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Please solve RD Sharma class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 9 maths textbook solution

Answers (1)


\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right)


Here, we use the basic concept of algebra       



Solution: Let


Solving for y we get

y=\pm \frac{\sqrt{400-16 x^{2}}}{3} \quad\quad \quad \quad .....(i)

Given that

\frac{d x}{d t}=-\frac{d y}{d t}

we have to calculate

→ Differentiating (i) with respect to t,we get

\begin{aligned} \frac{d y}{d t} &=\pm \frac{1}{3} \times \frac{1}{2 \sqrt{400-46 x^{2}}} \times-32 \times \frac{d x}{d t} \\ &=\mp \frac{16 x}{3 \sqrt{400-16 x^{2}}} \frac{d x}{d t} \end{aligned}

→ Substituting values

We get

\begin{aligned} 16 x=\pm 3 \sqrt{400-16 x^{2}}\\ \end{aligned}

Squaring the equation, we get

\begin{aligned} &256 x^{2}=9\left(400-16 x^{2}\right) \end{aligned}

Solving the equation we get

x^{2}=9 \\ x=\pm 3

→ Substituting in (i), we get

\begin{aligned} &y=\pm \frac{\sqrt{400-16 \times 9}}{3}=\pm \frac{16}{3}\\ &\text { So, }\\ &\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right) \end{aligned}

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Gurleen Kaur

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