Given:
d(SA)/dt = 10cm^2/min
d(d)/dt = ?
d = 11
SA = 4πr^2 = 4π(d/2)^2 = π*d^2
Calculation:
SA = π*d^2
d(SA)/dt = 2πd[d(d)/dt]
10 = 2π(11)[d(d)/dt]
10 = 22π[d(d)/dt]
5/(11π) = d(d)/dt
Looks like your mistake was the negative there somehow lol
If a snowball melts so that its surface area decreases at a rate of 10 cm^2/min, find the rate at which the diameter decreases when the diameter is 11 cm.
i got -5/11pi, but it says i'm wrong. where did i mess up?
3 answers
Remember that 5/(11π) is the rate that the diameter is decreasing! If it decreased at a rate of -5/(11π), then it's increasing at a rate of 5/(11π)!
Be very careful