Question

How do I find the equation of the tangent line to the curve y=2xcos(x) at the point (pi, -2pi)? The equation of this tangent line can be written in the form y=mx+b where m and b are?

I know how to do these when I'm given regular numbers, not trig stuff. You just find the derivative and plug in what they want, but I'm really confused with how to do this on a unit circle...

Answer 1

`m = -2`, `b=0`

Explanation:

Given your comment, I'll skip on the initial details: as you said, we must compute the derivative and plug the correct values. If you get confused with `pi`, remember that it is just a number, although "special", and it works exactly as any other number - plugging `5` or `pi` into an expression makes absolutely no difference. Let's see the computation:

We compute the derivative:

`f(x) = 2xcos(x) \implies f'(x) = 2cos(x)-2xsin(x)`

The derivative returns the slope of the tangent line for any given `x`, and we are interested in `x = \pi`. So, we simply plug `pi` in the derivative expression, i.e. we substitute every instance of `x` with `\pi`. The expression becomes

`f'(pi) = 2cos(pi)-2pisin(pi)`

We know that `cos(\pi) = -1` and `sin(pi)=0`. So, the expression becomes

`f'(pi) = 2*(-1)-2pi*0 = -2-0=-2`

So, the slope of the tangent line is `-2`.

At this point, we want the equation of a line passing through a given point with a given slope. There's a formula for cases like this:

`y-y_0 = m(x-x_0)`

In our case, `(x_0,y_0) = (\pi,-2\pi)` and `m = -2`. Again, we do nothing but substitute every variable with the corresponding value: the expression becomes

`y-(-2\pi) = (-2)(x-\pi)`

which we can rewrite into

`y+cancel(2\pi) = -2x+cancel(2\pi)`

And thus we have the answer `y = -2x`. This equation is written in the `y = mx+b` form, where `m=-2` and `b=0`.