What is the slope of the line normal to the tangent line of #f(x) = sec^2x-xcos(x-pi/4) # at # x= (15pi)/8 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = sec^2x-xcos(x-pi/4) # at # x= (15pi)/8  #?

Isabella Ackerman · Accepted Answer

#=> y = 0.063(x - (15pi)/8) - 1.08#

Interactive graph

The first thing we'll need to do is calculate #f'(x)# at #x = (15pi)/8#. Let's do this term by term. For the #sec^2(x)# term, note that we have two functions embedded within one another: #x^2#, and #sec(x)#. So, we'll need to use a chain rule here: #d/dx(sec(x))^2 = 2sec(x) * d/dx(sec(x))# #color(blue)(= 2sec^2(x)tan(x))# For the 2nd term, we'll need to use a product rule. So: #d/dx(xcos(x-pi/4)) = color(red)(d/dx(x))cos(x-pi/4) + color(red)(d/dxcos(x-pi/4))(x)# #color(blue)(= cos(x-pi/4) - xsin(x-pi/4))# You may wonder why we didn't use a chain rule for this part, since we have an #(x - pi/4)# inside the cosine. The answer is we implicitly did, but we ignored it. Notice how the derivative of #(x - pi/4)# is simply 1? Hence, multiplying that on doesn't change anything, so we do not write it out in calculations. Now, we put everything together: #d/dx(sec^2x-xcos(x-pi/4)) = color(violet)(2sec^2(x)tan(x) - cos(x-pi/4) + xsin(x-pi/4))# Watch your signs. Now, we need to find the slope of the line tangent to #f(x)# at #x = (15pi)/8#. To do this, we just plug this value into #f'(x)#: #f'((15pi)/8) = (2sec^2((15pi)/8)tan((15pi)/8) - cos((15pi)/8-pi/4) + (15pi)/8sin((15pi)/8-pi/4)) = color(violet)(~~-6.79)# However, what we want is not the line tangent to f(x), but the line normal to it. To get this, we just take the negative reciprocal of the slope above. #m_(norm) = -1/-15.78 color(violet)(~~0.015) # Now, we just fit everything into point slope form: #y = m(x-x_0) + y_0 #=> y = 0.063(x - (15pi)/8) - 1.08# Take a look at this interactive graph to see what this looks like! Hope that helped :)

Axel Alvarez · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = \sec^2(x) - x\cos(x - \frac{\pi}{4}) $ at $ x = \frac{15\pi}{8} $, we need to first find the slope of the tangent line at that point and then find the negative reciprocal of that slope to get the slope of the normal line.

The slope of the tangent line at $ x = \frac{15\pi}{8} $ can be found by taking the derivative of $ f(x) $ and evaluating it at that point.

$ f'(x) = 2\sec^2(x)\tan(x) + \cos(x - \frac{\pi}{4}) + x\sin(x - \frac{\pi}{4}) $

Evaluate $ f'(\frac{15\pi}{8}) $ to get the slope of the tangent line at $ x = \frac{15\pi}{8} $.

Then, take the negative reciprocal of this slope to find the slope of the normal line.