A particle moves along the curve $y = x^{2} + 2 x$ then the point on the curve such that $x$ and $y$ coordinates of the particle change with the same rate is

(1, 3)

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(1 / 2, 3 / 4)

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(- 1 / 2, - 3 / 4)

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(- 1, - 1)

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Solution

The correct option is B $(- 1 / 2, - 3 / 4)$
$y^{2} = 2 x + 2$
Differentiating both sides w.r.t, to $t$
$y = x^{2} + 2 x \frac{d y}{d t} = 2 x \frac{d x}{d t} + 2 \frac{d x}{d t} \frac{d y}{d t} = \frac{d x}{d t} \Rightarrow 1 = 2 x + 2 \Rightarrow x = - \frac{1}{2} y = - \frac{3}{4}$

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A particle moves along the curve y=x2+2x then the point on the curve such that x and y coordinates of the particle change with the same rate is

The correct option is B (−1/2,−3/4)y2=2x+2Differentiating both sides w.r.t, to ty=x2+2xdydt=2xdxdt+2dxdtdydt=dxdt⇒1=2x+2⇒x=−12y=−34

A particle moves along the curve $y = x^{2} + 2 x$ then the point on the curve such that $x$ and $y$ coordinates of the particle change with the same rate is

The correct option is B $(- 1 / 2, - 3 / 4)$
$y^{2} = 2 x + 2$
Differentiating both sides w.r.t, to $t$
$y = x^{2} + 2 x \frac{d y}{d t} = 2 x \frac{d x}{d t} + 2 \frac{d x}{d t} \frac{d y}{d t} = \frac{d x}{d t} \Rightarrow 1 = 2 x + 2 \Rightarrow x = - \frac{1}{2} y = - \frac{3}{4}$