TS EAMCET · Maths · Application of Derivatives
The height of a cone with semi vertical angle \(\pi / 3\) is increasing at the rate of 2 units \(/ \mathrm{min}\). The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is
- A \(\frac{1}{\sqrt{3}}\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\sqrt{3}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(r = h \tan(\pi/3) = \sqrt{3}h\) \(V = \frac{1}{3} \pi r^2 h\) \(\frac{dV}{dt} = \frac{1}{3} \pi (2r \frac{dr}{dt} h + r^2 \frac{dh}{dt})\) \(0 = 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}\) \(\frac{dr}{dt} = -\frac{r^2}{2rh} \frac{dh}{dt} = -\frac{r}{2h} \frac{dh}{dt}\)…
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