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

(C) 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 $6y = 7 - x^3$,dif

Vedclass · Accepted Answer

$\pi /2$

Vedclass · Answer

(C) 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 $6y = 7 - x^3$,differentiating with respect to $x$ gives $6 \frac{dy}{dx} = -3x^2$,which simplifies to $\frac{dy}{dx} = -\frac{x^2}{2}$. At the point $(1, 1)$,$m_2 = -\frac{1^2}{2} = -\frac{1}{2}$. Since $m_1 	imes m_2 = 2 	imes (-1/2) = -1$,the product of the slopes is $-1$. Therefore,the curves intersect at a right angle,and the angle between them is $\pi /2$.

What is the angle between the curves $y = x^2$ and $6y = 7 - x^3$ at the point $(1, 1)$?

Similar Questions

Find the equation of the tangent line to the curve $y=x^{2}-2x+7$ which is parallel to the line $2x-y+9=0$.

If all the normals drawn to the curve $y=\frac{1+3x^2}{3+x^2}$ at the points of intersection of $y=\frac{1+3x^2}{3+x^2}$ and $y=1$ pass through the point $(\alpha, \beta)$,then $3\alpha+2\beta=$

The slope of the normal to the curve $y=\frac{x}{x^2+1}$ at $x=-4$ is

If the lengths of the tangent,subtangent,normal and subnormal for the curve $y=x^2+x-1$ at the point $(1,1)$ are $a, b, c$ and $d$ respectively,then their increasing order is

The length of the normal to the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ at $\theta=\frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module