Polar to Parametric Equation?

Question

I'm struggling with this problem, I'm still only on part (a). I tried X=rcos(theta) Y=rsin(theta) but I don't think I'm doing it right.

Curve C has polar equation r=sin(${\theta}$)+cos(${\theta}$).

(a) Write parametric equations for the curve C.

$\left\{\begin{matrix} x= \\ y= \end{matrix}\right.$

(b) Find the slope of the tangent line to C at its point where ${\theta}$ = $\frac{\pi}{2}$.

(c) Calculate the length of the arc for 0 $\leq {\theta} \leq {\pi}$ of that same curve C with polar equation r=sin(${\theta}$)+cos(${\theta}$).

Answer 1

Hint: for (a), if you multiply by $r$ the conversion to Cartesian coordinates is not hard. Then you need to convert to parametric form. For (b) if you plug in $\theta=\frac {\pi}2$ you can find the $x,y$ coordinates of the point. Then use the Cartesian equations you got in (a) and take the derivative. For (c) you can use your usual Cartesian arc length, again finding the end points or you can use the arc length in polar coordinates $ds=\sqrt{(dr)^2+r^2(d\theta)^2}$

Answer 2

You can rewrite $x=r \cos \theta$ as $r=\frac{x}{\cos\theta}$ and plug that in. You immediately get $$x=\sin\theta\cos\theta+\cos^2\theta$$ Doing the same trick for $r=\frac{y}{\sin\theta}$ gives you $$y=\sin^2\theta+\sin\theta\cos\theta$$

From here on it's not hard - the slope of the tangent is $\frac{dy/d\theta}{dx/d\theta}$