Calculus question. Just want to check that I did this right.
Q. If a snowball melts so that its diameter d decreases at a rate of 2cm/minute, find the rate at which its surface area decreases when the diameter is 20cm.

First, I converted the diameter to radius.
r=d/2
r=20/2=10

then I took the derivative of the surface area formula.
SA=4(pi)(r^2)
SA'=4(pi)(r^2)'
SA'=8(pi)(r)

I then plugged in the radius
SA'(at r=10cm)=8(pi)(10)
=80(pi) cm/minute
Is this right?

Question

Calculus question. Just want to check that I did this right. 
Q. If a snowball melts so that its diameter d decreases at a rate of 2cm/minute, find the rate at which its surface area decreases when the diameter is 20cm. 

First, I converted the diameter to radius. 
r=d/2
r=20/2=10

then I took the derivative of the surface area formula.
SA=4(pi)(r^2)
SA'=4(pi)(r^2)'
SA'=8(pi)(r)

I then plugged in the radius
SA'(at r=10cm)=8(pi)(10)
                       =80(pi) cm/minute
Is this right?

dumbcow · Answer

not quite, you are not factoring in that diameter is decreasing at rate of 2cm/min
the rate they want is per time but you took derivative with respect to r not t so you can't just plug in radius
use this equation to help 
$$\frac{dA}{dt} = \frac{dA}{dr}*\frac{dr}{dt}$$
by multiplying the 2 rates, the "dr" cancels and you are left with desired rate of surface area per time
dr/dt is given as -1 cm/min
thus
dA/dt = -80pi sq.cm/ min

anonymous · Answer

thank you :). I got it!