Calculus Question

Question

If a snowball melts so that its surface area decreases at a rate of 0.3cm^2/min, find the rate at which the diameter decreases when the diameter is 13cm. 

Answer 1

For the equations let S = surface area, D = diameter and r = radius

S = 4πr² = 4π(D/2)² = πD² which means that D = √(S/π) where S is a function of time.

dD/dt = d√(S/π)/dt = (1/√π)d√S/dt

d√S/dt = [1/(2√S)](dS/dt) = [1/(2√S)](0.3) = 0.3/(2√S)

since we're given that dS/dt = 0.3

So dD/dt = (1/√π)d√S/dt = 0.3/(2√(πS))

We're asked for dD/dt when D = 13 cm

Since S = πD² this means πS = π²D² and √(πS) = πD

Substituting πD for √(πS) in the equation for dD/dt gives

dD/dt = 0.3/(2√(πS)) =0.3/(2πD)

At D = 13

dD/dt = 0.3/(2π(13)) = 0.3/(26π) = 3.67 × 10^-3 cm/min

Answer

The diameter is decreasing at a rate of 3.67 × 10^-3 cm/min

Answer 2

Hi Landon,

SA of a sphere=4πR²=4π(D/2)²=4πD²/4=πD²

^{dSA/dt=2πD(dD/dt) (chain rule)}

^{From the given, dSA/dt=-0.3 cm^2/min and d=13 cm}

^{dD/dt= (dSA/dt)/(2π*D)= (-0.3)/(26π) cm/min. (Plug into a calculator).}

^{. I hope this helps!}