15. expanding circle the area of a circle increases at a rate of 1 cm²/s.\na. how fast is the radius…

Explanation:

Step1: Recall the formula for the area of a circle

The area of a circle is (A=\pi r^{2}). Differentiate both sides with respect to time (t) using the chain - rule. (\frac{dA}{dt}=2\pi r\frac{dr}{dt}).

Step2: Solve for (\frac{dr}{dt}) in part (a)

We know that (\frac{dA}{dt} = 1\space cm^{2}/s) and (r = 2\space cm). From (\frac{dA}{dt}=2\pi r\frac{dr}{dt}), we can solve for (\frac{dr}{dt}). [ \begin{align*} \frac{dr}{dt}&=\frac{\frac{dA}{dt}}{2\pi r}\ \end{align*} ] Substitute (\frac{dA}{dt}=1) and (r = 2) into the formula: (\frac{dr}{dt}=\frac{1}{2\pi\times2}=\frac{1}{4\pi}\space cm/s).

Step3: Recall the formula for the circumference of a circle

The circumference of a circle is (C = 2\pi r), so (r=\frac{C}{2\pi}). Differentiate the area formula (A=\pi r^{2}) with respect to (t): (\frac{dA}{dt}=2\pi r\frac{dr}{dt}). Since (C = 2\pi r), when (C = 2\space cm), then (r=\frac{1}{\pi}\space cm).

Step4: Solve for (\frac{dr}{dt}) in part (b)

Using (\frac{dr}{dt}=\frac{\frac{dA}{dt}}{2\pi r}), and substituting (\frac{dA}{dt}=1) and (r=\frac{1}{\pi}) [ \begin{align*} \frac{dr}{dt}&=\frac{1}{2\pi\times\frac{1}{\pi}}\ &=\frac{1}{2}\space cm/s \end{align*} ]

Answer:

a. (\frac{1}{4\pi}\space cm/s) b. (\frac{1}{2}\space cm/s)