Question

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

Answer 1

curve y=sin x at(0,0)is

Answer 2

Given:-

y=sin x at(0,0)

To Find :-

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

Solution:-

$\rm \: Equation \: of \: line \: at \: (x_1 y_1) and \: having \: slope \: m \: is \\ \boxed{ \rm \: y - y_1 = m(x -x_1 )} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

First of all we have to find the slope of Normal

Slope of normal

$= \rm - \dfrac{1}{ \dfrac{dy}{dx} }$

Given

y = sinx

Differentiating both sides woth respects to x. we get

$\implies\rm \dfrac{dy}{dx} = \cos x \: \: ....(1)$

The given point is ( 0,0)

so, put it in equation ( 1)

$\implies\rm \dfrac{dy}{dx} \bigg]_{x = 0} = \cos 0 \degree = 1$

Slope of normal

$= \rm - \dfrac{1}{ \dfrac{dy}{dx} }$

$\rm = - \dfrac{1}{1} = - 1$

Slope ( m) = -1

so, equation of the normal at (0,0) and slope-1 is

$\rm\boxed{ \rm \: y - y_1 = m(x -x_1 )}$

$\implies \rm \: y - 0 = - 1(x - 0)$

$\implies \rm \: y = - x$

$\underline{\boxed{\implies \: \rm \: x + y = 0}}$

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

x + y = 0

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