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Need solution for RD Sharma maths class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 7

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\frac{d r}{d t}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec}


The relation between height h, radius r and semi vertical angle a is defined by

tan\: \alpha =\frac{r}{h}


\alpha=30^{\circ} \text { and } \frac{d \alpha}{d t}=2^{0} / \mathrm{sec}


Let, r be the radius, h be the height & α be the semi-vertical angle of cone

tan\: \alpha =\frac{r}{h} \quad \quad \quad \quad......(i)

\Rightarrow We\; have\; to\; find\; \frac{d r}{d t} \\ \Rightarrow Di\! f\! \! ferentiating \; (i) \; with \; respect \; to \; \mathrm{t}, we \;get \\ \sec ^{2} \alpha \frac{d \alpha}{d t}=\frac{1}{20} \frac{d r}{d t}

\begin{aligned} &\Rightarrow \text { Substituting values, We get }\\ &\begin{gathered} \sec ^{2} 30^{\circ} \times 2^{0}=\frac{1}{20} \frac{d r}{d t} \quad\quad\quad\quad\quad\left[\mathrm{~h}=20 \mathrm{~cm}, \alpha=30^{\circ} \text { and } \frac{d \alpha}{d t}=2^{0} / \mathrm{sec}\right] \\ \left(\sec ^{2} 30^{\circ}\right) \times 2^{\circ}=\frac{1}{20} \frac{d r}{d t} \\ \frac{4}{3} \times 2=\frac{1}{20} \frac{d r}{d t} \\ \frac{d r}{d t}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec} \end{gathered} \end{aligned}

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Gurleen Kaur

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