Polar to Parametric Equation? 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(${\theta}$).

 A: 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 $\frac{dy/d\theta}{dx/d\theta}$
A: 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 $ds=\sqrt{(dr)^2+r^2(d\theta)^2}$
A: Alternatively, you could recognize that, or any polar equation, $x = r \cos\theta $ and $y = r \sin \theta$. You also would need to know that $r^2=x^2+y^2$. This is because the radius is always equal to the distance from the origin to the x, y coordinate.
If you now tried to convert $r = \sin \theta + \cos \theta$, you could just multiply each side by $r$ getting you 
$$r^2 = r \sin \theta + r \cos \theta$$
which converts immediately to 
$$x^2 + y^2 = x + y$$
