What is the slope of the line normal to the tangent line of #f(x) = secx+cos^2(2x-pi) # at # x= (11pi)/8 #?

Answer 1

#-(2-sqrt(2))/(2[2-sqrt(2)-sqrt(2+sqrt(2))])=0.232#, to three significant figures

We get the slope (#m_t#) of the tangent line from the value of the first derivative #(df)/dx# at the given #x# value. The slope of the line normal to that we get from noting that the slopes of two perpendicular lines multiply to -1 - assuming that neither line is vertical.
Firstly note that we may immediately simplify the given function - #cosx# has period #2pi#, so #cos2x# has period #pi#, so #cos(2x-pi)=cos2x# and #f(x)=secx+cos^2 2x#.

Find the function derivative

Use the quotient rule for differentiation for the first term of #f# and the chain rule twice for the second term:
#(df)/dx=secxtanx+2cos2x*(-sin2x)*2# #(df)/dx=secxtanx-4sin2xcos2x#
Recall the identity #sin2x=2sinxcosx#
#(df)/dx=secxtanx-2sin4x#

Evalute the derivative

Now evaluate #(df)/dx# at the given #x#:
#(df)/dx((11pi)/8)=sec((11pi)/8)tan((11pi)/8)-2sin((11pi)/2)#
Recall that #cos(x+pi)=-cosx# (so #sec(x+pi)=-secx#). Then #sec((11pi)/8)=-sec((3pi)/8)# Similarly, recall that #tan(x+pi)=tanx#, so #tan((11pi)/8)=tan((3pi)/8)#. If these aren't easy to recall, think of the graphs of the functions.
As #sin(x+pi)=-sinx#, #sin((11pi)/2)=-sin(pi/2)=-1#
So #(df)/dx((11pi)/8)=-sec((3pi)/8)tan((3pi)/8)+2#
Now the trig values of the angle #(3pi)/8# are not standard ones, but they can be calculated.
#sin((3pi)/8)=sqrt(2+sqrt(2))/2#, as derived here: https://tutor.hix.ai
#cos((3pi)/8)=sqrt(2-sqrt(2))/2#, as derived here: https://tutor.hix.ai
Thus #sec((3pi)/8)=1/cos((3pi)/8)=2/sqrt(2-sqrt(2))# #tan((3pi)/8)=sin((3pi)/8)/cos((3pi)/8)=sqrt((2+sqrt(2))/(2-sqrt(2)))#
So #(df)/dx((11pi)/8)=-sec((3pi)/8)tan((3pi)/8)+2=-2/sqrt(2-sqrt(2))sqrt((2+sqrt(2))/(2-sqrt(2)))+2#
#(df)/dx((11pi)/8)=2-(2sqrt(2+sqrt(2)))/(2-sqrt(2))=2/(2-sqrt(2))(2-sqrt(2)-sqrt(2+sqrt(2)))#

This (complicated) value is the slope of the tangent line at the given point.

Calculate the normal slope

We now calculate the normal slope #m_n#via #m_nm_t=-1#.
#m_t=2-(2sqrt(2+sqrt(2)))/(2-sqrt(2))=2/(2-sqrt(2))(2-sqrt(2)-sqrt(2+sqrt(2)))#

so

#m_n=-1/m_t=-(2-sqrt(2))/(2[2-sqrt(2)-sqrt(2+sqrt(2))])#
Numerically, we calculate to three significant figures that #m_t=-4.31# and #m_n=0.232#.
Sanity check these answers by plotting the graphs of the function and these lines for the given #x# value graph{(y-secx-(cos(2x))^2)(y+4.308644x-16.49888)(y-0.232x+3.115689)=0 [2, 6, -5, 0]}
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the slope of the line normal to the tangent line of (f(x) = \sec(x) + \cos^2(2x - \pi)) at (x = \frac{11\pi}{8}), first find the derivative of the function, then evaluate it at the given (x)-value to find the slope of the tangent line. Then, take the negative reciprocal of this slope to find the slope of the line normal to the tangent line.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7