# Related rates, clarification sought

1. Apr 28, 2005

### ktpr2

Given, "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."

$$\frac{dD}{dt} = \frac{dD}{dA} \frac{dA}{dt}$$ were D is my diameter, A is (surface) area and t is time.

I relate D to A by, $$A = 4pi(\frac{D}{2})^2 = piD^2; A' = 2piD$$ and then solve,

$$\frac{dD}{dt} = 2piD * -1 cm^2/min =$$
$$2pi(10cm) * -1cm^2/min = -20 cm^3/min$$

I know I am wrong because I got volume, not length. When I solve for variables, like A, and differentiate, do i always place result where ever A shows up in the differentiation equation? In this case, dD/dA, so I should write 1/(2piD)?

The book give the answer as $$\frac{1}{20 cm/min}$$

2. Apr 28, 2005

### cepheid

Staff Emeritus
You solved for dA/dD

You want dD/dA

3. Apr 28, 2005

### ktpr2

yeah I was suspecting that. thanks.