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

1 Answer
Apr 14, 2017

y=.776x-.812

Explanation:

It helps to know some trig derivatives:

tutorial.math.lamar.edu

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