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

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   $\left ( -1,\frac{1}{3} \right )$