Velocity-Based Differentiation for B(s) in a Simple Homework Problem

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Homework Statement



I have nearly finished a problem. I am at the stage where I have:

[tex]B(s)\propto\frac{1}{s^{3}}[/tex]

[tex]v=\frac{ds}{dt}[/tex]

I want to find the rate of change of B(s), but expressed in terms of a velocity, rather than a displacement.

Homework Equations



So for displacement:

[tex]B'(s) \propto\frac{-3}{s^{2}}[/tex]

The Attempt at a Solution


How can I express this in terms of v instead? My guess would be that:

[tex]B'(s) \propto\frac{1}{v^{3}}[/tex]

but I'm not sure that's correct. This does seem like quite a simple problem but its really got me stumped.

Thanks in advance, Ewan
 
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Is the velocity constant?
Do you mean by rate of change

[tex]\frac{dB(s)}{dt}[/tex]?
And I think it would be better if you post whole question.
 
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azatkgz said:
Is the velocity constant?
Do you mean by rate of change

[tex]\frac{dB(s)}{dt}[/tex]?
And I think it would be better if you post whole question.

Yes the velocity is constant. The rest of the question isn't relevant to my problem, but I can post if you want. And yes by rate of change I mean:

[tex]\frac{dB(s)}{dt}[/tex]
 
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Never mind I managed to solve it (i think!).

[tex]B \propto \frac{-3}{s^{3}} \frac{dv}{dt}[/tex]

using substitution of s for v.
 
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