The angle between the curves y^2=x and x^2=y | vedclass

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(C) To find the angle between the curves $y^2=x$ and $x^2=y$ at the origin $(0,0)$:$1$. For the curve $y^2=x$, differentiating with respect to $x$ giv

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(C) To find the angle between the curves $y^2=x$ and $x^2=y$ at the origin $(0,0)$:$1$. For the curve $y^2=x$, differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{2y}$. At the origin $(0,0)$, the slope is undefined, which means the tangent is the $y$-axis $(x=0)$.$2$. For the curve $x^2=y$, differentiating with respect to $x$ gives $2x = \frac{dy}{dx}$. At the origin $(0,0)$, the slope is $\frac{dy}{dx} = 0$, which means the tangent is the $x$-axis $(y=0)$.$3$. The angle between the $x$-axis and the $y$-axis is $\frac{\pi}{2}$ radians. Therefore, the angle between the two curves at the origin is $\frac{\pi}{2}$.

The angle between the curves $y^2=x$ and $x^2=y$ at the origin is:

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