What is the slope of the line normal to the tangent line of #f(x) = secx-cos3x # at # x= (pi)/12 #?

1 Answer
Mar 6, 2018

Slope of normal: #-((sqrt2+sqrt6)/(11-sqrt3))#

Explanation:

#f(x) = secx - cos3x#

since we know that the derivative of #secx = tanxsecx#
#f'(x) = tanxsecx - (3)(-sin3x)#
#f'(x) = tanxsecx +3sin3x#

if #x = pi/12#, then
#f(pi/12) = sec(pi/12) - cos3(pi/12)#
#f(pi/12) = 4/(sqrt2 + sqrt6) - sqrt2/2#
therefore, #f(pi/12)= (6+sqrt3)/(sqrt2+sqrt6)#

therefore, the point at which the normal passes is #(pi/12, (6+sqrt3)/(sqrt2+sqrt6))#

#f'(pi/12) = tan(pi/12)sec(pi/12) + 3sin3(pi/12)#
#f'(pi/12) = (2-sqrt3)(4/(sqrt2+sqrt6)) + (3sqrt2)/2#
^ this simplifies to:
#f'(pi/12) = (11-sqrt3)/(sqrt2+sqrt6)#

therefore, the slope of the tangent is #(11-sqrt3)/(sqrt2+sqrt6)#
since the slope of the normal is the negative reciprocal to the slope of a line, the slope of the normal will be:
Slope of normal: #-((sqrt2+sqrt6)/(11-sqrt3))#

Side note:

  • I simplified a few things and skipped those steps purposely so that this question doesn't run forever

  • I would say the hardest part about this question would be the simplification such as figuring out what #tan(pi/12)# is or what #sec(pi/12)# is