Question

the slope of the line normal to the curve f(x） 2cos4x at x= pi/12 is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\:f(x) = 2 \: cos4x$

Let we assume that,

$\rm :\longmapsto\:y = 2 \: cos4x$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} 2 \: cos4x$

We know,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \:\dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx} \: f(x) }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =2 \: \dfrac{d}{dx} \: cos4x$

We know,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \:\dfrac{d}{dx}cosx = - \: sinx }}}$

Using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: 2 \: sin \: 4x \: \dfrac{d}{dx} \:4x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: 8 \: sin \: 4x \: \dfrac{d}{dx} \:x$

We know,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \:\dfrac{d}{dx}x = \: 1 }}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: 8 \: sin \: 4x \: \times 1$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: 8 \: sin \: 4x \:$

Therefore,

Slope of tangent is given by

$\rm :\longmapsto\:\dfrac{dy}{dx}_{ \: \: x \: = \: \dfrac{\pi}{12} } = - \: 8 \: sin \: {\bigg(\dfrac{4\pi}{12} \bigg) }\:$

$\rm :\longmapsto\:\dfrac{dy}{dx}_{ \: \: x \: = \: \dfrac{\pi}{12} } = - \: 8 \: sin \: {\bigg(\dfrac{\pi}{3} \bigg) }\:$

$\rm :\longmapsto\:\dfrac{dy}{dx}_{ \: \: x \: = \: \dfrac{\pi}{12} } = - \: 8 \: \times \: {\bigg(\dfrac{ \sqrt{3} }{2} \bigg) }\:$

$\rm :\longmapsto\:\dfrac{dy}{dx}_{ \: \: x \: = \: \dfrac{\pi}{12} } = - 4 \sqrt{3}$

Therefore,

$\rm :\longmapsto\:Slope \: of \: normal = \dfrac{ - 1}{Slope \: of \: tangent}$

$\rm :\longmapsto\:Slope \: of \: normal = \dfrac{ - 1}{\dfrac{dy}{dx}_{ \: \: x \: = \: \dfrac{\pi}{12} }}$

$\rm :\longmapsto\:Slope \: of \: normal = \dfrac{ - 1}{ - 4 \sqrt{3} }$

$\bf :\longmapsto\:Slope \: of \: normal = \dfrac{ 1}{ 4 \sqrt{3} }$

Additional Information :-

Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

$\red{ \boxed{ \sf{ \: Slope \: of \: tangent \: = \: \bigg(\dfrac{dy}{dx} \bigg)_P}}}$

and

$\red{ \boxed{ \sf{ \: Slope \: of \: normal\: = \: \frac{ - 1}{\bigg(\dfrac{dy}{dx} \bigg)_P} }}}$

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined

4. Two lines having slope M and m are parallel, iff M = m

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.