# 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(${\theta}$).

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That sounds like a good plan. So you get something like $x = r\cos\theta = (\sin\theta + \cos\theta)\cos\theta$, right? – Dylan Moreland Dec 14 '11 at 23:30
@DylanMoreland Correct, but I don't know what to do from there. – Kyle V. Dec 15 '11 at 0:26

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 $ds=\sqrt{(dr)^2+r^2(d\theta)^2}$

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So regarding part (a); I multiplied by r and now I have X=cos^2(θ)+sinθcosθ and Y=sin^2(θ)+sinθcosθ. So would this be Cartesian form? How do I convert it to parametric? – Kyle V. Dec 15 '11 at 0:01
@StickFigs It's already parametric! The parameter is $\theta$. – Dylan Moreland Dec 15 '11 at 0:45
@DylanMoreland Excellent, Thanks! – Kyle V. Dec 15 '11 at 1:12
@StickFigs: When you multiply by $r$ you get $x^2+y^2=x+y$. It looks like you multiplied by $\cos \theta$ and $\sin \theta$ which is more effective for your purpose. – Ross Millikan Dec 15 '11 at 3:41

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 $\frac{dy/d\theta}{dx/d\theta}$

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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

$$x^2 + y^2 = x + y$$

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