Find the angle of intersection of the curves | vedclass

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(A) Given curves are $y^{2}=x$ and $x^{2}=y$. Solving these equations,we substitute $y=x^{2}$ into $y^{2}=x$,which gives $(x^{2})^{2}=x \Rightarrow x^

Vedclass · Accepted Answer

$\frac{\pi}{2}, 	an^{-1}(\frac{3}{4})$

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(A) Given curves are $y^{2}=x$ and $x^{2}=y$. Solving these equations,we substitute $y=x^{2}$ into $y^{2}=x$,which gives $(x^{2})^{2}=x \Rightarrow x^{4}-x=0$. $x(x^{3}-1)=0$,so $x=0$ or $x=1$. For $x=0$,$y=0$ and for $x=1$,$y=1$. The points of intersection are $(0,0)$ and $(1,1)$. For $y^{2}=x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx}=1$,so $\frac{dy}{dx}=\frac{1}{2y}$. For $x^{2}=y$,differentiating with respect to $x$ gives $\frac{dy}{dx}=2x$. At $(0,0)$,the slope of $y^{2}=x$ is undefined (vertical) and the slope of $x^{2}=y$ is $0$ (horizontal). Thus,the angle of intersection is $\frac{\pi}{2}$. At $(1,1)$,the slope $m_{1}$ of $y^{2}=x$ is $\frac{1}{2(1)}=\frac{1}{2}$ and the slope $m_{2}$ of $x^{2}=y$ is $2(1)=2$. The angle $	heta$ between them is given by $	an 	heta = |\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}| = |\frac{2-1/2}{1+(2)(1/2)}| = |\frac{3/2}{2}| = \frac{3}{4}$. Thus,$	heta = 	an^{-1}(\frac{3}{4})$. The angles of intersection are $\frac{\pi}{2}$ and $	an^{-1}(\frac{3}{4})$.

Find the angle of intersection of the curves $y^{2}=x$ and $x^{2}=y$.

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