Hello, SuperTyphoon!

5. A circle is inscribed in a square. The circumference of the circle is increasing at a constant

rate of 6 in/sec. As the circle expands, the square expands to maintain tangency.

A. Find the rate at which the perimeter of the square is increasing. Code:

*-------*-*-*-------*
| * * |
| * * |
|* *|
| |
* *
2r * * - - - - *
* r *
| |
|* *|
| * * |
| * * |
*-------*-*-*-------*
2r

The radius of the circle is $\displaystyle r.$

The side of the square is $\displaystyle 2r.$

The circumference of the circle is: .$\displaystyle C \:=\:2\pi r$

Differentiate with respect to time: .$\displaystyle \frac{dC}{dt} \:=\:2\pi\!\cdot\!\frac{dr}{dt}$

We are told that $\displaystyle \frac{dC}{dt} = 6$

So we have: .$\displaystyle 6 \:=\:2\pi\!\cdot\!\frac{dr}{dt} \quad\Rightarrow\quad\frac{dr}{dt} \:=\:\frac{3}{\pi}$ in/sec. .[1]

The perimeter of the square is: .$\displaystyle P \:=\:4(2r) \:=\:8r$

Differentiate with respect to time: .$\displaystyle \frac{dP}{dt} \:=\:8\!\cdot\!\frac{dr}{dt}$ .[2]

Substitute [1] into [2]: .$\displaystyle \frac{dP}{dt} \:=\:8\left(\frac{3}{\pi}\right) \:=\:\boxed{\frac{24}{\pi}\text{ in/sec}}$

B. When the area of the circle is 25π inē,

find the rate on increase of the area between the circle and the square.

The area of the circle is: .$\displaystyle A_c \:=\:\pi r^2$

If the area is $\displaystyle 25\pi$ inē, we have: .$\displaystyle 25\pi \:=\:\pi r^2\quad\Rightarrow\quad r \:=\:5$ .[3]

The area of the square is: .$\displaystyle A_s \:=\:(2r)^2 \:=\:4r^2$

The difference of the areas is: .$\displaystyle D \:=\:A_s - A_c \:=\:4r^2 - \pi r^2\quad\Rightarrow\quad D \:=\:(4-\pi)r^2$

Differentiate with respect to time: .$\displaystyle \frac{dD}{dt} \:=\:2(4-\pi)r\!\cdot\!\frac{dr}{dt}$ .[4]

Substitute [1] and [3] into [4]: .$\displaystyle \frac{dD}{dt} \:=\:2(4-\pi) (5)\left(\frac{3}{\pi}\right) \:=\:\boxed{\frac{30(4-\pi)}{\pi}\text{ in}^2\text{/sec}}$