What is the normal of f(x)=3sin(2x),\at x= pi/8 ?

The normal of f(x)=3sin(2x),\at x= pi/8 is f(x)=-1/(3sqrt(2))x+3/(sqrt(2))+pi/(24sqrt(2))

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normal of f(x)=3sin(2x),\at x= pi/8

normal of

f (x) = 3 sin (2 x), at x = \frac{π}{8}

f (x) = - \frac{1}{3 2} x + \frac{3}{2} + \frac{π}{24 2}

Solution steps

Find the tangent point:

(\frac{π}{8}, \frac{3}{2})

Find the slope of

f (x) = 3 sin (2 x) : \frac{df ( x )}{d x} = 6 cos (2 x)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 32

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{3 2}

Find the line with slope m=

- \frac{1}{3 2}

and passing through

(\frac{π}{8}, \frac{3}{2}) : f (x) = - \frac{1}{3 2} x + \frac{3}{2} + \frac{π}{24 2}

f (x) = - \frac{1}{3 2} x + \frac{3}{2} + \frac{π}{24 2}

What is the normal of f(x)=3sin(2x),\at x= pi/8 ?
The normal of f(x)=3sin(2x),\at x= pi/8 is f(x)=-1/(3sqrt(2))x+3/(sqrt(2))+pi/(24sqrt(2))