If a snowball melts so that its surface area decreases at a rate of 9 cm2/min, find the rate at which the diameter decreases when the diameter is 12 cm.

Question

zepdrix · Answer

Hmmmmmmmmmmmmmm

So we'll assume the snowball is spherical :)
The Surface Area of a sphere is given as:
$$\huge A=4 \pi r^2$$
I think.. right?

zepdrix · Answer

They gave us some information.. they told us the instantaneous rate of change of the Surface Area:
$$\huge A'=-9cm^2/\min$$

zepdrix · Answer

Negative since it's DECREASING! :O

zepdrix · Answer

So they told us that when the diameter (d) is a particular value, find d'

zepdrix · Answer

So we want to rewrite our Area formula is terms of DIAMETER, not RADIUS.
So how can we replace the r^2 with d's?

anonymous · Answer

totally missed that part^

anonymous · Answer

r = d/2

anonymous · Answer

since you have A' you can use that to get d' im guessing

zepdrix · Answer

$$\huge A=4\pi(r)^2=4\pi\left(\frac{ d }{ 2 }\right)^2$$$$\huge =\pi(d)^2$$

zepdrix · Answer

Yes! true story! :D

anonymous · Answer

a' = 2pi d

zepdrix · Answer

Just don't forget to square the entire substitution you make when you plug in the d/2! the 1/2 get's squared also! :D cancelling out with the 4.

anonymous · Answer

-9/2pi = d

zepdrix · Answer

Remember, we're differentiating with respect to TIME. so not only do we get an A', but we should also get a d' when we differentiate d.

anonymous · Answer

uhm. yes..

anonymous · Answer

so do you just get 2d * d' in this case

anonymous · Answer

and then divide it.. because it's given?

zepdrix · Answer

yes looks good! c:
And from that point, we're left with 3 variables (2 of which they TOLD US!) and we can solve for d' from there! :D

anonymous · Answer

it's.. magic.

zepdrix · Answer

hah XD you know it cool cat!

anonymous · Answer

you should get a trophy

zepdrix · Answer

Pshhhh i got a medal :D good enough!