1. Polar coordinates.

Let C be the curve in the plane defined by $\displaystyle r=\sin \theta, 0 \leq \theta \leq 2 \pi$.
a). Draw a sketch of C.
My sketch looks like a figure of 8.

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

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

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

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

Therefore:

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

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

Hence:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I'm pretty sure this is right. I would just like someone to confirm my answer.

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

Therefore:

$\displaystyle \frac{dy}{d \theta}=3 \sin^2 \theta$
Unfortunately not

Lucky you are : it does not change anything and your results are correct !

3. I mean $\displaystyle \frac{dy}{d \theta}=3\sin^2 \theta cos \theta!$

4. Your results are correct anyway !

5. yay!

MATH WIN!

Thankyou kindly running gag.