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

Length of Curve (Calc 2) question

  • #1
So the length of a curve given by a vector r is:

L = [tex] \int |r'(t)|dt[/tex]

If the vector is given by r(t) = < t, t^2/srt2, t^3/3> on an interval from 0 to 1, find the length

I am incredibly confused here, as most of the examples we've used have been using the cos^2+sin^2 = 1 identity.

I took the derivative of r(t), and got

r'(t) = 1, 2t/sqrt2, t^2>

then i set up the integral using the magnitude of that equation, and end up with L = [tex] \int (1^2 + (2t/sqrt2)^2 + (t^2)^2)dt [/tex]

I get this feeling like I need to set the t value equal to something in order to make the magnitude work out for an integral that would work out to integrating something like dt itself.

Any help would be much appreciated
Thank you
Twiztidmxcn
 

Answers and Replies

  • #2
6
0
You forgot the squareroot and the limits, but the rest looks fine. You just have to calculate

[tex] L = \int \limits_a^b | r'(t) |\mathrm{d} t = \int \limits_0^1 \sqrt{1^2 + (\sqrt{2}t)^2 + (t^2)^2 } \mathrm{d} t [/tex]
 
  • #3
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
Just to help you a bit further:
Note that:
[tex]1^{2}+(\sqrt{2}t)^{2}+(t^{2})^{2}=1+2t^{2}+t^{4}=(1+t^{2})^{2}[/tex]
 

Related Threads for: Length of Curve (Calc 2) question

  • Last Post
Replies
2
Views
3K
Replies
13
Views
3K
Replies
7
Views
132
Replies
3
Views
15K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
300
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
2K
Top