What is the equation of the line that is normal to #f(x)=sin^2x+e^xcotx # at # x=pi/3#?

Answer 1

#y=.776x-.812#

It helps to know some trig derivatives:

First, take the derivative of #f(x)#. I suggest rewriting #sin^2x# as #sinxsinx# so you can use the product rule. You will have to use the product rule on #e^xcotx# too.

#f'(x)=sinxcosx+sinxcosx +e^xcot(x)-csc^2(x)e^x#

Simplify:

#f'(x)=2sinxcosx+e^xcot(x)-csc^2(x)e^x#

Plug in #x=pi/3#:

#f'(pi/3)=2sin(pi/3)cos(pi/3)+e^(pi/3)cot(pi/3)-csc^2(pi/3)e^(pi/3)=-1.288#

That's the slope of the tangent line but we need the slope perpendicular to that which is simply the negative reciprocal.

#-1/-1.288=-.776#

We also need the point at #x=pi/3# so we plug into the initial equation:

#f(pi/3)=sin^2(pi/3)+e^(pi/3)(cot(pi/3))=2.395#

Now we can use point-slope form to find the equation of the normal line:

#y-(pi/3)=.776(x-2.395)#

#y=.776x-1.859+pi/3#

#y=.776x-.812#

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Answer 2

The equation of the line that is normal to f(x) = sin^2x + e^xcotx at x = pi/3 is y = -sqrt(3) + sqrt(3)(x - pi/3).

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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.

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