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

Answers (1)


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} .

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