Question

What is the equation of the normal line of `f(x)=x/sinx` at `x=pi/6`?

Answer 1

`y-pi/3=sqrt3/(pi-2sqrt3)(x-pi/6)`

Explanation:

First find the point that the normal line will pass through on the curve.

`f(pi/6)=(pi/6)/sin(pi/6)=(pi/6)/(1/2)=pi/3`

So the tangent line will pass through the point `(pi/6,pi/3)`.

To find the slope of the normal line, we will first have to find the slope of the tangent line. The slope of the tangent line at a point is given through the function's derivative.

To differentiate `x/sinx`, we will use the quotient rule. This gives

`f'(x)=((d/dxx)sinx-x(d/dxsinx))/sin^2x`

Since `d/dxx=1` and `d/dxsinx=cosx`, we see that

`f'(x)=(sinx-xcosx)/sin^2x`

So the slope of the tangent line at `x=pi/6` is

`f'(pi/6)=(sin(pi/6)-pi/6cos(pi/6))/sin^2(pi/6)=(1/2-pi/6(sqrt3/2))/(1/2)^2`

`color(white)(f'(pi/6))=(1/2-(pisqrt3)/12)/(1/4)=4(1/2-(pisqrt3)/12)=2-(pisqrt3)/3`

`color(white)(f'(pi/6))=2-pi/sqrt3=(2sqrt3-pi)/sqrt3`

The slope of the normal line is the opposite reciprocal of the slope of the tangent line, since the two lines are perpendicular. Thus the slope of the normal line is

`-1/((2sqrt3-pi)/sqrt3)=sqrt3/(pi-2sqrt3)`

So the normal line has slope `sqrt3/(pi-2sqrt3)` and passes through `(pi/6,pi/3)`. Using point-slope form the equation of the line is

`y-y_1=m(x-x_1)`

`=>y-pi/3=sqrt3/(pi-2sqrt3)(x-pi/6)`

Graphed for reference:

graph{(y-x/sinx)(y-pi/3-sqrt3/(pi-2sqrt3)(x-pi/6))=0 [-5.69, 6.8, -1.57, 4.673]}