{dt}$

$4=4 r^2\frac{dr}{dt}$

$\frac{dr}{dt}= \frac{1}{r^2}=\frac}{1}{(10)^2}=0.01$ cm per second

Now ,we are given that surface area of sphere

$S=4\pir^2$

Differentiate w.r.t time then we get

$\frac{dS}[dt}=8\pir\frac{dr}{dt}$

$\frac{dS}{dt}=8\pi\times 20\times 2$

$\frac{dS}{dt}=320$cm per second