Question

Find the polar equation that has the same graph as the equation in x and y:
(x^2 + y^2)tan^-1 (y/x) = ay
where a > 0 (cochleoid, or Oui-ja board curve)

HELP! I don't even know what this means...

Answer 1

Cochleoid? Hmm I guess they're telling us this traces out some interesting shape that has a specific name. :o

We'll do the same thing as before :)
Using these two identities to go from cartesian to polar:\[\large x=r \cos \theta \qquad\qquad\qquad y=r \sin \theta\]

\[\Large \left(r^2\cos^2\theta+r^2\sin^2\theta\right)\arctan\left(\frac{r \sin \theta}{r \cos \theta}\right)=a r \sin \theta\]

There are a few nice simplifications.
Have an idea of how we might simplify it from here?

Answer 2

We could pull out an r^2 and cancel out the r's in the arctan to get \[r^2(\cos^2 \theta + \sin^2 \theta)\arctan\frac{ \sin \theta }{ \cos \theta } = arsin \theta\] Then, cos^2 + sin^2 = 1 and sin/cos = tan
So, we get \[r^2 = ar \sin \theta\] and divide both sides by r to get \[ r = a \sin \theta\]

Answer 3

Woops, one small boo boo,\[\Large \arctan\left(\tan \theta\right) \qquad\ne\qquad 1\]

Answer 4

It equals theta doesn't it?

Answer 5

yes good :)

Answer 6

THANK YOU!