What is the angle between the curves y = x^2 | vedclass

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(C) For the curve $y = x^2$,the slope $m_1$ is given by $\frac{dy}{dx} = 2x$. At the point $(1, 1)$,$m_1 = 2(1) = 2$. For the curve $6y = 7 - x^3$,dif

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$\pi /2$

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

What is the angle between the curves $y = x^2$ and $6y = 7 - x^3$ at the point $(1, 1)$?

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