Answer to Question #121255 in Calculus for kylie

Question

A round pizza is baking in the oven. As the pizza gets warmer, its radius increases at a rate of 0.01 centimeters per minute. When the diameter of the pizza is 50 centimeters, what is the rate that the pizza's area is increasing?
Your solution myst include a sketch of the situation and labeling all given quantities. Leave your final answer in terms of pi.

Answer 1

The area of the round pizza is $A=\pi r^2$ , where $r$ is he radius.

When the diameter is $50$ cm, the radius of the pizza is $r=\frac{50}{2}=25$ cm.

Differentiate $A$ with respect to $t$ as,

$\frac{dA}{dt}=\frac{d}{dt}(\pi r^2)$

$=\pi (2r)\frac{dr}{dt}$

$=2\pi r\frac{dr}{dt}$

Plug $r=25$ and $\frac{dr}{dt}=0.01$ to find the rate of increase of area as,

$\frac{dA}{dt}=2\pi(25)(0.01)$

$=50\pi(0.01)$

$=\frac{1}{2}\pi$

Therefore, the rate of increase of area is $\frac{dA}{dt}=\frac{1}{2}\pi=0.5\pi$ $cm^2/min$