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provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 26

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Answer: \left ( 1,\frac{5}{3} \right )  and \left ( -1,\frac{1}{3} \right )

Hint: Here we use the equation of curve.

Given: As particle moves along the curve   y=\frac{2}{3} x^{3}+1

Solution: Equation of curve is   y=\frac{2}{3} x^{3}+1

Differentiating the above equation,

\frac{d y}{d t}=\frac{d\left(\frac{2}{3} x^{3}+1\right)}{d t}

      =\frac{d\left(\frac{2}{3} x^{3}+1\right)}{d t}+\frac{d(1)}{d t}=\frac{2}{3} \times 3 x^{2} \frac{d x}{d t} .......…(i)

When y-coordinate is changing twice as fast as x coordinate

\frac{d y}{d t}=2 \frac{d x}{d t} …......(ii)

Equating equation (i) and (ii)

2 x^{2} \frac{d x}{d t}=2 \frac{d x}{d t} 

x=\pm 1

When x=1; y=\frac{2}{3}(1)^{3}+1

 \begin{aligned} &y=\frac{2+3}{3} \\\\ &y=\frac{5}{3} \end{aligned}

 When x=-1; y=\frac{2}{3}(-1)^{3}+1


\begin{aligned} &y=\frac{-2+3}{3} \\\\ &y=\frac{1}{3} \end{aligned} 

\left ( 1,\frac{5}{3} \right )    and   \left ( -1,\frac{1}{3} \right )

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