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 ?

1 Answer
Jul 10, 2018

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