What is the slope of the line normal to the tangent line of #f(x) = sec^2x-sin^2x # at # x= (pi)/12 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = sec^2x-sin^2x # at # x= (pi)/12  #?

Charles Anctil · Accepted Answer

#y=(-74/11-128sqrt(3)/33)x+15/2-15/4sqrt(3)+37/66*pi+32/99*pi*sqrt(3)# #f'(x)=2sec^2(x)tan(x)-2sin(x)cos(x)# #f'(pi/12)=111/2-32sqrt(3)# so the slope of the normal line is given by #-1/(111/2-32sqrt(3))=-74/11-128sqrt(3)/33# #f(pi/12)=15/8*(sqrt(3)-1)^2#

Elijah Abram · Answer

undefined To find the slope of the line normal to the tangent line of $f(x) = \sec^2x - \sin^2x$ at $x = \frac{\pi}{12}$, we first need to find the derivative of $f(x)$ and evaluate it at $x = \frac{\pi}{12}$ to get the slope of the tangent line. Then, we can find the negative reciprocal of this slope to get the slope of the line normal to the tangent line. The derivative of $f(x)$ is $f'(x) = 2\sec^2x\tan x + 2\sin x\cos x$, and evaluating it at $x = \frac{\pi}{12}$ yields $f'\left(\frac{\pi}{12}\right) = 2\sec^2\left(\frac{\pi}{12}\right)\tan\left(\frac{\pi}{12}\right) + 2\sin\left(\frac{\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)$. After calculating this, we take the negative reciprocal of the result to find the slope of the line normal to the tangent line.