Find the equation of the normal to the curve | vedclass

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

Question

(C) Given the curve $y = x^3$. First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 3x^2$. At the point $P(1, 1)$,the slo

Vedclass · Accepted Answer

$x + 3y - 4 = 0$

Vedclass · Answer

(C) Given the curve $y = x^3$. First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 3x^2$. At the point $P(1, 1)$,the slope of the tangent is $m_t = 3(1)^2 = 3$. The slope of the normal $m_n$ is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{m_t} = -\frac{1}{3}$. Using the point-slope form of the equation of a line,$y - y_1 = m_n(x - x_1)$,we get: $y - 1 = -\frac{1}{3}(x - 1)$. Multiplying both sides by $3$: $3(y - 1) = -(x - 1)$. $3y - 3 = -x + 1$. Rearranging the terms gives the equation of the normal: $x + 3y - 4 = 0$.

Find the equation of the normal to the curve $y = x^3$ at the point $P(1, 1)$.

Similar Questions

Find the equation of the normal to the curve $y=x^3-3x$ which is parallel to the line $2x+18y=9$.

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

The equation of the normal drawn to the curve $y^3=4 x^5$ at the point $(4,16)$ is

The normal to the curve $x=a(1+\cos \theta)$,$y=a \sin \theta$ at $\theta$ always passes through a fixed point. Find the fixed point.

The points on the curve $y=\frac{2}{3} x^3+\frac{1}{2} x^2$,where the tangents make equal angles with coordinate axes are

Vedclass Test Series

Exam Paper Generator

Online Exam Module