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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) 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 $x = y^2$,differenti

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{3}{4}ight)$

Vedclass · Answer

(D) 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 $x = y^2$,differentiating with respect to $x$ gives $1 = 2y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{2y}$.At the point $(1, 1)$,$m_2 = \frac{1}{2(1)} = \frac{1}{2}$.The angle $	heta$ between the two curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{2 - 1/2}{1 + 2(1/2)} ight| = \left| \frac{3/2}{2} ight| = \frac{3}{4}$.Therefore,$	heta = 	an^{-1}\left(\frac{3}{4}ight)$.

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

Similar Questions

At any point for the curve $3y^2 = (x+5)^3$,if $ST$ represents the length of the subtangent and $SN$ represents the length of the subnormal,then $9(ST)^2 = $

The length of the subtangent to the curve $x^{2} y^{2}=a^{4}$ at $(-a, a)$ is

What is the angle of inclination of the tangent to the curve $y = (x - 1)(x - 2)$ at the point $(1, 0)$ with the $x$-axis?

If $\theta$ denotes the acute angle between the curves $y=10-x^2$ and $y=2+x^2$,at a point of the intersection,then $|\tan \theta|$ is equal to

Find the points on the curve $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ at which the tangents are parallel to the $x$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module