-1
$\begingroup$

So I have just entered 11th grade and started limits on my own but my Physics textbook has an equation which I don't understand, I suspect it uses integration which I haven't learned yet. So can someone explain this equation to me:-

The question is:-

Find the value of $n$ (by using dimensional analysis): $$\int \frac{dx}{\sqrt{2ax-x^2}} = a^n \sin^{-1} [\frac{x}{a} -1] $$

The equation looks similar to $$ v^2-u^2 = 2ax $$ But I don't understand what $dx$ means.

$\endgroup$

closed as off-topic by Kyle Kanos, ACuriousMind, Rob Jeffries, Martin, Danu May 26 '15 at 9:55

  • This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ dx is just the variable with which to integrate with respect to. like $\frac{dy}{dx}$ is differentiating y with respect to x. Here, you integrate with respect to x also. $\endgroup$ – Weasel May 24 '15 at 11:50
  • 6
    $\begingroup$ I think, this is rather a question for math.SE, as it is concerded only with mathematical concepts. $\endgroup$ – Sebastian Riese May 24 '15 at 12:00
  • $\begingroup$ It would help you would just give a general understanding, you don't have to go too deep. $\endgroup$ – Abhishek Mhatre May 24 '15 at 12:05
  • $\begingroup$ Perhaps you should ask the question here : artofproblemsolving.com/community/… $\endgroup$ – user37026 May 24 '15 at 12:49
  • $\begingroup$ It's a calculus formula. This is not the right place to learn about how calculus works. $\endgroup$ – Rob Jeffries May 25 '15 at 9:54
1
$\begingroup$

Clearly $a$ has the same dimension of $x$ (see the argument of root or of $\sin^{-1}$) so the left member is dimensionless (ratio between dimension of x: remember that differential dx count in dimensional calculus!), and the second member too has to be dimensionless: so n=0.

$\endgroup$
1
$\begingroup$

Integration is finding the area under a curve that isn't necessarily straight. If you have a velocity time graph and find the area under it, this gives you the distance travailed. If you have a acceleration-time graph the area under it is the change in velocity. There are several techniques to integration, which I will not go into here. As mentioned in the comments $dx$ tells you, you are integrating w.r.t. $x$ it is a label rather then a physical quantity in this sense. When doing dimensional analysis you can simply give the dx a dimension of length then and ignore the integral sign $\int$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.