Question

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`

Explanation:

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`