AP EAMCET · Maths · Application of Derivatives
If the surface area of a spherical bubble is increasing at the rate of \(4 \mathrm{sq}. \mathrm{cm} / \mathrm{sec}\), then the rate of change in its volume (in cubic \(\mathrm{cm} / \mathrm{sec}\) ) when its radius is 8 cms is
- A \(8\)
- B \(12\)
- C \(15\)
- D \(16\)
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
\(A = 4\pi r^2\) \(\frac{dA}{dt} = 8\pi r \frac{dr}{dt}\) \(4 = 8\pi (8) \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{16\pi}\) \(V = \frac{4}{3}\pi r^3\) \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\) \(\frac{dV}{dt} = 4\pi (8)^2 \left(\frac{1}{16\pi}\right)\)…
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