Question

The surface area of a melting snowball decreases at a rate of3.8cm2/min. Find the rate at which its diameter decreases when the diameter is13cm. (Round your answer to three decimal places if required)

Answer 1

Answer:

Step-by-step explanation:

This is a pretty basic related rates problem. I'm going to go through this just like I do in class when I'm teaching it to my students.

We see we have a snowball, which is a sphere. We are talking about the surface area of this sphere which has a formula of

$S=4\pi r^2$

In the problem we are given diameter, not radius. What we know about the relationship between a radius and a diameter is that

d = 2r so

$\frac{d}{2}=r$ Now we can have the equation in terms of diameter instead of radius. Rewriting:

$S=4\pi(\frac{d}{2})^2$ which simplifies to

$S=4\pi(\frac{d^2}{4})$ and a bit more to

$S=\pi d^2$ (the 4's cancel out by division). Now that is a simple equation for which we have to find the derivative with respect to time.

$\frac{dS}{dt}=\pi*2d\frac{dD}{dt}$ Now let's look at the problem and see what we are given as far as information.

The rate at which the surface area changes is -3.8, and we are looking for $\frac{dD}{dt}$, the rate at which the diameter is changing, when the diameter is 13. Filling in:

$-3.8=\pi(2)(13)\frac{dD}{dt}$ and solving for the rate at which the diameter is changing:

$-\frac{3.8}{26\pi}=\frac{dD}{dt}$ and divide to get

$\frac{dD}{dt}=-.459\frac{cm}{min}$ Obviously, the negative means that the diameter is decreasing.