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Calculate the length of the given curve part 2

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Homework Statement



[itex]\vec r (t) = \vec u +t \vec v for -4 \le t \le 6[/itex] where u and v are constant vectors and v not equal 0

Do i proceed with the usual calculation of differentiating r wrt t, find the magnitude and then integrate between limits? Looks too simple for that. Thanks
 

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  • #2
Simon Bridge
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http://www.ucl.ac.uk/Mathematics/geomath/level2/fvec/fv4.html

but in this case it indeed looks like:

[tex]\frac{d\vec r}{dt}=\vec v[/tex]

... think what differentiation means. How [itex]\vec r[/itex] changes with time is it grows in the [itex]\vec v[/itex] direction and the rate of that growth is the magnitude.

The intergral will be the area between the path mapped out by [itex]\vec r(t)[/itex] and ... something.

Of course, I'm only guessing that you have been asked to differentiate and integrate that function ... you didn't actually say.

If we consider r to be a displacement, then [itex]\vec u = \vec r(0)[/itex] and [itex]\vec v[/itex] is, indeed, the velocity.
It is not clear what the area under the displacement-time graph represents.
It's easier to think about if r is a velocity, u is initial velocity, and v is the acceleration.
 
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http://www.ucl.ac.uk/Mathematics/geomath/level2/fvec/fv4.html

but in this case it indeed looks like:

[tex]\frac{d\vec r}{dt}=\vec v[/tex]

... think what differentiation means. How [itex]\vec r[/itex] changes with time is it grows in the [itex]\vec v[/itex] direction and the rate of that growth is the magnitude.

The intergral will be the area between the path mapped out by [itex]\vec r(t)[/itex] and ... something.

Of course, I'm only guessing that you have been asked to differentiate and integrate that function ... you didn't actually say.

If we consider r to be a displacement, then [itex]\vec u = \vec r(0)[/itex] and [itex]\vec v[/itex] is, indeed, the velocity.
It is not clear what the area under the displacement-time graph represents.
It's easier to think about if r is a velocity, u is initial velocity, and v is the acceleration.
It says I have to calculate the length of the curve for the given problem.
 
  • #4
LCKurtz
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Homework Statement



[itex]\vec r (t) = \vec u +t \vec v for -4 \le t \le 6[/itex] where u and v are constant vectors and v not equal 0

Do i proceed with the usual calculation of differentiating r wrt t, find the magnitude and then integrate between limits? Looks too simple for that. Thanks
It is very simple. Just do it.
 
  • #5
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It is very simple. Just do it.
I get [itex]10 \vec v[/itex] by differentiating r(t) wrt t , getting its magnitude which is v and then integrating between the limits wrt t...
 
  • #6
LCKurtz
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I get [itex]10 \vec v[/itex] by differentiating r(t) wrt t , getting its magnitude which is v and then integrating between the limits wrt t...
That is confusing. You don't get [itex]10\vec v[/itex] as the derivative if that is what you mean. And if you mean that is the answer it isn't correct either because the answer must be a scalar.
 
  • #7
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Well yes I realise the answer is a scalar because I know the vector v is a constant as stated in the problem hence 10 times a constant?

r(t)= u + tv.
dr(t)= v.

Magnitude = (v^2)^(1/2) =v. Then integrate this between the limits from -4 to 6....
 
  • #8
LCKurtz
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Well yes I realise the answer is a scalar because I know the vector v is a constant as stated in the problem hence 10 times a constant?

r(t)= u + tv.
dr(t)= v.

Magnitude = (v^2)^(1/2) =v. Then integrate this between the limits from -4 to 6....
You need to use different notation for a vector [itex]\vec v[/itex] and its magnitude [itex]|\vec v| = v[/itex]. Otherwise it is difficult to tell whether you are using the vector or its magnitude. You are using v for both above.
 
  • #9
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You need to use different notation for a vector [itex]\vec v[/itex] and its magnitude [itex]|\vec v| = v[/itex]. Otherwise it is difficult to tell whether you are using the vector or its magnitude. You are using v for both above.
[itex]d \vec r (t)= \vec v[/itex]. Therefore the magnitude is [itex]| \vec v| = v[/itex]. Then

[itex]\int_{-4}^{6} v dt = 10v[/itex]...?

That is what I did . Correct?
 
  • #10
LCKurtz
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[itex]d \vec r (t)= \vec v[/itex]. Therefore the magnitude is [itex]| \vec v| = v[/itex]. Then

[itex]\int_{-4}^{6} v dt = 10v[/itex]...?

That is what I did . Correct?
Yes. That is much better and it is correct.
 
  • #11
Simon Bridge
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Neato :)
I'm a big fan of understanding the problem.

If r(t) is a position vector, then the thingy's motion has speed |v| in the direction v/|v| and passes through position u at t=0. Since v is a constant vector, the path is a straight line. We want to know how far the thingy has traveled between t=-4 and t=6 units.

Since it's linear, that's just the final position subtracted from the initial position - or the speed multiplied by the elapsed time.

Of course, that's probably not mathematically rigorous.
 
  • #12
Simon Bridge
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Neato :)
I'm a big fan of understanding the problem.

If r(t) is a position vector, then the thingy's motion has speed |v| in the direction v/|v| and passes through position u at t=0. Since v is a constant vector, the path is a straight line. We want to know how far the thingy has traveled between t=-4 and t=6 units.

Since it's linear, that's just the final position subtracted from the initial position - or the speed multiplied by the elapsed time.

Of course, that's probably not mathematically rigorous.
 
  • #13
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good insight! Thanks
 

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