7. a spherical balloon is expanding at a rate of ( 60pi \text{ in}^3/\text{sec} ). how fast is the surface…

7. a spherical balloon is expanding at a rate of ( 60pi \text{ in}^3/\text{sec} ). how fast is the surface area of the balloon expanding when the radius of the balloon is 4 inches? ( v=\frac{4}{3}pi r^3 ) and ( a = 4pi r^2 ).

7. a spherical balloon is expanding at a rate of ( 60pi \text{ in}^3/\text{sec} ). how fast is the surface area of the balloon expanding when the radius of the balloon is 4 inches? ( v=\frac{4}{3}pi r^3 ) and ( a = 4pi r^2 ).

Answer

Explanation:

Step1: Differentiate volume formula

Given (V=\frac{4}{3}\pi r^{3}), differentiate with respect to (t): (\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}). We know (\frac{dV}{dt} = 60\pi) and (r = 4). Substitute values: (60\pi=4\pi(4)^{2}\frac{dr}{dt}). Solve for (\frac{dr}{dt}): (\frac{dr}{dt}=\frac{60\pi}{16\pi}=\frac{15}{4}).

Step2: Differentiate surface - area formula

Given (A = 4\pi r^{2}), differentiate with respect to (t): (\frac{dA}{dt}=8\pi r\frac{dr}{dt}). Substitute (r = 4) and (\frac{dr}{dt}=\frac{15}{4}): (\frac{dA}{dt}=8\pi\times4\times\frac{15}{4}). Calculate: (\frac{dA}{dt}=120\pi).

Answer:

The surface area of the balloon is expanding at a rate of (120\pi\space in^{2}/sec).