# Homework Help: Cal III S reparametrization problem

1. Mar 21, 2010

### crims0ned

Question:
reparametrize the curve r(t)=<cos(t)-tcos(t), sin(t)-tsin(t)> : t=0

I know I need to integrate the mag from zero to t but it's not working.

got down to the integral of sqrt((t^2)-2t+1) and after a u-sub and a theta sub I got the integral of sec(theta). Then got natural log(sec(theta)+tan(theta)) but if i back sub in I get ln(0) first of all and then a bunch of other nasty problems just i try to solve for t in terms of s.

*********
I actually got sqrt((t^2)-2*t+2)

the equation for the reparametrization should be s=diffident integral( sqrt((dx/dt)^2)+(dy/dt)^2).

I first get sqrt((sin(t)+(t-1)cos(t))^2 + (-sin(t)(1-t)-cos(t))^2)

= sqrt(((t^2)-2t+2)(sin(t)^2+cos(t)^2))........ sin(t)^2+cos(t)^2 becomes 1

now i'm left with strictly the integral of sqrt((t^2)-2t+2) evaluated from zero to t

I completed the square and got sqrt((t-1)^2 + 1) then u-subbed u=t-1

so integral sqrt(u^2 + 1) i then i used a trig sub to get to just the integral of sec^3(theta)

so now i have the integral of sec^3(theta) which equals (1/2)sqrt((t-1)^2 + 1)*(t-1) + (1/2)ln(t + sqrt((t-1)^2 + 1) - 1)

and I have to evaluate this from zero to t

...okay and here's my real problem. This question began and a reparametrization so I have to solve for t in terms of s. Other then this being some algebra I haven't worked in a while, I think I can solve it but is there a trig i.d. i missed in the beginning or something? because I don't think a s-parametrization should be this complicated, but maybe i'm wrong.

Last edited: Mar 21, 2010
2. Mar 21, 2010

### Dick

The integral of sqrt(t^2-2*t+1) shouldn't be that hard if you notice t^2-2*t+1=(t-1)^2.

3. Mar 21, 2010

### gabbagabbahey

$$\sqrt{t^2-2t+1}=\sqrt{(t-1)^2}=|t-1|=\left\{\begin{array}{lr}t-1, & t\geq1 \\ 1-t, & t \leq1\end{array}\right.$$

You shouldn't have much trouble integrating that.

Edit Dick beat me to it

Last edited: Mar 21, 2010
4. Mar 21, 2010

### Staff: Mentor

Gabba2hey, you omitted a 2 in the first radical...
$$\sqrt{t^2-\bold{2}t+1}=\sqrt{(t-1)^2}=|t-1|=\left\{\begin{array}{lr}t-1, & t\geq1 \\ 1-t, & t \leq1\end{array}\right.$$

5. Mar 21, 2010

### gabbagabbahey

Good eye, just a typo though.

6. Mar 21, 2010

### gabbagabbahey

It's better to simply add a new reply rather than edit your original response. If Mark hadn't replied, I never would have noticed your edited post (and I'm not sure Dick would have either), and you may never have gotten further replies.

I think its best if you post the entire original problem, word for word. (In a new reply!) So that we can see exactly what you are being asked to do.

I can't speak for Dick or Mark, but I'm not sure exactly what you mean be s-reparameterization. Is there some specific property that you would like your new paramater $s$ to satisfy, or is this just some random reparameterization?

7. Mar 21, 2010

### crims0ned

we've been using s reparametrizations just as kind of random parametrization i guess. pretty much it seems to be finding the arc-length and evaluating it from some number (t sub not) to an indefinite t. And what ever we come up with we then solve for t is terms of s and substitute it back into the original equation.

8. Mar 21, 2010

### gabbagabbahey

Do you mean that your new parameter $$s(t)[/itex] is supposed to represent the arclength from $\textbf{r}(0)$ to $\textbf{r}(t)$? So, [tex]s=\int_0^t ||\textbf{r}'(\overline{t})||d\overline{t}$$

Last edited: Mar 21, 2010
9. Mar 21, 2010

### crims0ned

the equation is this $$s= \int \sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2} dt$$ evaluated from some starting t to t

10. Mar 21, 2010

### gabbagabbahey

Okay, and your original curve $\textbf{r}(t)$ is ____ ?

11. Mar 21, 2010

### crims0ned

Yes, that's it, just the magnitude of dr/dt. Are you familiar with these types of problems? Because all of the ones I've worked pier to this and everything in my calculus book always out really clean.

12. Mar 21, 2010

### crims0ned

the original problem is $$r(t)=<cos(t)-tcos(t), sin(t)-tsin(t)> ; t=o$$

13. Mar 21, 2010

### crims0ned

there seems to be only an $$i$$ and $$j$$ component

14. Mar 21, 2010

### gabbagabbahey

Okay, so you seem to have correctly gotten to the point where you are trying to evaluate

$$\int_{-1}^{t-1} \sqrt{u^2+1}du$$

Right?

From here, try using integration by parts once...what do you get?

15. Mar 21, 2010

### crims0ned

once i got to $$\int_{-1}^{t-1} \sqrt{u^2+1}du$$ i trig subbed and got $$\int sec^3(\theta) d\theta$$ but can i change the limits along with the variable and still solve for $$s$$?

16. Mar 21, 2010

### gabbagabbahey

Well, if $u=\tan\theta$, then your limits on $\theta$ will be $\tan^{-1}(-1)$ and $\tan^{-1}(t-1)$.

Personally, I think that using integration by parts is easier than trying to integrate $\int\sec^3\theta d\theta$, but it's up to you.