A circular ink blot on a piece of paper increases its area at the rate \(4mm^{2}/s\). Find the rate of the radius of the blot when the radius is 8mm. \([\pi = \frac{22}{7}]\).
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Correct Answer: Option B
Explanation:
Given: \(\frac{\mathrm d A}{\mathrm d t} = 4 mm^{2}/s\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
\(A = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
\(\implies 4 = 2\pi r \times \frac{\mathrm d r}{\mathrm d t}\)
\(\frac{\mathrm d r}{\mathrm d t} = \frac{4}{2\pi r} = \frac{4 \times 7}{2 \times 22 \times 8}\)
= \(0.07954 mm/s \approxeq 0.08 mm/s\)
Given: \(\frac{\mathrm d A}{\mathrm d t} = 4 mm^{2}/s\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
\(A = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
\(\implies 4 = 2\pi r \times \frac{\mathrm d r}{\mathrm d t}\)
\(\frac{\mathrm d r}{\mathrm d t} = \frac{4}{2\pi r} = \frac{4 \times 7}{2 \times 22 \times 8}\)
= \(0.07954 mm/s \approxeq 0.08 mm/s\)