Question

a circle is inscribed in a square. the circumference of the circle is increading at a constant rate of 6 inches per second. As the circle expands, the aquare expands to maintain the condition of tangency. find th rate at which the perimeter of the square is increasing

Answer 1

Answer:

The rate at which Perimeter of the square is increasing is $\frac{24}{\pi} \ in/secs$.

Step-by-step explanation:

Given:

Circumference of the circle = $2\pi r$

Rate of change of in circumference = 6 in/secs

We need to find the rate at which the perimeter of the square is increasing

Solution:

Now we know that;

$\frac{d(2\pi r)}{dt} =6\\\\2\pi\frac{dr}{dt}=6\\\\\frac{dr}{dt}=\frac{6}{2\pi}\\\\\frac{dr}{dt}=\frac{3}{\pi}$

Now we know that;

side of the square= diameter of the circle

side of the square = $2r$

Now Perimeter of the square is given by 4 times length of the side.

$P=4\times 2r =8r$

Now we need to find the rate at which Perimeter is increasing so we will find the derivative of perimeter.

$\frac{dP}{dt}= \frac{d(8r)}{dt}\\\\\frac{dP}{dt}= 8\times\frac{dr}{dt}$

But $\frac{dr}{dt} =\frac{3}{\pi}$

So we get;

$\frac{dP}{dt}= 8\times\frac{3}{\pi}\\\\\frac{dP}{dt}= \frac{24}{\pi}\ in/sec$

Hence The rate at which Perimeter of the square is increasing is $\frac{24}{\pi} \ in/secs$.