Length of Curve (Calc 2) question

  • #1
twiztidmxcn
43
0
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
MrSeaman
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
10,089
135
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]
 

Suggested for: Length of Curve (Calc 2) question

Replies
5
Views
428
  • Last Post
Replies
8
Views
450
  • Last Post
Replies
32
Views
356
  • Last Post
Replies
3
Views
493
  • Last Post
Replies
8
Views
342
  • Last Post
Replies
16
Views
521
Replies
5
Views
287
Replies
7
Views
996
  • Last Post
Replies
12
Views
949
Replies
7
Views
616
Top