Answer with Step-by-step explanation:
We are given that
Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]
Where r= Radius of sphere in ft
a.[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]
