The radius of a circle increases at a rate of 0.5\(cms^{-1}\). Find the rate of...
The radius of a circle increases at a rate of 0.5\(cms^{-1}\). Find the rate of change in the area of the circle with radius 7cm. \([\pi = \frac{22}{7}]\)
- A. 11\(cm^{2}s^{-1}\)
- B. 22\(cm^{2}s^{-1}\)
- C. 33\(cm^{2}s^{-1}\)
- D. 44\(cm^{2}s^{-1}\)
Correct Answer: B. 22\(cm^{2}s^{-1}\)
Explanation
With radius = 7cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7^{2}\)
= \(154cm^{2}\)
The next second, radius = 7.5cm, \(Area = \pi r^{2} = \frac{22}{7} \times 7.5^{2}\)
= \(176cm^{2}\)
Change in area = \((176 - 154)cm^{2} = 22cm^{2}\)
\(\therefore\) The rate of increase = \(22cm^{2}s^{-1}\)
OR
\(Area (A) = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
Given \(\frac{\mathrm d r}{\mathrm d t} = 0.5\)
\(\frac{\mathrm d A}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d t}\)
\(\frac{\mathrm d A}{\mathrm d t} = 2\pi r \times 0.5 = 2 \times \frac{22}{7} \times 7 \times 0.5\)
= \(22cm^{2}s^{-1}\)