The angle of intersection of curves y = x^2 a | vedclass

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(C) Given curves are $y = x^2$ and $6y = 7 - x^3$.For the curve $y = x^2$,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.At point $(1

Vedclass · Accepted Answer

$\pi /2$

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(C) Given curves are $y = x^2$ and $6y = 7 - x^3$.For the curve $y = x^2$,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.At point $(1, 1)$,the slope $m_1 = 2(1) = 2$.For the curve $6y = 7 - x^3$,differentiating with respect to $x$,we get $6 \frac{dy}{dx} = -3x^2$,which simplifies to $\frac{dy}{dx} = -\frac{x^2}{2}$.At point $(1, 1)$,the slope $m_2 = -\frac{1^2}{2} = -\frac{1}{2}$.Since $m_1 	imes m_2 = 2 	imes (-\frac{1}{2}) = -1$,the product of the slopes is $-1$.Therefore,the curves intersect at a right angle,which is $\frac{\pi}{2}$.

The angle of intersection of curves $y = x^2$ and $6y = 7 - x^3$ at the point $(1, 1)$ is:

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