Tips For How To Understand Application of Integrals In Physics

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Since I'm trying to make sense of the application of a concept, I figured I'd post this thread here. If this belongs in the homework forum, I sincerely apologize.

For the longest time I would do integral problems in Calculus. I figured I understood how to do them because after I finished working problems, I would look in the back of my book and see my answers match what's there.

However, going into Physics, I see I really don't understand the whole IDEA behind integrals. When I say this, I mean that I'm never sure how to setup my integral for problems or even WHEN to use an integral.

For example, in one of my E. Science books, they posed this example question:

Derive the expression for the charge accumulated at the upper terminal of the element for t>0

and at this point the book had given me 2 ratios in differential form:

v = dw/dq

and

i = dq/dt



When the book told me to "derive the expression", this confused me because I can't with faith know exactly what to differentiate with respect to; that is, I'd just be guessing at things to differentiate with respect to until something lines up with what the book says.

Is something like this example problem a thing that only makes sense if you work out as many problems as possible or is it something that I can read around to get a better understanding of?

I do many problems in my 1st time physics class and I notice many times I don't know the direction to take with a problem unless I've seen it done before. So, is it something sort of like that to understand applications of integrals in physics?
 

Answers and Replies

  • #2
Pengwuino
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Take for example the simplest integral problem we know of in physics, a constant acceleration in kinematics.

So let's say you want your velocity at any given time. Your acceleration is [tex]{{dv} \over {dt}} = a[/tex].

Simple enough right? So let's say I want to figure out the velocity after a certain time. Well, if you integrate with respect to dt, what you're doing is summing up all the changes in velocity over some range of time (keep the integral as some area under the curve on a graph). You know some small change in time, dt, multiplied by the constant acceleration gives you some small change in velocity, dv. The integral is summing up all the contributions. Of course, this holds in general. You could then integrate upwards again to find your position with respect to time following the same idea. You'd find your basic kinematic [tex]x_f(t) = x_i(t) + v_o(t) + .5at^2[/tex].

In that case, the problem was obvious, you know you want to integrate over time. You know you want position. Simple. This problem is simple as well! What are they looking for? They explicitly say "Derive the expression for the charge accumulated". In other words they want the charge, Q. So Q is our X (position) in this analogy. Now if you could find [tex]dQ = f(t)dt[/tex], you could integrate immediately and find the answer right? Well, from what you're telling us, the relationship between dQ and dt is 'i'.

In other words, [tex]I = {{dQ} \over {dt}}[/tex]

However, is 'I' constant? If it is, then the integration is trivial. Infact you'll get the exact same form as a velocity in the kinematic equations. However, 'I' is probably not constant, we need the rest of the information from the problem to decide what to do next.
 
  • #3
tiny-tim
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Hi Meadman23! :wink:
Derive the expression for the charge accumulated at the upper terminal of the element for t>0
I'll just add this to what Pengwuino :smile: has said …

the word "derive" has nothing to do with "derivative" …

we differentiate to find the derivative

we derive a derivation

it just means "deduce" or "find" :smile:
 
  • #4
Pengwuino
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Ouch how did I miss that! Thanks tiny-tim.
 
  • #5
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Hi Meadman23! :wink:

the word "derive" has nothing to do with "derivative" …

we differentiate to find the derivative

we derive a derivation

it just means "deduce" or "find" :smile:
That REALLY helps put things into perspective. Seeing the words "derive" is part of the reason I had such difficulty figuring these out...

However, is 'I' constant? If it is, then the integration is trivial. Infact you'll get the exact same form as a velocity in the kinematic equations. However, 'I' is probably not constant, we need the rest of the information from the problem to decide what to do next.
Hmm. The rest of the problem says that no charge exists for all "t < 0", but it mentions at "t = 0" a 5A charge begins to flow.

In general, are the limits of your integral going to be explicitly (or really close to) defined or do you have to do some deep thinking to understand how to set them up?

The integral is summing up all the contributions. Of course, this holds in general. You could then integrate upwards again to find your position with respect to time following the same idea. You'd find your basic kinematic.
Honestly I can't really understand how to find my "basic kinematic". I feel that if I start at the very beginning I can understand how to get to this step, so I went to a website and found instructions for how to relate velocity to acceleration and this is what they showed me:
Remembering that the acceleration is defined by the derivative
[URL]http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity_2.gif[/URL]

we can apply the Fundamental Theorem of Calculus to write this relationship in the form
[URL]http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity_3.gif[/URL]

If we pick call the initial time and the final time , then this integral has the form
[URL]http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity_6.gif [Broken][/URL]
Now what is making me lose my mind is trying to understand this step:
If we pick call the initial time and the final time , then this integral has the form
[URL]http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity_6.gif [Broken][/URL]
If I'm integrating everything, why does the left side stay as "a(t) dt" instead of becoming "a(t)T"???
 
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  • #6
Pengwuino
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You have to integrate the function. What if a(t) was something ridiculous like [tex]a(t)= 3cos(5t+9) - {{t^2} \over {e^{3t}}}+ bicycle[/tex]. You have to integrate that function as a function of t. For the typical physics kinematics you see in your first semester course, you always assume constant acceleration so that a is constant. This part is the generalization.

There is occasion that you do have to have a sort of intuition on what the domain of your integration will be. Typically a problem will start at t=0, which should be expected because that's a good start right? However you'll quickly learn how your integration limits are handled. If for example, your question asks how much charge gets accumulated, that kinda should sound like "how much charge gets accumulated if you let the thing go on forever", in other words letting the upper bound go to t -> infinity. When you're dealing with time, you usually go from 0 to infinity because problems typically don't ask "find out the charge after 5 seconds". Though if it did, it's obvious what the integration limit should be right?

Volume integrals are where the real trouble starts. If you know you have to take the integral of something in 3-dimensional space, you could start having problems. A lot of problems typically do have you integrate towards infinity which in cartesian is coordinates is pretty easy. You just imagine a 3d graph and you know you need to go from negative infinity on each axis to positive infinity on each axis. You integrate over all space. However, you'll see problems that have finite integrals more often. So instead of integrating over all space, you might have to integrate over a finite box. Then you need to start really understanding what your integration limits are meant to be.
 
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I must say i feel with Meadman23.

In all the googleing about setting differential equations in physics, all every reference talks about is the most simple kinematic equations, and stop there.

I have a some examples where i couldn't put an analogy to the kiematic equations, and my equations were incorrect.

Here are some:

1.What is the change of water density on the depth h (from the water surface)? Coefficient of water compressibility is [tex]\chi[/tex] and water density on the surface is [tex]\rho_0[/tex].

2. A wagon with mass m is travelling on a rail without friction on a horizontal rail. A continuous stream of water is hitting the back side of the wagon. After the hit, the water falls to the ground. The pipe through which the water is coming form has a cross section sized S and it throws [tex]\Phi[/tex] [m^3] of water per second. What is the speed of the wagon after s_0. The wagon was standing still at the beginning.

and so an...

I can't see why isn't there a set of examples that relate mathematical with physical calculus (i can already see accusations that they are the same thing, and that i don't know enough mathematics :) ).
 
  • #8
Pengwuino
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They're both the same. Physics just requires the added skill of interpretation of problems. We use the tools of differential equations to solve our physical problems.
 

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