Arc Length in Three Dimensions Question

In summary: Why do you say it's wrong?I said it was wrong because I had a feeling it was wrong. I don't have a calculator with me, so I can't check it. However, I just checked it in wolfram alpha and it looks like it's correct. Thank you very much for your help! In summary, the arc length of r(t) = <tsin(t), tcos(t), 3t> from 0 to 2pi is found by first taking the derivative of the function and then calculating the magnitude of the derivative. This simplifies using double angle trigonometric identities and results in the integral of the square root of t^2+10. Using a trigonometric
  • #1
RJLiberator
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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?
 
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  • #2
RJLiberator said:

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?
 
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  • #3
Yes, I have. I will attempt this tomorrow morning and report back. This sounds promising.

Thank you.
 
  • #4
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
RJLiberator said:
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.
 
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  • #6
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
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
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
RJLiberator said:
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.
 
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  • #10
Fantastic. That was a great problem. I wasn't expecting it to be that in depth. Thank you for your guidance.
 

Related to Arc Length in Three Dimensions Question

What is arc length in three dimensions?

Arc length in three dimensions refers to the distance along the surface of a three-dimensional object between two points on a curve.

How is arc length calculated in three dimensions?

To calculate arc length in three dimensions, you must first determine the equation of the curve. Then, you can use calculus to integrate the arc length formula: L = ∫√(1 + (f'(x))^2) dx, where f'(x) is the derivative of the curve's equation.

What is the difference between arc length and arc distance?

Arc length refers to the distance along a curve, while arc distance refers to the straight line distance between two points on a curve. In three dimensions, arc length takes into account the curvature of the surface, while arc distance does not.

Why is arc length important in three-dimensional space?

Arc length is important in three-dimensional space because it allows us to measure the distance along a curve, which is necessary for many real-world applications such as finding the shortest path between two points on a curved surface or calculating the surface area of a three-dimensional object.

What are some real-world examples of using arc length in three dimensions?

Arc length in three dimensions has many practical applications, such as calculating the distance a roller coaster travels along its track, determining the length of a curved wire needed for a specific design, and finding the shortest path for a spacecraft to travel between two planets.

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