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Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 14

Answers (1)

Answer:

\frac{d r}{d t}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec}

Hint:

The volume of sphere, the radius r is defined by

\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad \quad .....(i) \end{aligned}

Given:

\begin{aligned} V=288 \pi \mathrm{cm}^{3} \end{aligned}

Solution:

\begin{aligned} &288 \pi=\frac{4}{3} \pi r^{3}\\ &\text { Solving for } r^{r}, \text { we get } r=6 \mathrm{~cm} \end{aligned}

\begin{aligned} &\rightarrow \text { Given that } \frac{d V}{d t}=4 \pi \mathrm{cm}^{3} / \mathrm{sec}\\ &\rightarrow \text { Differentiating (i) with respect to t, we get } \end{aligned}

\begin{aligned} &\frac{d V}{d t}=\frac{4}{3} \pi r^{2} \times \frac{d r}{d t}\\ &\rightarrow \text { Substituting values, we get }\\ &4 \pi=4 \pi \times 6^{2} \times \frac{d r}{d t}\\ &\frac{d r}{d t}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec} \end{aligned}

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Gurleen Kaur

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