How to compute the integral of r(t) in three dimensions?

  • #1
RJLiberator
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Homework Statement


r(t) = <cos(t), 2t, sin(t)> from 0 to 2pi (included).
Compute the integral from 0 to 2pi of r(t).

Homework Equations




The Attempt at a Solution


[/B]
We haven't learned this in class yet, so I am getting a head. My idea is to simply put r(t) in and integrate term by term so that the final answer is
sin(t), t^2, -cos(t) and evaluate it from 0 to 2pi which gets
<0, 4pi^2, -1> - < 0, 0, -1>
which results in
<0, 4pi^2, 0>

And then I take the magnitude? Eh, answer should not be a vector I assume since this is the area of the space.
 
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  • #2
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Looks correct apart from the third component. Not the most typical integral with multiple dimensions, but why not... if r(t) would be a velocity, you would now have the displacement, for example.
 
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  • #3
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Homework Statement


r(t) = <cos(t), 2t, sin(t)> from 0 to 2pi (included).
Compute the integral from 0 to 2pi of r(t).

Homework Equations




The Attempt at a Solution


[/B]
We haven't learned this in class yet, so I am getting a head. My idea is to simply put r(t) in and integrate term by term so that the final answer is
sin(t), t^2, -cos(t) and evaluate it from 0 to 2pi which gets
<0, 4pi^2, -1> - < 0, 0, -1>
which results in
<0, 4pi^2, -2>

And then I take the magnitude? Eh, answer should not be a vector I assume since this is the area of the space.
A vector result is fine. If you were integrating an acceleration vector over time, for example, your result would be a change in velocity vector v(2Pi) - v(0).
Since seem to not want a vector, and r(t) is usually reserved notation for a position vector, perhaps you actually want the line integral that computes the arc length of the space curve parametrized by r(t). In that case, the idea is to partition the interval [0, 2Pi], and take the length of one tangent vector to the curve |r'(t)| for each interval, and sum them to approximate the length of the curve. The Riemann integral is formed by taking the least upper bound of all such sums (or the limit as the maximum width of each partition approaches 0). As the interval lengths get smaller, the sum of the lengths of the tangent vectors should more closely approximate the curve, as long as the curve is well-behaved. This gives the integral form [tex]\int_0^{2\pi} ||r'(t)|| \, dt.[/tex] This is the integral that you will want to evaluate, but of course you must first find the scalar function ||r'(t)|| to integrate.
Without further context, I cannot say which of these your book is requesting you do.
 
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  • #4
LCKurtz
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Homework Statement


r(t) = <cos(t), 2t, sin(t)> from 0 to 2pi (included).
Compute the integral from 0 to 2pi of r(t).

Homework Equations




The Attempt at a Solution


[/B]
We haven't learned this in class yet, so I am getting a head. My idea is to simply put r(t) in and integrate term by term so that the final answer is
sin(t), t^2, -cos(t) and evaluate it from 0 to 2pi which gets
<0, 4pi^2, -1> - < 0, 0, -1>
which results in
<0, 4pi^2, -2>

And then I take the magnitude? Eh, answer should not be a vector I assume since this is the area of the space.
The answer to what shouldn't be a vector? You haven't told us what problem you are purportedly working. Given that you now wonder if you should take the magnitude, I'm guessing that the integral you are working is likely not appropriate for whatever problem you are trying to solve. State the original problem.
 
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  • #5
RJLiberator
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The answer to what shouldn't be a vector? You haven't told us what problem you are purportedly working. Given that you now wonder if you should take the magnitude, I'm guessing that the integral you are working is likely not appropriate for whatever problem you are trying to solve. State the original problem.
This is all the information I'm given:
 

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  • #6
RJLiberator
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Looks correct apart from the third component. Not the most typical integral with multiple dimensions, but why not... if r(t) would be a velocity, you would now have the displacement, for example.
I see the minor calculation error and have now fixed it (or it should be fixed :p). Thank you.
 
  • #7
RJLiberator
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A vector result is fine. If you were integrating an acceleration vector over time, for example, your result would be a change in velocity vector v(2Pi) - v(0).
Since seem to not want a vector, and r(t) is usually reserved notation for a position vector, perhaps you actually want the line integral that computes the arc length of the space curve parametrized by r(t). In that case, the idea is to partition the interval [0, 2Pi], and take the length of one tangent vector to the curve |r'(t)| for each interval, and sum them to approximate the length of the curve. The Riemann integral is formed by taking the least upper bound of all such sums (or the limit as the maximum width of each partition approaches 0). As the interval lengths get smaller, the sum of the lengths of the tangent vectors should more closely approximate the curve, as long as the curve is well-behaved. This gives the integral form [tex]\int_0^{2\pi} ||r'(t)|| \, dt.[/tex] This is the integral that you will want to evaluate, but of course you must first find the scalar function ||r'(t)|| to integrate.
Without further context, I cannot say which of these your book is requesting you do.
Thank you for this thoughtful reply. I have since posted the original question which does include the unit tangent, however, I am not sure if that is part of what they are asking to integrate.

Your description makes clear sense to me that it can be a vector (acceleration to velocity) or perhaps their asking us to integrate the unit tangent.

Thank you. I await the next reply from someone.
 
  • #8
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Thank you for this thoughtful reply. I have since posted the original question which does include the unit tangent, however, I am not sure if that is part of what they are asking to integrate.

Your description makes clear sense to me that it can be a vector (acceleration to velocity) or perhaps their asking us to integrate the unit tangent.

Thank you. I await the next reply from someone.
The integral presented is indeed the integral of a vector quantity with respect to t, which results in another vector quantity, so your original interpretation is the right one. They may just have asked for the unit tangent as a preparatory exercise for later on in the text.
 
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  • #9
RJLiberator
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Thank you.
That does make sense to me now due to your earlier clarification/example.

I appreciate the help.
 

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