A circle with radius 5cm has its radius increasing at the rate of 0.2m/s. What will be the corresponding increase in the area?
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Correct Answer: Option A
Explanation:
Area of the circle (A) = \(\pi r^{2}\)
\(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
\(r = 5cm\); \(\frac{\mathrm d r}{\mathrm d t} = 0.2\)
\(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} . \frac{\mathrm d r}{\mathrm d t}\)
\(\frac{\mathrm d A}{\mathrm d t} = 2\pi \times 5 \times 0.2 = 2\pi\)
Area of the circle (A) = \(\pi r^{2}\)
\(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
\(r = 5cm\); \(\frac{\mathrm d r}{\mathrm d t} = 0.2\)
\(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} . \frac{\mathrm d r}{\mathrm d t}\)
\(\frac{\mathrm d A}{\mathrm d t} = 2\pi \times 5 \times 0.2 = 2\pi\)