Related Rates Question

[JaiLit51853]

Hi, I got a set of review questions that I don't the answers to - would appreciate it if someone could check my work for this question and give me advice if I went wrong:

If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decreases when the diameter is 10 cm.

My work:
A = 4πr^2
A = 4π(d/2)^2
A = 4πd^2/4
A = πd^2

dA/dt = π(2d)(dd/dt)
-1 cm^2/min = 2π(10 cm)(dd/dt)
-0.1 cm/min = 2π(dd/dt)
dd/dt = -0.016 cm/min ................................edited

Thanks so much!

 

[Romsek]

looks good til the last line. How can cm be the units of \(\displaystyle \dfrac{dd}{dt}\) ?

 

[JaiLit51853]

Oh, that makes sense. It's still cm/min from the last line - thanks!

 

[HallsofIvy]

You might have noticed that it is NOT a good idea to use "d" to mean "diameter" when you are using "d" as part of the derivative! Use "D" or "x" or something else instead.