#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 13 maths textbook solution

Answer: $x=-\frac{1}{2}$  and $y=-\frac{3}{4}$

Hint: Here we use the equation of curve are $y=x^{2}+2x$ .

Given: Given as particle moves along the curve $y=x^{2}+2x$ .

Solution: As to find the points at which the curve are the x and y coordinates of the particle changing at the same rate $y=x^{2}+2x$

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

When x and y co-ordinate of the particle are changing at the same rate

\begin{aligned} &\frac{d y}{d t}=\frac{d x}{d t} \\\\ &\frac{d y}{d x}=\frac{d t}{d t} \\\\ &\frac{d y}{d x}=1 \end{aligned}

Substituting value from eq (i)

So, $2x+2=1$

\begin{aligned} &2 x=-1 \\\\ &x=-\frac{1}{2} \end{aligned}

and  $y=x^{2}+2x$

\begin{aligned} &y=\frac{1}{4}-1 \\\\ &y=-\frac{3}{4} \end{aligned}

$x=-\frac{1}{2}$  and  $y=-\frac{3}{4}$ .