Question

a circle is growing, its radius increasing by 2 mm per second. find the rate at which the area is changing at the moment when the radius is 16 mm.
when the radius is 16 mm, the area is changing at approximately
(round to the nearest thousandth as needed.)

Explanation:

Step1: Recall the formula for the area of a circle

The area formula of a circle is \(A = \pi r^{2}\).

Step2: Differentiate the area formula with respect to time \(t\)

Using the chain - rule \(\frac{dA}{dt}=\frac{dA}{dr}\cdot\frac{dr}{dt}\).
Differentiate \(A = \pi r^{2}\) with respect to \(r\): \(\frac{dA}{dr}=2\pi r\).
We are given that \(\frac{dr}{dt} = 2\) mm/s.

Step3: Substitute the values of \(r\) and \(\frac{dr}{dt}\) into the formula for \(\frac{dA}{dt}\)

When \(r = 16\) mm and \(\frac{dr}{dt}=2\) mm/s, substitute into \(\frac{dA}{dt}=2\pi r\cdot\frac{dr}{dt}\).
\(\frac{dA}{dt}=2\pi\times16\times2\).
\(\frac{dA}{dt}=64\pi\).
Calculate \(64\pi\approx64\times3.14159 = 201.062\)

Answer:

\(201.062\)