7. a spherical balloon is expanding at a rate of 60π in³/sec. how fast is the surface area of the balloon…

Explanation:

Step1: Differentiate volume formula

Given (V=\frac{4}{3}\pi r^{3}), differentiate with respect to (t) (time). Using the chain - rule (\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}). We know (\frac{dV}{dt} = 60\pi) in³/sec. So, (60\pi=4\pi r^{2}\frac{dr}{dt}).

Step2: Solve for (\frac{dr}{dt})

Divide both sides of (60\pi = 4\pi r^{2}\frac{dr}{dt}) by (4\pi r^{2}). (\frac{dr}{dt}=\frac{60\pi}{4\pi r^{2}}=\frac{15}{r^{2}}). When (r = 4) inches, (\frac{dr}{dt}=\frac{15}{4^{2}}=\frac{15}{16}) in/sec.

Step3: Differentiate surface - area formula

Given (A = 4\pi r^{2}), differentiate with respect to (t) (time). Using the chain - rule (\frac{dA}{dt}=8\pi r\frac{dr}{dt}).

Step4: Substitute (r = 4) and (\frac{dr}{dt}=\frac{15}{16})

(\frac{dA}{dt}=8\pi\times4\times\frac{15}{16}). First, (8\times4\times\frac{15}{16}=\frac{8\times4\times15}{16}=\frac{480}{16} = 30).

Answer:

(30\pi) in²/sec