Question

The radius of a circle is decreasing at a rate of 6.56.56, point, 5 meters per minute. At a certain instant, the radius is 121212 meters. What is the rate of change of the area of the circle at that instant (in square meters per minute)

Answer 1

Answer:

The rate of change of the area of the circle is  approximately $-490.09 \ m^2/min$.

Step-by-step explanation:

Given:

$\frac{dr}{dt} = -6.5 \ m/min$

radius $r =12 \ m$

We need to find the rate of change of the area of the circle at that instant.

Solution:

Now we know that;

Area of the circle is given by π times square of the radius.

framing in equation form we get;

$A= \pi r^2$

to find the rate of change of the area of the circle at that instant we need to take the derivative on both side.

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

Substituting the given values we get;

$\frac{dA}{dt}= 2\times \pi \times 12 \times -6.5\\\\\frac{dA}{dt}\approx -490.09 \ m^2/min$

Hence The rate of change of the area of the circle is  approximately $-490.09 \ m^2/min$.