Question

If the rate of change in the circumference of a circle is $0.3$ cm/s, then the rate of change in the area of the circle when the radius is $5$ cm is

• A. $1.5$ sq cm/s
• B. $0.5$ sq cm/s
• C. $5$ sq cm/s
• D. $3$ sq cm/s

Hint:

Chain rule shortcut: $\dfrac{dA}{dt} = \dfrac{dA}{dr}\cdot\dfrac{dr}{dt} = 2\pi r\cdot\dfrac{dr}{dt}$. Find $\dfrac{dr}{dt}$ from the circumference rate first.

Answer 1

The Correct Option is A

Let's solve the problem step-by-step:

1. We know that the circumference \(C\) of a circle is given by the formula: \(C = 2\pi r\)
2. The rate of change of the circumference, \(\frac{dC}{dt}\), is given as \(0.3 \, \text{cm/s}\).
3. Differentiate the formula for circumference with respect to time \(t\) to find the rate of change of the radius: \(\frac{dC}{dt} = 2\pi \frac{dr}{dt}\)
4. Substitute the given rate of change of circumference: \(0.3 = 2\pi \frac{dr}{dt}\)
5. Solve for \(\frac{dr}{dt}\): \(\frac{dr}{dt} = \frac{0.3}{2\pi}\)
6. Calculate the rate of change of the area. The area \(A\) of a circle is given by: \(A = \pi r^2\)
7. Differentiate the area with respect to time \(t\): \(\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt}\)
8. Substitute \(r = 5 \, \text{cm}\) and \(\frac{dr}{dt} = \frac{0.3}{2\pi}\) into the equation for \(\frac{dA}{dt}\): \(\frac{dA}{dt} = 2\pi \times 5 \times \frac{0.3}{2\pi}\)
9. Simplify: \(\frac{dA}{dt} = 5 \times 0.3 = 1.5 \, \text{sq cm/s}\)

Therefore, the rate of change in the area of the circle when the radius is \(5\) cm is \(1.5 \, \text{sq cm/s}\).