This doesn't seem to be working? Relativity

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SUMMARY

The discussion focuses on evaluating the derivative of the force vector \(\vec{F} = \frac{d}{dt} \frac{m \vec{v}}{\sqrt{1 - v^2/c^2}}\) to determine the acceleration \(a = \frac{dv}{dt}\) of a particle. The user attempted to apply the quotient rule while treating mass \(m\) and the denominator as constants, leading to incorrect results. The correct approach requires recognizing that \(v^2\) cannot be treated as a constant during differentiation, as highlighted by the relationship \(\frac{dv}{dt} = \frac{\vec{a} \cdot \vec{v}}{v}\).

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  • Understanding of calculus, specifically the quotient rule for derivatives.
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  • Basic understanding of the relationship between velocity, acceleration, and force in physics.
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  • Study the application of the quotient rule in calculus, particularly in physics contexts.
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Evaluate the derivative of [tex]\vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}}[/tex] to find the acceleration [tex]a = \frac {dv}{dt}[/tex] of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

[tex]\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}[/tex]
[tex]\frac {F}{m} (1 + \frac {v^2}{3c^2})[/tex]
 
Last edited:
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IntegrateMe said:
Evaluate the derivative of [tex]\vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}}[/tex] to find the acceleration [tex]a = \frac {dv}{dt}[/tex] of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

[tex]\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}[/tex]
[tex]\frac {F}{m} (1 + \frac {v^2}{3c^2})[/tex]
How can you treat v2 as a constant?
 
Remember the fact that:

[tex]\frac{dv}{dt}=\frac{\vec{a}\cdot \vec{v}}{v}[/tex]
 
Last edited:

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