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

Arc Length in Three Dimensions Question

  • #1
RJLiberator
Gold Member
1,095
63

Homework Statement


Find the arc length of r(t)= <tsin(t), tcost(t), 3t> from 0 to t to 2pi (inclusive)

Homework Equations


Integral from 0 to 2pi of the magnitude of r'(t) dt

The Attempt at a Solution


1. Must find the derivative of the function.
Using the product rule a few times, the derivative becomes
<tcos(t)+sin(t), -tsin(t)+cos(t), 3>

2. Now take the magnitude
Well, this is a bit lengthy, but it simplifies pretty nice with the double angle trig identity. (I hope what I did is correct, please check my work in this step.

(tcos(t)+sin(t))^2 = t^2cos^2t+2tsin(t)*cos(t)+sin^2t
Simplifies to: t^2cos^2t+tsin(2t)+sin^2t

(-tsin(t)+cos(t))^2 = t^2sin^2t+(-2tsin(t)*cos*(t))+cos^2t
simplifies to: t^2sin^2t-tsin(2t)+cos^2t

3^2 = 9

So, when we add everything the double angles cancel out. The cos^2t+sin^2t is equal to 1. We factor a t^2 out of the t^2cos^2t and t^2sin^2t equations and this becomes t^2*1.

We are left with 9+1+t^2
t^2+10, beautiful!

3. Integrate from 0 to 2pi the square root of (t^2+10) dt

Well, here's where I am stuck.
I feel like I've done everything properly up to this point. But this integral cannot be taken easily (at least with calc 3 tools).

Did I do something wrong? Any hints on the integral?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement


Find the arc length of r(t)= <tsin(t), tcost(t), 3t> from 0 to t to 2pi (inclusive)

Homework Equations


Integral from 0 to 2pi of the magnitude of r'(t) dt

The Attempt at a Solution


1. Must find the derivative of the function.
Using the product rule a few times, the derivative becomes
<tcos(t)+sin(t), -tsin(t)+cos(t), 3>

2. Now take the magnitude
Well, this is a bit lengthy, but it simplifies pretty nice with the double angle trig identity. (I hope what I did is correct, please check my work in this step.

(tcos(t)+sin(t))^2 = t^2cos^2t+2tsin(t)*cos(t)+sin^2t
Simplifies to: t^2cos^2t+tsin(2t)+sin^2t

(-tsin(t)+cos(t))^2 = t^2sin^2t+(-2tsin(t)*cos*(t))+cos^2t
simplifies to: t^2sin^2t-tsin(2t)+cos^2t

3^2 = 9

So, when we add everything the double angles cancel out. The cos^2t+sin^2t is equal to 1. We factor a t^2 out of the t^2cos^2t and t^2sin^2t equations and this becomes t^2*1.

We are left with 9+1+t^2
t^2+10, beautiful!

3. Integrate from 0 to 2pi the square root of (t^2+10) dt

Well, here's where I am stuck.
I feel like I've done everything properly up to this point. But this integral cannot be taken easily (at least with calc 3 tools).

Did I do something wrong? Any hints on the integral?
You use a trig substitution on the integral. Something like t=sqrt(10)*tan(u). Have you done stuff like that?
 
Last edited:
  • #3
RJLiberator
Gold Member
1,095
63
Yes, I have. I will attempt this tomorrow morning and report back. This sounds promising.

Thank you.
 
  • #4
RJLiberator
Gold Member
1,095
63
How's this:

Let u = sqrt(10)*tan(theta)
dtheta = sqrt(10)sec^2(theta)

The integral is from 0 to 2pi the square root of 10+t^2
then you get the integral of the square root of 10+10tan^2(theta) under the radical
You can factor the 10 and the tan^2(theta)+1 becomes sec^2(theta) which simplifies to just sec after it is out of the radical

so now you are left with
sqrt(10)*sec^2theta/sqrt(10)sec(theta)
some cancelations make it the integral of sec(theta) from 0 to 2pi

answer: ln(sec(theta)+tan(theta)) from 0 to 2pi

1-1=0

well... that's not good.
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
618
How's this:

Let u = sqrt(10)*tan(theta)
dtheta = sqrt(10)sec^2(theta)

The integral is from 0 to 2pi the square root of 10+t^2
then you get the integral of the square root of 10+10tan^2(theta) under the radical
You can factor the 10 and the tan^2(theta)+1 becomes sec^2(theta) which simplifies to just sec after it is out of the radical

so now you are left with
sqrt(10)*sec^2theta/sqrt(10)sec(theta)
some cancelations make it the integral of sec(theta) from 0 to 2pi

answer: ln(sec(theta)+tan(theta)) from 0 to 2pi

1-1=0

well... that's not good.
That's pretty sloppy. This might go better if you didn't scramble variables. t=sqrt(10)*tan(u). dt=sqrt(10)*sec^2(u)du. You are going to wind up with a sec^3(u) integral, which is challenging, but you can always look it up. And u doesn't go from 0 to 2pi. t goes from 0 to 2pi.
 
  • #6
RJLiberator
Gold Member
1,095
63
Thanks for the hints.

I was doing the previous (sloppy) work in my head earlier :p.

Now I've got a pen and paper:

So after the tangent substitution, my integral becomes
10*sec^3(u)

The problem states that we can use table integrals to help solve our problems, so I assume that this is what he means. I apply the table integral of sec^3(u)

10*[1/2*sec(u)*tan(u)+12*ln|sec(u)+tan(u)|]

That seems to be correct. Now I just need to figure out the bounds, for some reason I am struggling with this (it's been a while). I'll keep working on it and report back.
 
  • #7
RJLiberator
Gold Member
1,095
63
I previously let
t = sqrt(10)*tan(u)

to change my bounds, i altered this equation to
tan^-1(t/sqrt(10)) = u

I then plugged in 0 and 2pi for t.
0 works out nicely, however, 2pi does not work out.

Something seems to be wrong :/
 
  • #8
RJLiberator
Gold Member
1,095
63
Ah. I can use the right triangle trigonometry.

Tan(u) = t/sqrt(10)
so sec(u)= sqrt(10+t^2)/sqrt(10)

And then plug in the bounds via this distinction.

The 2pi however is making the answer messy.

My final answer is
10*[pi*sqrt(10+4pi^2)/10+1/2*ln|(sqrt(10+4pi^2)+2pi)/sqrt(10)|
 
  • #9
Dick
Science Advisor
Homework Helper
26,258
618
Ah. I can use the right triangle trigonometry.

Tan(u) = t/sqrt(10)
so sec(u)= sqrt(10+t^2)/sqrt(10)

And then plug in the bounds via this distinction.

The 2pi however is making the answer messy.

My final answer is
10*[pi*sqrt(10+4pi^2)/10+1/2*ln|(sqrt(10+4pi^2)+2pi)/sqrt(10)|
Looks ok to me.
 
  • #10
RJLiberator
Gold Member
1,095
63
Fantastic. That was a great problem. I wasn't expecting it to be that in depth. Thank you for your guidance.
 

Related Threads on Arc Length in Three Dimensions Question

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
5K
Replies
6
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
2
Views
981
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
1K
Top