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Calculating a value that depends on itself.

by tomyuey938
Tags: depends
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tomyuey938
#1
Nov11-10, 04:26 AM
P: 14
Hi,
I'm calculating the velocity of something that's stretching, and as it gets longer, the velocity decreases.
So to calculate the velocity at time (t):
v(t) = A / (B L(t))

A change in the distance over a small time dt will be given by:
dL=v(t) dt

So I guess the distance is:
L= integral from 0 to t of v(t) dt ?

How can I go about solving such an equation? Are itterative methods required? Can anyone give me some keywords to help my search on google? I don't really know where to start.

Thanks a lot for your time.
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tiny-tim
#2
Nov11-10, 06:34 AM
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hi tomyuey938!
Quote Quote by tomyuey938 View Post
Hi,
I'm calculating the velocity of something that's stretching, and as it gets longer, the velocity decreases.
So to calculate the velocity at time (t):
v(t) = A / (B L(t))

A change in the distance over a small time dt will be given by:
dL=v(t) dt
i'm confused

do you mean dL/dt = A/BL ?

if so just move the terms around: L dL = A dt/B
tomyuey938
#3
Nov11-10, 09:01 AM
P: 14
Hi tiny-tim,

Thanks for your reply.

Well, I'm trying to calculate the value of v(t) at a given value of (t). So I don't think your re-arrangement is relevant in this case, since I need v(t)=something.
But I don't know the value of L, since L is the integral of v(t) dt up to that time (t).

Does that make sense? Please do let me know if you'd like me to explain anything more, or if you think I'm mis-understanding something (which is quite possible!).

Thanks and with regards.

maverick_starstrider
#4
Nov11-10, 09:50 AM
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P: 1,164
Calculating a value that depends on itself.

Hi tomyuey. tiny-tim's approach is indeed correct (although if you're curious the practice of solving an equation which is dependent on itself uses what's called a SELF-CONSISTENT approach). However, for this case [tex]\frac{dL}{dt}=v[/tex] so we can plug that in to get

[tex]v=\frac{dL}{dt} = \frac{A}{B L} \rightarrow L dL = \frac{A}{B}dt[/tex]

Integrating one gets

[tex]\frac{L^2}{2}=\frac{At}{B}\rightarrow L=\sqrt{\frac{2At}{B}}[/tex]
tomyuey938
#5
Nov11-10, 05:56 PM
P: 14
Hi Starstrider,

Thanks for your reply.

So in order to get an expression for v(t), I simply differentiate both sides by t to give:

dL/dt = SQRT(2A/B) (1/2) t^(-1/2)

Thanks also for the "self-consistent" term. This will be helpful in the future.

Thank you both so much for your help. I really do appreciate it.

Edit: but this suggests as t->0 the speed becomes infinite. Is this really correct, or have I made a mistake?
maverick_starstrider
#6
Nov11-10, 06:55 PM
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P: 1,164
L equals what you said not dL/dt
Citan Uzuki
#7
Nov12-10, 11:51 AM
P: 274
maverick_starstrider's approach is missing a constant of integration. It should be:

[tex]\frac{L^2}{2}=\frac{At}{B}+C \Rightarrow L = \sqrt{\frac{2At}{B} + 2C}[/tex]

Where 2C is the square of the length at t=0. Now, if the object has zero length at t=0 (as MS implicitly assumes), then since the velocity is inversely proportional to the length, one would indeed expect the velocity to be infinite at t=0.


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