Question

Angle between $y^2=x$ and $x^2=y$ at the origin is.

Answer 1

Step 1: Calculate the slope of the curve $y^2=x$ at the origin.

The curve given here is $y^2=x$ .

Differentiating it with respect to $x$.

$\begin{array}{c}{y}^2=x\\ 2y\frac{dy}{dx}=1\\ \frac{dy}{dx}=\frac{1}{2y}\end{array}$

Apply $\left(0,0\right)$ in $\frac{dy}{dx}=\frac{1}{2y}$.

Therefore,

$\begin{array}{c}{\left(\frac{dy}{dx}\right)}_{(0,0)}=\frac{1}{2\left(0\right)}\\ {\left(\frac{dy}{dx}\right)}_{(0,0)}=\infty \end{array}$

Step 2: Calculate the slope of the curve $x^2=y$ at the origin.

The curve given here is $x^2=y$ .

Differentiating it with respect to $x$.

$\begin{array}{c}{x}^2=y\\ 2x=\frac{dy}{dx}\\ \Rightarrow \frac{dy}{dx}=2x\end{array}$

Apply $\left(0,0\right)$ in $\frac{dy}{dx}=2x$.

Therefore,

$\begin{array}{c}{\left(\frac{dy}{dx}\right)}_{(0,0)}=2\left(0\right)\\ {\left(\frac{dy}{dx}\right)}_{(0,0)}=0\end{array}$

Step 3: Calculate the angle between the curves.

Know that the slope of the curves $y^2=x\text{\hspace{0.17em}and\hspace{0.17em}}{x}^2=y$ at the origin is $\infty \text{\hspace{0.17em}and\hspace{0.17em}}0$respectively.

Therefore,

$\begin{array}{c}\tan \theta =\frac{0-\infty }{1+\left(0\right)\left(\infty \right)}\\ \tan \theta =\infty \end{array}$

Apply tan inverse on both sides.

$\begin{array}{c}{\tan }^{-1}\left(\tan \theta \right)={\tan }^{-1}\left(\infty \right)\\ \theta ={\tan }^{-1}\left(\tan \frac{\pi }{2}\right)\\ \theta =\frac{\pi }{2}\end{array}$

Hence, the angle between $y^2=x$ and $x^2=y$ at the origin is $\frac{\pi }{2}$.