Cal III S reparametrization problem

[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.

 

[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.

 

[gabbagabbahey]

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.

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

You shouldn't have much trouble integrating that.

Edit Dick beat me to it

 

[Mark44]

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

You shouldn't have much trouble integrating that.

Edit Dick beat me to it

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

 

[gabbagabbahey]

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

Good eye, just a typo though.

 

[gabbagabbahey]

*********
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.

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?

 

[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.

 

[gabbagabbahey]

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.

Do you mean that your new parameter [tex]s(t)[/itex] is supposed to represent the arclength from $\textbf{r}(0)$ to $\textbf{r}(t)$?

So, $$s=\int_0^t ||\textbf{r}'(\overline{t})||d\overline{t}$$

 

[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

 

[gabbagabbahey]

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

 

[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.

 

[crims0ned]

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

 

[crims0ned]

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

 

[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?

 

[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 $$s$$?

 

[gabbagabbahey]

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 $$s$$?

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.