
#1
Jan2213, 08:23 PM

P: 142

In my dynamics class we are solving questions using integration. But I am having trouble deciding which variable I should do the integration with. for example consider the following question.
If an object is dropped from rest the acceleration is given by [tex] a = 9.81(1  \frac {v^2}{10000}) [/tex] what is velocity after 7 seconds. I know that [tex] a = \frac{dv}{dt}[/tex] so [tex] dv = a \cdot dt [/tex] here I chose time as the variable I am integrating with respect to because I am given time in my question. [tex] \int^v_0 dv = \int^t_09.81(1  \frac {v^2}{10000}) [/tex] [tex] v = 9.81t (1  \frac {v^2}{10000}) [/tex] solving the quadratic at t = 7 I get v = 50.9 but the book integrates with respect to v as in [tex] dt = \frac{dv}{a} [/tex] putting in the values and solving this I get 59.6 as my answer which is correct. What confuses me is the choice between whether to isolate dv and integrate w.r.t dt or the other way around. Same problem arises when I am using other equations especially [tex] a = v \frac{dv}{ds}[/tex] Is there some sort of rule or trick to figuring this out because my math class is way behind my dynamics class and my dynamics class isn't going to slow down. 



#2
Jan2213, 09:18 PM

P: 4,570

Hey wahaj.
It doesn't really matter how you separate the differentials, just as long as it keeps the relationship consistent: in other words, it keeps the relationship of the system the same under a transformation of modifying the differentials. Recall that mathematics deals with the nature of looking at ways to transform things but in a way to not change the nature and the information of the system itself. As long as the nature and information of the system and its properties stay the same under a transformation, then the transformation is allowed. 



#3
Jan2213, 09:23 PM

P: 142

then how come I am getting different answers when I use different differentials.




#4
Jan2213, 09:25 PM

P: 4,570

How to know which differential to chooseIts the same reason that dividing by zero is not allowed and also why cancelling denominator terms that have zero is not allowed either (and why dividing by zero is not allowed at all). 



#5
Jan2213, 09:33 PM

P: 142

I showed what I did in the first post. I skipped a few calculation steps though. Technically it shouldn't matter which differential I use to integrate with as you said but it doesn't seem to work this way. I can't really tell whether the transformation is consistent or not in an exam so I can't tell which transformation is right.




#6
Jan2213, 09:37 PM

P: 4,570

You can't do what you did in the first post since v is not constant with t.
When one variable changes with another, then you need to factor that in when integrating. Do you understand this? 



#7
Jan2213, 09:40 PM

P: 142

you mean I can't take integrals like this?
[tex] \int^v_0 dv = \int^t_09.81(1  \frac {v^2}{10000}) [/tex] This is exactly how its done in my book. 



#8
Jan2213, 09:46 PM

P: 4,570

If you have integrals like the above where an integral has two terms and one term is dependent on the other (i.e. nonzero derivative) then it means that you either have to substitute the explicit value for one of them in terms of what is being integrated to or to use a numerical scheme to get an approximation.
So a substitution would be getting v(t) and substituting that function in terms of a function of t and then calculating the integral. Only when one something is a constant with respect to another variable (and hence its derivative is 0) can you do these kinds of integrations. 



#9
Jan2213, 09:57 PM

P: 142

To be honest I can't do any kind of integration. In my math class we have just gotten to the Riemann sum and starting to understand the basics of integration. I am using a list of calculated integrals at the back of my book to do my work. But I understand pretty much all I know except for differentials. However I think I understand what you are saying. If I have something like
x = ay and z = xy, to be able to integrate the second equationI have to first do this z = ay^{2} but in the question I posted as an example I only have one equation to work with 



#10
Jan2213, 10:07 PM

P: 4,570

What you need to do is get all the v's on side and all the t's (and constants) on the other and then do the integration.
We call this kind of DE a separable DE (because we can separate both variables on different sides of the equality). If you do this and integrate both sides you will get the proper answer and you won't have the issue that I described above. 



#11
Jan2213, 10:17 PM

P: 142

Now I understand why the book does it the way it does. And I think this will help me solve more questions because now I know which differential belongs with which variables. Thanks for helping me out. You really saved me here. just one more question, by DE do you mean differential equation? because from what I know those are some hard equations to solve for a first year



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