What is the slope of the line normal to the tangent line of #f(x) = cscx-sec^2x +1 # at # x= (17pi)/12 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = cscx-sec^2x +1 # at # x= (17pi)/12  #?

William Abdalla · Accepted Answer

#y=-1(6*sqrt(6))*x+1/432*(432+864*sqrt(2)+17*sqrt(6)*pi)# We get #f'(x)=-cot(x)*csc(x)-sec(x)*tan(x)# so #f'(17*pi/12)=6*sqrt(6)# and the slope of the normal line is given by #m_N=-1/(6*sqrt(6))# and #f(17/12*pi)=1+2sqrt(2)# From the equation #1+2sqrt(2)=-1/(6*sqrt(6))*17/12*pi+n# we get #n=1/432*(432+864*sqrt(2)+17*sqrt(6)*pi)#

Isaiah Allen · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = \csc(x) - \sec^2(x) + 1 $ at $ x = \frac{17\pi}{12} $, we first find the derivative of $ f(x) $ using the quotient rule and the chain rule. Then, we find the slope of the tangent line by evaluating the derivative at $ x = \frac{17\pi}{12} $. Finally, we find the slope of the normal line by taking the negative reciprocal of the slope of the tangent line.

The derivative of $ f(x) $ is:

$$ f'(x) = -\csc(x) \cot(x) + 2\sec(x) \tan(x) $$

Evaluating $ f'(x) $ at $ x = \frac{17\pi}{12} $:

$$ f'\left(\frac{17\pi}{12}\right) = -\csc\left(\frac{17\pi}{12}\right) \cot\left(\frac{17\pi}{12}\right) + 2\sec\left(\frac{17\pi}{12}\right) \tan\left(\frac{17\pi}{12}\right) $$

Then, find the slope of the normal line:

$$ \text{Slope of normal line} = -\frac{1}{{f'\left(\frac{17\pi}{12}\right)}} $$