If the radius of the circle changes at the rate of \(\rm-\frac{2}{\pi}\ m/sec\), at what rate does the circle's area change when the radius is 10 m?

Concept:

• The area of a circle with r units, is: πr² sq. units.
• The rate of change of the value of a function f(x) with respect to a variable t, is given by: \(\rm \frac{d}{dt}f(x)\).

• Chain Rule of Derivatives: \(\rm \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\).

• \(\rm \frac{d}{dx}x^n=nx^{n-1}\).

Calculation:

Let's say that the radius of the circle is r meters and let t be the time in seconds.

It is given that \(\rm \frac{dr}{dt}=-\frac{2}{\pi}\ m/sec\).

Now, Area of the circle: A = πr2 m².

Using the chain rule of derivatives:

\(\rm \frac{dA}{dt}=\frac{dA}{dr}\times \frac{dr}{dt}\)

⇒ \(\rm \frac{dA}{dt}=2\pi r\times\frac{-2}{\pi}\ m^2/sec\)

⇒ \(\rm \frac{dA}{dt}=-4r\ m^2/sec\)

When the radius is 10 m, the rate of change of area will be: -40 m²/sec.

NOTE: A negative value of the rate of change indicates a decrease.