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

Question

The equation of the normal to the curve y = sin x at (0,0) is (A) x = 0 (B) x + y = 0 (C) y = 0 (D) x -y = 0

Tardigrade Admin · Accepted Answer

Since (dy/dx) = cos x, therefore, slope of tangent at (0,0) = cos 0 = 1 and hence slope of normal at (0,0) is - 1. ∴ Equation of normal is y-0 = - 1 (x-0) ⇒ x + y = 0

Q. The equation of the normal to the curve $y = s in x$ at $(0, 0)$ is

$x = 0$

$x + y = 0$

$y = 0$

$x - y = 0$

Since $\frac{d y}{d x} = cos x$ ,
therefore, slope of tangent at $(0, 0) = cos 0 = 1$ and hence slope of normal at $(0, 0)$ is $- 1$ .
$∴$ Equation of normal is $y - 0 = - 1 (x - 0)$
$\Rightarrow x + y = 0$

Q. The equation of the normal to the curve y=sinx at (0,0) is

x=0

x+y=0

y=0

x−y=0