Question

the area of a circle increases at a rate of 1 cm²/s.
a. how fast is the radius changing when the radius is 5 cm?
b. how fast is the radius changing when the circumference is 3 cm?

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 \(r = 5\space cm\). From \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\), we can solve for \(\frac{dr}{dt}\).

$$ \frac{dr}{dt}=\frac{\frac{dA}{dt}}{2\pi r} $$

Substitute \(\frac{dA}{dt}=1\) and \(r = 5\) into the formula: \(\frac{dr}{dt}=\frac{1}{2\pi\times5}=\frac{1}{10\pi}\space cm/s\)

Step3: Recall the formula for the circumference of a circle

The circumference of a circle is \(C = 2\pi r\), so \(r=\frac{C}{2\pi}\). The area formula \(A=\pi r^{2}=\pi(\frac{C}{2\pi})^{2}=\frac{C^{2}}{4\pi}\)
Differentiate \(A\) with respect to \(t\): \(\frac{dA}{dt}=\frac{2C}{4\pi}\frac{dC}{dt}\). But we also know from the original area - radius relation \(\frac{dA}{dt}=2\pi r\frac{dr}{dt}\), and since \(C = 2\pi r\) (so \(r=\frac{C}{2\pi}\)), \(\frac{dA}{dt}=C\frac{dr}{dt}\)

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

We know that \(\frac{dA}{dt}=1\space cm^{2}/s\) and \(C = 3\space cm\). From \(\frac{dA}{dt}=C\frac{dr}{dt}\), we solve for \(\frac{dr}{dt}\)

$$ \frac{dr}{dt}=\frac{\frac{dA}{dt}}{C} $$

Substitute \(\frac{dA}{dt}=1\) and \(C = 3\) into the formula: \(\frac{dr}{dt}=\frac{1}{3}\space cm/s\)

Answer:

a. \(\frac{1}{10\pi}\space cm/s\)
b. \(\frac{1}{3}\space cm/s\)