# Need help with Curves and parameterizations

I'm having some trouble solving a couple of problems:

• I know this one must be pretty easy but can't find the way to solve it. I need to find the arc length of a curve described by $r=1- \theta ; 1\leq \theta \leq 2.$

From my notes, this should be solved with $$\int_{C} f(\sigma (t)) \left \|\sigma '(t)\right \| dt$$ I would use $\sigma (t)= (1-t, t)$ since that describes $C$, but what $F$ am I supposed to use?

• Again, I feel like this should be really easy, but can't figure it out:
$$C = {(x,y,z,): y = 1 - x^2 ; x + y + z = 1 ; x,y \geq 0}$$

a) I need to find a regular parametrization of $C$ that starts in $(0,1,0)$ and ends in $(1,0,0).$

b) I need to find $\int_{C} F. ds$ with $F = (2x, y, -x).$

My problem here is finding the parametrization. I think I could use $$1 \leq z \leq x + y - 1 , 0 \leq x \leq a(x,y,z) , 0 \leq y \leq b(x,y,z).$$

I can't really decide what those $a$ and $b$ should be. Once that is done, I think solving b) should be pretty easy.

Anyway, I'll be grateful for any pointers on how to solve this and similar problems, since I can't seem to grasp the concepts behind most of this.

EDIT 1: Regarding problem 1, I think I should do $$\int_{1}^{2} \left \|\frac{\partial (1-\theta ; \theta )}{\partial \theta } \right \| d\theta = \sqrt{2}$$ and that should be the length. I had actually tried that but didn't seem right, but after looking around a little bit more and finding nothing, I think this is it.

EDIT 2: EDIT 1 was wrong. I had to use $$\int_{a}^{b} \sqrt{(r(\theta))^{2} + (\frac{\partial{r(\theta)}}{\partial{\theta}})^{2}}d\theta$$ with $r(\theta)=1-\theta$ which gives me $$\int_{1}^{2} \sqrt{2-2\theta+(\theta)^{2}} d\theta$$ which is pretty hard to solve. I'm guessing I still have something wrong.

Regarding problem 2: I wrote $$\sigma(t) = (t, 1-t^{2}, t^{2}-t) \in C^1$$ and also $\sigma'(t) \neq (0,0,0)$, noting that $\sigma(0)=(0,1,0)$ and $\sigma(1)=(1,0,0).$
So point a) is done. Now I needed to solve $$\int_{C} F. ds \Rightarrow \int_{0}^{1} F(\sigma(t))\sigma'(t)dt$$ I got $$\int_{0}^{1} (2t,2-2t^2, -t)(1,-2t, 2t-1)dt = \frac{-1}{6}$$ This one I think is right, but I'm not sure what the result means. Does F goes through $\sigma$ the other way around?

EDIT 3: The integral from the second excercise should be $$\int_{0}^{1} (2t,1-t^2, -t)(1,-t,2t-1) dt = \frac{1}{3}$$ which is a lot nicer. Thanks J.M.
Still not sure about the first one though.

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A tip on the first one: personally, I don't bother to remember the special arclength formula for polar coordinates; what I do remember is the general formula $$s=\int\sqrt{f^\prime(t)^2+g^\prime(t)^2}\mathrm dt$$ for a parametrically defined curve $x=f(t),\qquad y=g(t)$ and then use the usual polar-to-Cartesian conversion to derive a parametric representation from a polar representation... – J. M. May 12 '11 at 0:39
For the second one: what does "regular" mean? – J. M. May 12 '11 at 0:40
@J.M. Regular mens that $\sigma \in C^1$ and $\sigma'(t) \neq (0,0,0)$. It must have a different name in english, but I don't know it. I made some advances, but took some time to write it, I'm still very new to latex. What's the usual polar to Cartesian conversion? – Bananas May 12 '11 at 1:02
Yes, so your parametrization ought to be $x=(1-\theta)\cos\,\theta,\qquad y=(1-\theta)\sin\,\theta$ then, no? Then you can use the general arclength formula... – J. M. May 12 '11 at 1:26
@hhh: If $r=\rho(\theta)$ is the polar equation for some curve, then $(\rho(\theta)\cos\,\theta\quad \rho(\theta)\sin\, \theta)^T$ is a parametric representation, 'no? – J. M. Jun 2 '11 at 3:58

Let $x=(\Theta-1)$,

$dx= d\Theta$,

$r=(1-\Theta)$,

$\frac{dr}{d\Theta}=-1$,

and $s= \int\sqrt{(-1)^2 +(1-\Theta)^2}d\Theta$

We know that $$\int\sqrt{x^2+a^2}dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln(x+\sqrt{x^2+a^2})+C$$.

With substitution of $x$ for $(\Theta-1)$ and taking the limits between 1 and 2, we get the answer $\frac{\sqrt{2}}{2}0.5\ln(1 +\sqrt{2})$ which comes to 1.1477936.

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You want to find the arc-length of the curve given in polar form by the equation $r = 1- \theta$ for $1 \le \theta \le 2.$ Well, if you have a polar equation, say $r = f(\theta)$, and you want to find the arc-length of the resulting curve for $\theta_1 \le \theta \le \theta_2$ then you need to use the following formula:

$$s = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^{\!\! 2}} \, d\theta \, .$$

In the case where $r = 1-\theta$ and $1 \le \theta \le 2$, we have:

$$s = \int_1^2 \sqrt{(1-\theta)^2+(-1)^2} \, d\theta \, ,$$

$$s = \int_1^2 \sqrt{\theta^2-2\theta+2} \ d\theta \, ,$$

$$s = \int_1^2 \sqrt{(\theta-1)^2+1} \ d\theta \, ,$$

$$s = \int_0^1 \sqrt{\psi^2+1} \ d\psi \, ,$$

$$s = \frac{1}{2}\left[\psi\sqrt{\psi^2+1} + \sinh^{-1} \psi \right]_0^1 \, ,$$

$$s = \frac{1}{2}\left(\sqrt{2} - \ln\left(\sqrt{2}-1\right) \right) .$$

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Convention is to take $r$ positive. So if you take $r < θ -1$ in the interval $1≤θ≤2$ radius stays positive. Arc length computation is pretty straightforward, as others also point out.

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

$r$ can be negative also. Sketch the Lemniscate of Bernoulli for example. $\sin \theta, \cos\theta$ when correspondingly negative yields a proper graph.

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