What is the equation of the normal line of f(x)= secxtanxf(x)=secxtanx at x = pi/8x=π8?

1 Answer
Apr 16, 2017

y-.488=1.454(x-pi/8)y.488=1.454(xπ8)

Explanation:

It helps to know some trig derivatives:

tutorial.math.lamar.edu

First, take the derivative of f(x)f(x). You will need to use product rule :

f'(x)=secx(sec^2(x)) + tanx(secxtanx)

Simplify:

f'(x)=sec^3x+secxtan^2x

Plug in pi/8

f'(pi/8)=sec^3(pi/8)+sec(pi/8)tan^2(pi/8)~~1.454

This is the slope of the tangent line but we need the slope of the normal line. Take the negative reciprocal of 1.454 to get -.688.

Now we need a point. Plug in pi/8 into the initial equation:

f(pi/8)=sec(pi/8)tan(pi/8)~~.448

Now we have the point (pi/8,.448) and the slope 1.454, plug into point-slope form:

y-.488=1.454(x-pi/8)