Surface area snowball

Question

If a snowball melts so that its surface area decreases at a rate of 4 cm²/min, find the rate at which the diameter decreases when the diameter is 9 cm

Answer 1

Let's assume a snowball is a sphere. The formula for the surface area of a sphere is 4πr² but since the question wants diameter, lets substitute d/2 = r so:

A_s(d) = 4π(d/2)² = 4πd²/4 = πd²

Since we need to take the derivative, let's use a different variable for diameter (too many d's) so how about "w" for width:

A_s(w) = πw²

To find the rate, take the derivative with respect to time:

dA_s/dt = 2πw•(dw/dt) using the chain rule.

We are looking for dw/dt so let's solve for that:

dw/dt = (dA_s/dt)/(2πw)

Now, plug in the numbers

dA_s/dt = 4

w = 9

So: dw/dt = (4)/(2π(9)) = 2/(9π) ≈ 0.07 cm/min

Since it is decreasing, the proper way to denote it is dw/dt ≈ -0.07 cm/min

Answer 2

Hi Hamoton! Thanks for asking me a good question! As long you know basic calculus, you should be able to understand my steps!

I hope you know, by now, that the volume of the sphere is V = (4/3) * π * r ^ 3.

Note that we can find the surface area by differentiating the volume equation with respect to radius (because we know that differentiating volume (cm ^ 3) yields area (cm ^ 2)) or from memory that SA = 4 * π * r ^ 2.

We already know from the problem that the rate of the surface area that is changing is -4 cm ^ 2 / min, and we specifically want to know that rate of change of radius when it is 4.5 cm (diameter / 2 = radius, right?)!

Because we want to know the rate of change of SA with respect to time (note the units of the -4 cm ^ 2 / min is in area per time, so the derivative we are about to take must be with respect to time)!

d(SA) / dt = 4π * d/dt(r ^ 2) = 4π * 2r * dr/dt ⇒ -4 = 4π * 2 * (4.5) * dr/dt ⇒ dr/dt = -0.0354 cm ^ 2 / min

However, this is the rate of change of radius with respect to time, and recall that:

d = 2r ⇒ d(d) / dt = 2 * dr/dt ⇒ -0.0707 cm ^ 2 / min

Therefore, the diameter is decreasing at a rate of 0.0707 cm ^ 2 / min <<< ANSWER

Hope this helped!