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need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 15

Answers (1)

Answer: \left ( 1,1 \right )  and \left ( -1,-1 \right )

Hint: Here we use the equation of curve y=x^{3}

Given: As particle moves along the curve  y=x^{3}

Solution: Differentiate the above equation with respect to t ,

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

 

When y-coordinate change three times more rapidly than x-coordinate that is

\frac{d y}{d t}=3 \frac{d x}{d t} …(ii)

 

Then equating (i) and (ii)

\begin{aligned} &3 x^{2} \frac{d x}{d t}=3 \frac{d x}{d t} \\\\ &x^{2}=1 \Rightarrow x=\pm 1 \end{aligned}

 

 When x=1

y=x^{3}=(1)^{3}=y=1

 

When x=-1

y=x^{3}=(-1)^{3}=y=-1

 

\left ( 1,1 \right ) and \left ( -1,-1 \right ) .

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