Integrating both sides of an equation question

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  • Thread starter TomServo
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  • #1
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I am studying a paper and a math step like this was used:

[tex]
dt'=(1+\frac{h}{2}sin^2(\theta))dr \\
\int^{t1}_t dt'=\int^d_0 (1+\frac{h}{2}sin^2(\theta))dr \\
where\\
h=h(t-\frac{r}{c}-\frac{r}{c}cos(\theta))
[/tex]

This seems wrong because it seems to me that you're not doing the same thing to both sides of the equation. You ought to integrate both sides with respect to t', correct? Is there some change of variables I'm not seeing here? Or in infinitesimal equations like that, can you really just integrate each term with respect to the differential factor's corresponding variable?
 

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  • #2
mathman
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You ought to integrate both sides with respect to t', correct?
No.

What needs to be specified is the relationship between t' and r, so that the corresponding limits on the integrals can be defined.
 
  • #3
FactChecker
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The entire right side goes from t to t1, but r only goes from 0 to d. The limits of the right side are the limits of r.
Consider a simple equation like y=2x.
dy=2dx.
01dy = ∫01/22dx
 
  • #4
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Actually you are correct, you can't just arbitrarily integrate both sides of an equation with respect to different variables any more than you can differentiate the two sides of an equation with respect to different variables or multiply the two sides by different numbers. This is a question that arises in every calc 1 class because it does look wrong. There is actually a change of variable involved. What you are actually saying is that t is a function of r: [tex]t = f(r)[/tex]. Then [tex]\frac{{dt}}{{dr}} = f'(r)[/tex] and [tex]\int {\frac{{dt}}{{dr}}dr = \int {dt = } \int {f'(r)dr} } [/tex]. The limits are found from the original functional relationship between the two variables.
 
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  • #5
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No.

What needs to be specified is the relationship between t' and r, so that the corresponding limits on the integrals can be defined.

t'=t+r/c-(r/c)cos(theta)
 
  • #6
281
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Actually you are correct, you can't just arbitrarily integrate both sides of an equation with respect to different variables any more than you can differentiate the two sides of an equation with respect to different variables or multiply the two sides by different numbers. This is a question that arises in every calc 1 class because it does look wrong. There is actually a change of variable involved. What you are actually saying is that t is a function of r: [tex]t = f(r)[/tex]. Then [tex]\frac{{dt}}{{dr}} = f'(r)[/tex] and [tex]\int {\frac{{dt}}{{dr}}dr = \int {dt = } \int {f'(r)dr} } [/tex]. The limits are found from the original functional relationship between the two variables.

I think that is what I first considered and wound up with a factor that had its numerator and denominator opposite of what the paper had.
 
  • #7
281
9
Actually you are correct, you can't just arbitrarily integrate both sides of an equation with respect to different variables any more than you can differentiate the two sides of an equation with respect to different variables or multiply the two sides by different numbers. This is a question that arises in every calc 1 class because it does look wrong. There is actually a change of variable involved. What you are actually saying is that t is a function of r: [tex]t = f(r)[/tex]. Then [tex]\frac{{dt}}{{dr}} = f'(r)[/tex] and [tex]\int {\frac{{dt}}{{dr}}dr = \int {dt = } \int {f'(r)dr} } [/tex]. The limits are found from the original functional relationship between the two variables.

NM, I found out what I was doing wrong. Thanks for confirming that my initial hunch on what was going on was correct, even if I botched the execution.
 

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