Question

What is the slope of the line normal to the tangent line of `f(x) = sec^3x+sin(2x-(3pi)/8)` at `x= (11pi)/8`?

Answer 1

Slope of the normal is 1/0.8104=1.2339

Explanation:

`f(x)=sec^3x+sin(2x-(3pi)/8)`

`x=(11pi)/8`

`f'(x)=3sec^2xsecxtanx+cos(2x-(3pi)/8)xx2`

`f'(x)=3sec^3xsecx+2cos(2x-(3pi)/8)`

`f'((11pi)/8)=3sec^3((11pi)/8)tan((11pi)/8)+2cos(2((11pi)/8)-(3pi)/8)`

`=3sec^3(pi/8)(-tan(pi/8)+2cos((19pi)/8))`

`=-3sec^3(pi/8)tan(pi/8)+2cos((3pi)/8))`

`=-0.8104`
Slope of the tangent is -0.8104

Slope of the normal is 1/0.8104=1.2339