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 #?

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Answer 1 · 2018-07-10T12:49:28

Slope of the normal is 1/0.8104=1.2339

#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