Graphing a Cotangent

[Jason76]

Any mistakes so far?

\(\displaystyle y = 2\cot(x)\)

Given formula: \(\displaystyle y = A\cot(Bx - C) + D\)

1. Identify the Letters:

\(\displaystyle A = 2, B = 1, C = 0, D = 0\)

2. Find the x intercept

Set \(\displaystyle y = 0\), solve for \(\displaystyle x\) by eventually using an inverted \(\displaystyle \arctan\) because \(\displaystyle \cot = \dfrac{1}{\tan}\)

\(\displaystyle 0 = 2\cot(x)\)

\(\displaystyle 0 = 2\dfrac{1}{\tan(x)}\)

\(\displaystyle 0 = \dfrac{1}{\tan(x)}\)

\(\displaystyle \arctan(0) = \arctan(\dfrac{1}{\tan(x)}\)

\(\displaystyle 0 = \dfrac{1}{x}\)

3. Locate Vertical Asymptotes

a)

\(\displaystyle 0 = 2\cot(x)\)

\(\displaystyle 0 = 2\dfrac{1}{\tan(x)}\)

\(\displaystyle 0 = \dfrac{1}{\tan(x)}\)

\(\displaystyle \arctan(0) = \arctan(\dfrac{1}{\tan(x)})\)

\(\displaystyle 0 = \dfrac{1}{x}\)

b)

\(\displaystyle \pi = 2\cot x\)

\(\displaystyle \pi = 2\dfrac{1}{\tan(x)}\)

\(\displaystyle \dfrac{\pi}{2} = \dfrac{1}{\tan(x)}\)

\(\displaystyle \arctan(\dfrac{\pi}{2}) = \arctan(\dfrac{1}{\tan(x)})\)

\(\displaystyle 88.7 = \dfrac{1}{x}\)

 

[MarkFL]

The roots will occur where:

\(\displaystyle \cos(x)=0\)

\(\displaystyle x=\dfrac{\pi}{2}+k\pi=\dfrac{\pi}{2}(2k+1)\) where \(\displaystyle k\in\mathbb{Z}\)

The vertical asymptotes will occur where:

\(\displaystyle \sin(x)=0\)

\(\displaystyle x=k\pi\)

 

[Jason76]

As far as using \(\displaystyle \arctan\), you can use it to find the extra points, but not the \(\displaystyle x\) intercept (because it leads to an undefined answer). In order to use \(\displaystyle \arctan\) you have to convert the \(\displaystyle \cot\) to \(\displaystyle \tan\) knowing that \(\displaystyle \cot = \dfrac{1}{\tan}\). Found this out (and figured it out on my own) from the math adjunct professor.

On the other hand, you can get extra points on the tangent function and the \(\displaystyle x\) intercept using \(\displaystyle \arctan\) (but you can also just divide parts into two)

As far as the \(\displaystyle \cot\) graph goes, again, If you need the \(\displaystyle x\) intercept, then just divide the area between the vertical asymptotes. Vertical asymptotes are found by setting the original equation to \(\displaystyle \pi\) and \(\displaystyle 0\) respectively (in seperate equations).

As far as asymptotes go, you can find one (the \(\displaystyle \pi\) one) via \(\displaystyle \arctan\), but not the \(\displaystyle 0\) one. So probably to find both asymptotes, a different way is needed. Any ideas?