hi everyone what a grat place to assist and learn - i really would appreciate if anyone can find the integral of ( 1- (x^/a^)* . dt where ^ = power 2 and * = power3/2 - By the way i know the equation is with respect to 't' and yet inside the brackets is x - take x = velocity and t = time I know i need to change the velocity variable to match time t so it can be integrated with respect to time - but how ? - if you show me the steps you use i would be very grateful. Im stuck on this.
is this your integral? \[\int\left(1-\frac{x^2}{a^2}\right)^{3/2}\mathrm dt\]
is \(a\) acceleration?
Hello UnkleRhaukus - thank you for your response- Yes that is my integral. The a is not acceleration it is just a constant - its is in fact c for the speed of light. :)
\[\int\left(1-\frac{x^2}{a^2}\right)^{3/2}\mathrm dt\] It's very confusing having \(x\) as velocity if \(y\) is position \[\int\left(1-\frac{1}{a^2}\left(\frac{\mathrm dy}{\mathrm dt}\right)^2\right)^{3/2}\mathrm dt\], hmm
Where did you get this question?
The question is a form of physics formula for acceleration under special relativity - \[a = f/m ( 1- \frac{^{v2} }{ ^{c2} } ) ^{3/2}\] - it want to find the velocity on the left hand side with respect to time - So i write it as a = v/t = RHS Then to to find the velocity on right hand side i multiply both sides by t to get the integral you wrote already but i just left out the f/m in the equation since they are just constants of integration
i mean to find the velocity on the LEFT HAND SIDE with respect to time - i integrate
\[\int\limits_{?}^{?} dv = f/m \int\limits_{?}^{?} ( 1- \frac{ ^{v2} }{ c2 } ) ^{3/2} dt\] that is the actual integral
maybe some more like this ?\[\int \frac{\mathrm dv}{\left( 1- \frac{ {v^2} }{ c^2 } \right) ^{3/2}} = \frac fm \int dt\]
Yes i tried in that form but that \[\frac{ ^{v2} }{ ^{c2} }\] is bothering since i need the \[^{c2}\] out of the brackets before i can integrate it in standard form that you suggested just now
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