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:[tex]\frac{dr}{dt}=0.01 [/tex]cm per second[tex]\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[/tex]Volume of sphere is increasing at a constant rate [tex]\frac{dV}{dt}=4[/tex] cubic centimeters per secondWe have to find the rate of  radius at which increasing when r= 10 cmDifferentiating w.r.t time[tex]\frac{dV}{dt}=\frac{4}{3}\pi\cdot3r^2\frac{dr}{dt}[/tex][tex]4=4 r^2\frac{dr}{dt}[/tex][tex]\frac{dr}{dt}= \frac{1}{r^2}=\frac}{1}{(10)^2}=0.01 [/tex] cm per second Now ,we are given that surface area of sphere [tex] S=4\pir^2[/tex]Differentiate w.r.t time then we get [tex]\frac{dS}[dt}=8\pir\frac{dr}{dt}[/tex][tex]\frac{dS}{dt}=8\pi\times 20\times 2[/tex][tex]\frac{dS}{dt}=320 [/tex]cm per second