MHT CET · Maths · Application of Derivatives
The volume of a ball is increasing at the rate of \(4 \pi \mathrm{cc} / \mathrm{sec}\). The rate of increase of the radius, when the volume is \(288 \pi \mathrm{cc}\), is
- A \(\frac{1}{6} \mathrm{~cm} / \mathrm{sec}\)
- B \(\frac{1}{36} \mathrm{~cm} / \mathrm{sec}\)
- C \(6 \mathrm{~cm} / \mathrm{sec}\)
- D \(36 \mathrm{~cm} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{36} \mathrm{~cm} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & V=\frac{4}{3} \pi r^3 \\ & \Rightarrow 288 \pi=\frac{4}{3} \pi r^3 \\ & \Rightarrow r=6 \mathrm{~cm}\end{aligned}\)
\(\begin{aligned} & \quad \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \therefore \quad \frac{\mathrm{dV}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \cdot \frac{\mathrm{dr}}{\mathrm{dt}} \\ & \Rightarrow 4 \pi=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & \therefore \quad \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{\mathrm{r}^2}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec}\end{aligned}\)
\(\begin{aligned} & \quad \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \therefore \quad \frac{\mathrm{dV}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \cdot \frac{\mathrm{dr}}{\mathrm{dt}} \\ & \Rightarrow 4 \pi=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & \therefore \quad \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{\mathrm{r}^2}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec}\end{aligned}\)
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