Question

The acute angle between the curves $y^2=x\mathrm{and}{x}^2=y$ at (1, 1) is .

• A. ${\tan }^{-1}\frac{4}{3}$
• B. ${\tan }^{-1}\frac{3}{4}$
• C. $90°$
• D. $45°$

Answer 1

The correct option is B ${\tan }^{-1}\frac{3}{4}$

The given curves are $y^2=x\mathrm{and}{x}^2=y$.
Let $\theta$ be the acute angle between the
given curves.
Let $m_1\mathrm{and}{m}_2$be the slope of the given curves $y^2=x\mathrm{and}{x}^2=y$ at (1, 1)
Differentiating both sides w.r.t. x,
we get
$2y\frac{\mathrm{dy}}{\mathrm{dx}}=1\mathrm{and}2x=\frac{\mathrm{dy}}{\mathrm{dx}}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2y}\mathrm{and}\frac{\mathrm{dy}}{\mathrm{dx}}=2x\Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(1,1)}=\frac{1}{2}\mathrm{and}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(1,1)}=2\Rightarrow m_1=\frac{1}{2}\mathrm{and}{m}_2=2\tan \theta =\left(\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left(\frac{\frac{1}{2}-2}{1+\frac{1}{2}\times 2}\right|=\left(\frac{-\frac{3}{2}}{2}\right|=\frac{3}{4}\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{4}\right)\therefore \mathrm{The}\mathrm{acute}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{curves}\mathrm{is}{\tan }^{-1}\left(\frac{3}{4}\right).$