• Support PF! Buy your school textbooks, materials and every day products Here!

Cal III S reparametrization problem

  • Thread starter crims0ned
  • Start date
  • #1
17
0
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:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
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
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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.
[tex]\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.[/tex]

You shouldn't have much trouble integrating that.:wink:

Edit Dick beat me to it:cry:
 
Last edited:
  • #4
33,485
5,174
[tex]\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.[/tex]

You shouldn't have much trouble integrating that.:wink:

Edit Dick beat me to it:cry:
Gabba2hey, you omitted a 2 in the first radical...
[tex]\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.[/tex]
 
  • #5
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Gabba2hey, you omitted a 2 in the first radical...
[tex]\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.[/tex]
Good eye, just a typo though.
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
6
*********
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 [itex]s[/itex] to satisfy, or is this just some random reparameterization?
 
  • #7
17
0
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
gabbagabbahey
Homework Helper
Gold Member
5,002
6
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 [itex]\textbf{r}(0)[/itex] to [itex]\textbf{r}(t)[/itex]?

So, [tex]s=\int_0^t ||\textbf{r}'(\overline{t})||d\overline{t}[/tex]
 
Last edited:
  • #9
17
0
the equation is this [tex]s= \int \sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2} dt [/tex] evaluated from some starting t to t
 
  • #10
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Okay, and your original curve [itex]\textbf{r}(t)[/itex] is ____ ?
 
  • #11
17
0
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
17
0
the original problem is [tex] r(t)=<cos(t)-tcos(t), sin(t)-tsin(t)> ; t=o [/tex]
 
  • #13
17
0
there seems to be only an [tex] i [/tex] and [tex] j [/tex] component
 
  • #14
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Okay, so you seem to have correctly gotten to the point where you are trying to evaluate

[tex]\int_{-1}^{t-1} \sqrt{u^2+1}du[/tex]

Right?

From here, try using integration by parts once...what do you get?
 
  • #15
17
0
once i got to [tex]\int_{-1}^{t-1} \sqrt{u^2+1}du[/tex] i trig subbed and got [tex]\int sec^3(\theta) d\theta[/tex] but can i change the limits along with the variable and still solve for [tex] s [/tex]?
 
  • #16
gabbagabbahey
Homework Helper
Gold Member
5,002
6
once i got to [tex]\int_{-1}^{t-1} \sqrt{u^2+1}du[/tex] i trig subbed and got [tex]\int sec^3(\theta) d\theta[/tex] but can i change the limits along with the variable and still solve for [tex] s [/tex]?
Well, if [itex]u=\tan\theta[/itex], then your limits on [itex]\theta[/itex] will be [itex]\tan^{-1}(-1)[/itex] and [itex]\tan^{-1}(t-1)[/itex].

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

Related Threads on Cal III S reparametrization problem

  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
11K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
10K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
836
Replies
4
Views
653
Top