# Polar coordinates.

• Apr 28th 2009, 10:54 AM
Showcase_22
Polar coordinates.
Quote:

Let C be the curve in the plane defined by $r=\sin \theta, 0 \leq \theta \leq 2 \pi$.
Quote:

a). Draw a sketch of C.
My sketch looks like a figure of 8.

Quote:

b). Find the co-ordinates of the four points at which the tangent to C is vertical.

Hint: The formula $\frac{ dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d \theta}}$ may be helpful.
I know that:

$x=r \cos \theta \Rightarrow \ x=\sin^2 \theta \cos \theta$

$y=r \sin \theta \Rightarrow \ y=\sin^3 \theta$

Therefore:

$\frac{dx}{d \theta}=2 \sin \theta \ cos^2 \theta -\sin^3 \theta$

$\frac{dy}{d \theta}=3 \sin^2 \theta$

Hence:

$\frac{dy}{dx}=\frac{ 3 \sin^2 \theta}{ 2 \sin \theta \cos^2 \theta- \sin^3 \theta}=\frac{3 \sin \theta}{2 \cos^2 \theta-\sin^2 \theta}$

I want when $\frac{dy}{dx}= \infty$. This will be when the denominator of my expression is equal to 0.

$2 \cos^2 \theta=1- \cos^2 \theta$

$\cos^2 \theta+2 \cos^2 \theta-1=0$

$3 \cos ^2 \theta=1 \Rightarrow \ \boxed{\cos \theta= \pm \frac{1}{\sqrt{3}}}$

Since I have an expression for $\cos \theta$, it is useful to write x and y in terms of $\cos \theta$.

$x=r \cos \theta \Rightarrow \ x=\sin^2 \theta \cos \theta \Rightarrow \ x=(1-\cos^2 \theta) \cos \theta$

$y=r \sin \theta \Rightarrow \ y=\sin^3 \theta=(1-\cos^2 \theta)\sqrt{1-\cos^2 \theta}$

Therefore when $\cos \theta=\frac{1}{\sqrt{3}}$ we have:

$x=\left( 1- \frac{1}{3} \right) \frac{1}{\sqrt{3}}=\left( \frac{2}{3} \right) \frac{1}{\sqrt{3}}=\frac{2}{3 \sqrt{3}}$

$y=\left( 1-\frac{1}{3} \right) \sqrt{1-\frac{1}{3}}=\frac{2}{3}\sqrt{\frac{2}{3}}=\pm \left( \frac{2}{3} \right)^{\frac{3}{2}}$

The first two points are $\left( \frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$ and $\left( \frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$.

It's also apparent that the next x coordinate is $-\frac{2}{3 \sqrt{3}}$ with the same y coordinates as before.

The other two points are $\left( -\frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$ and $\left( -\frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$.

The four points are (listed together for clarity) $\left( -\frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$, $\left( -\frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$, $\left( \frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$ and $\left( \frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right)$.

I'm pretty sure this is right. I would just like someone to confirm my answer.
• Apr 28th 2009, 11:30 AM
running-gag
Quote:

Originally Posted by Showcase_22
$y=r \sin \theta \Rightarrow \ y=\sin^3 \theta$

Therefore:

$\frac{dy}{d \theta}=3 \sin^2 \theta$

Unfortunately not (Shake)

Lucky you are : it does not change anything and your results are correct ! (Nod)
• Apr 28th 2009, 11:31 AM
Showcase_22
I mean $\frac{dy}{d \theta}=3\sin^2 \theta cos \theta!$
• Apr 28th 2009, 11:40 AM
running-gag
Your results are correct anyway !
• Apr 28th 2009, 11:41 AM
Showcase_22
yay!

MATH WIN!

Thankyou kindly running gag. (Cool)