What is the equation of the normal line of #f(x)= cotx# at #x = pi/8#?

1 Answer
Timber Lin
Jan 10, 2018

#y-(sqrt(2)+1)=(2-sqrt(2))/4(x-pi/8)#

Explanation:

To find the normal line, first find the slope of the tangent line of #f(x)=cotx# at #x=pi/8#.

This can be found by finding the derivative of #f(x)=cotx#, which is #f'(x)=-(cscx)^2#

proof

#f'(pi/8)# is the slope of the tangent line of #f(x)=cotx# at #x=pi/8#.
#f'(pi/8)=-(csc(pi/8))^2#
#f'(pi/8)=-(2/sqrt(2-sqrt(2)))^2# csc(pi/8)
#f'(pi/8)=-4/(2-sqrt(2))#

the slope of the normal like is the opposite reciprocal of that, which would be:
#-1/(-4/(2-sqrt(2)))#

#=1/(4/(2-sqrt(2)))#

#=(2-sqrt(2))/4#

the point the normal line passes through is #(pi/8,f(pi/8))#
#f(pi/8)=cot(pi/8)#
#=sqrt(2)+1# (cot(pi/8)

use point-slope form:
#y-(sqrt(2)+1)=(2-sqrt(2))/4(x-pi/8)#

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