Due to a work conference, I've fallen a bit behind in my Calculus course and am performing practice problems out of the book to try and catch up on the missed material. One such questions is as follows:
Find all points on the curve $y=cot (x)$, $0<x<pi$, where the equation of the tangent line is parallel to the line $y=-x$. Put final answer in point-slope form.
Can anyone elaborate on this? I can visually see on a graphing calculator the lines formed by $cot(x)$ and $(-x)$. Being between $0<x<pi$ (pi being 3.14...) I assume then I'm looking at the $cot(x)$ line falling just right of the origin?
We have $y'(x)=-\csc^2(x)$, and since this is the slope of the tangent line at the point $(x,y(x))$ and parallel lines have the same slopes, we have to solve the equation
$$-\csc^2(x) = -1$$
That's equivalent to $\sin^2(x) = 1$, so we need to find all values of $x$ that give $\sin(x) = 1$ or $\sin(x) = -1$. These have the form $x=\frac{\pi}{2}+k\pi$, where $k\in\mathbb{Z}$. For all these points, $\cos(x) = 0$, so $y=\cot(x)=0$.
Finally, the equations of the tangent lines in point-slope form are given by
$$y-0=-\Bigg(x-\Bigg(\frac{\pi}{2}+k\pi\Bigg)\Bigg), k\in\mathbb{Z}$$