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need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 27

Answers (1)


Answer: \left ( 2,4 \right )

Hint: Here we use the concept of equation curve  y^{2^{}}=8x

Given: A point on the curve  y^{2^{}}=8x .

Solution: Differentiating the above equation,

\begin{aligned} &\frac{d\left(y^{2}\right)}{d t}=\frac{d(8 x)}{d t} \\\\ &2 y \frac{d y}{d t}=8 \frac{d x}{d t} \\\\ &\frac{y}{4} \frac{d y}{d t}=\frac{d y}{d t} \end{aligned}..........(i)

\frac{d y}{d t}=\frac{d x}{d t}    ..........(ii)

Let use equation (i) and (ii)

\begin{aligned} &\frac{y}{4} \frac{d y}{d t}=\frac{d y}{d t} \\\\ &y=4 \end{aligned}

When y^{2}=8 x \Rightarrow(4)^{2}=8 x


Hence the point on the curve at which the abscissa and coordinate change at the same rate is \left ( 2,4 \right ) .

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