So, yesterday…yes, Christmas…I proved (or perhaps, better said “derived”) the Quadratic Formula. The fun didn’t stop there. 🙂
My oldest son (visiting for the holiday), seemed to enjoy what I was doing…so I figured I’d blow his mind with how polar equations act in graphs, first showing him a simple \(r = 2\sin \theta\) one. After a bit discussing that, I thought I’d blow his mind more and create the equivalent of \(x = 1\), which is \(r = \frac{1}{{\cos \theta }}\).
Having a brain cramp, I thought I’d shift it one by adding 1 to it (\(r = \frac{1}{{\cos \theta }} + 1\)), but that results in something quite weird. You see, in a \(g(x) = f(x) + 1\) world, \(g(x)\) is just \(f(x)\) shifted up 1. Not so in the polar equation world. 🙂
The green line is \(r = \frac{1}{{\cos \theta }}\). The red lines are what happened when I added one to it (\(r = \frac{1}{{\cos \theta }} + 1\)):
But, why would we want to stop having fun there? Could we convert the polar equation to a Cartesian one? What would…
\[r = \frac{1}{{\cos \theta }} + 1\]
…be with \(x\) and \(y\)?
To help our transformation, let’s remember three things:
\[\begin{gathered}
{r^2} = {x^2} + {y^2} \hfill \\
x = r\cos \theta \hfill \\
y = r\sin \theta \hfill \\
\end{gathered} \]
Ready?!
Start with our original:
\[r = \frac{1}{{\cos \theta }} + 1\]
Get rid of the \(cos()\) in the denominator by multiplying both sides by \(\cos \theta\):
\[(r = \frac{1}{{\cos \theta }} + 1) \times \cos \theta \]
Which, multiplied out, is:
\[r\cos \theta = 1 + \cos \theta \]
Remember this?:
\[x = r\cos \theta \]
That means we can immediately transform what we have to:
\[x = 1 + \cos \theta \]
That’s great! However, how do we get rid of that pesky \(\cos \theta\) on the right?
Well, you may recall that:
\[\cos = \frac{{adjacent}}{{hypotenuse}}\]
What value is adjacent in an xy world? \(x\)
What value is hypotenuse? \(r\)
But \(r\) doesn’t do us much good…since it is a polar value. However, do you remember this?:
\[{r^2} = {x^2} + {y^2}\]
Therefore:
\[r = \sqrt {{x^2} + {y^2}} \]
(\(r\) is always positive.)
So…and it’s really about to get messy…replacing \(\cos \theta\) with adjacent divided by hypotenuse means:
\[x = 1 + \cos \theta \]
Becomes:
\[x = 1 + \frac{x}{{\sqrt {{x^2} + {y^2}} }}\]
Which is the same as:
\[x – 1 = \frac{x}{{\sqrt {{x^2} + {y^2}} }}\]
We want to end up with \(y\) alone on the left. To do that, we divide both sides by \(x-1\) and multiply both sides by \({\sqrt {{x^2} + {y^2}} }\), giving us:
\[\sqrt {{x^2} + {y^2}} = \frac{x}{{x – 1}}\]
Now we square both sides to get rid of the square root on the left:
\[{x^2} + {y^2} = {(\frac{x}{{x – 1}})^2}\]
Then, we move \(x^2\) to the right side of the equation by subtracting it from both sides:
\[{y^2} = {(\frac{x}{{x – 1}})^2} – x^2\]
And take the square root of each side so that it is \(y\) versus \(y^2\):
\[y = \pm \sqrt {{{(\frac{x}{{x – 1}})}^2} – {x^{^2}}} \]
Voila!
That was easy, eh? 🙂
But, is it right?
Let’s check it’s graph and see how it compares. The green line is \(x=1\) and the red is our new Cartesian equation:
Okay, to save you scrolling, here is our original beside it:
Did we do it?!
Well, it’s time to watch the rest of Die Hard 2. That’s a Christmas movie too, right? 🙂
P.S. You can see the polar graph here on Desmos, and the Cartesian one here.
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