Question

The area of a circle is expanding at a rate of 5 inches^2/sec. At what rate is the radius changing when the radius is 8 inches.

Answer 1

\[\frac{\mathrm{d}A}{\mathrm{d}t}=5 \ inch^2/s \]

\[A=\pi r^2\]
\[\frac{\mathrm{d}A}{\mathrm{d}t}=2\pi r \frac{\mathrm{d}r}{\mathrm{d}t} \]
\[\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{\mathrm{d}A}{\mathrm{d}t}\frac{1}{2 \pi r}\]
then you just plug in your values for dA/dt and r ^_^

does that make sense?

Answer 2

\[ \frac{dr }{ dt}= 5(\frac{ 1 }{ 2 \pi (8) })\]


is this the answer

Answer 3

yup yup yup ^_^