Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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(${\theta}$).

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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 $\frac{dy/d\theta}{dx/d\theta}$

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.