The formula for the volume of a sphere is S = 4 3 πr3, where r (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun. (a) Suppose r = 1 (t + 1)2 − 1 15 , where t is time in minutes. Use the chain rule, dS dt = dS dr · dr dt , to find the rate at which the snowball is melting. dS dt = (b) Use the results from part (a) to find the rate (in ft3/min) at which the volume is changing at t = 1 min. (Round your answer to four decimal places.) ft3/min

Answer with Step-by-step explanation: We are given thatVolume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]Where r= Radius of sphere in fta.[tex]r=\frac{1}{(t+1)^2}-\frac{1}{15}[/tex][tex]\frac{dr}{dt}=\frac{-2}{(1+t)^3}[/tex][tex]\frac{dS}{dt}=\frac{dS}{dr}\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}[/tex][tex]\frac{dS}{dt}=4\pi(\frac{1}{(1+t)^2}-\frac{1}{15})^2\times \frac{-2}{(1+t)^3}[/tex][tex]\frac{dS}{dt}=\frac{-8\pi}{(1+t)^3}(\frac{1}{(1+t)^2}-\frac{1}{15})^2[/tex]b.Substitute t=1 Then, we get [tex]\frac{dS}{dt}=\frac{-8\pi}{8}(\frac{1}{4}-\frac{1}{15})^2[/tex][tex]\frac{dS}{dt}=-0.1056 ft^3/min[/tex]