The equation of the normal to the curve y = s | vedclass

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

Question

(C) Given the curve $y = \sin x$. First,find the derivative $\frac{dy}{dx} = \cos x$. The slope of the tangent at $(0, 0)$ is $\left(\frac{dy}{dx}\rig

Vedclass · Accepted Answer

$x + y = 0$

Vedclass · Answer

(C) Given the curve $y = \sin x$. First,find the derivative $\frac{dy}{dx} = \cos x$. The slope of the tangent at $(0, 0)$ is $\left(\frac{dy}{dx}ight)_{(0,0)} = \cos(0) = 1$. The slope of the normal at $(0, 0)$ is $m = -\frac{1}{	ext{slope of tangent}} = -\frac{1}{1} = -1$. The equation of the normal passing through $(0, 0)$ with slope $m = -1$ is given by $y - y_1 = m(x - x_1)$. Substituting the values: $y - 0 = -1(x - 0)$. This simplifies to $y = -x$,or $x + y = 0$.

The equation of the normal to the curve $y = \sin x$ at the point $(0, 0)$ is

Similar Questions

If the curve $y = ax^{2} + bx + c, x \in R,$ passes through the point $(1, 2)$ and the tangent line to this curve at the origin is $y = x$,then the possible values of $a, b, c$ are:

Find the equation of all lines having slope $2$ and being tangent to the curve $y+\frac{2}{x-3}=0.$

The length of the subtangent to the curve $\sqrt{x} + \sqrt{y} = 3$ at the point $(4, 1)$ is

The equations of straight lines which are both tangent and normal to the curve $27x^2 = 4y^3$ are

For the curve $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$,the lengths of the tangent and the sub-tangent at point $\theta$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module