Question about a derivative in a Physics formula

[Soontir C'boath]

Ok, it is my second day of class in Physics and I noticed it is a bit different than High School physics in that it has some calculus in it now. I never learned derivatives since I am only taking a Calculus class now so bear with me.

In class, we had a formula in which it was for the derivative of 't', it is (btw I don't know how to insert that vertical line) vdt= (v+at)dt= vt+(1/2)at^2

My question after canceling out the derivative is, why does it have a coeffiecient of 1/2? I'm sure it has to do with the 't' that is multiplying with acceleration but I'd like an explanation.

Is it because by taking the derivative, there has to be a coefficient that must be the reciprocal of the exponet?

Also in the second part, v turns into (v+at) because of dt correct? That it has to assume the same value but it must include time, right?

[SCRawl]

The expression on the right actually looks it was integrated, not differentiated. I'm sure that I'm not the one to try and give a lesson in the kinematic equations and calculus (well, I could, but I'd be dredging up physics I haven't used in 10 years).

In the example you have, though, it looks like you've integrated velocity with respect to time. I don't know if that helps.

Your high school definitely didn't do you any favours by denying you calculus, by the way. Is this typical of US public school? I attended public school in Canada, and we went up to rudimentary integral calculus as well.

[Soontir C'boath]

Oy, then I guess I mixed the terms up. I didn't quite well understand it in class and as I said I never have taken calculus before.

My high school physics class had no calculus in it whatsover. Our AP Physics class was B which didn't have any either.

[Kuroneko]

Depending on the level of calculus expected in your physics class, if the following makes little sense, you should probably drop it and wait until you have understanding of basic calculus. The "vertical line" is the symbol that indicates an integral, which is the reverse of differentiation, up to a constant. Actually, integrals (there is more than one type) are defined in completely different ways, and it just so happens that it reverses differentiation, a result known as the fundamental theorem of calculus, but nevermind the details right now--thinking the two as reversing each other is probably the most accessible view for a beginner.

My question after canceling out the derivative is, why does it have a coeffiecient of 1/2? I'm sure it has to do with the 't' that is multiplying with acceleration but I'd like an explanation.

First, one must understand derivatives. The derivative of a function f(x) is the rate of change of that function. Pick an x to compute the rate of change at, and pick a point x+h near x (h is small, possibly negative), and compute the slope between (x,f(x)) and (x+h,f(x+h): [f(x+h)-f(x)]/[(x+h)-x]. Letting h be smaller and smaller, this slope may tend to a specific value--if it does, that is the value of the derivative f'(x). Or, in limit notation: f'(x) = lim_{h->0}[ (f(x+h)-f(x))/h ]. Let's try computing the derivative of f(t) = t². Now, [(t+h)²-t²]/h = [t²+2th+h²-t²]/h = 2t+h. As h gets smaller and smaller, 2t+h tends toward 2t. Thus, the derivative of t² is 2t. Consequently, the integral of 2t must be t²+A, for some arbitrary constant A (one can verify easily that the derivative of t²+A is 2t), so the integral of t must be t²/2, up to a constant.

In class, we had a formula in which it was for the derivative of 't', it is (btw I don't know how to insert that vertical line) vdt= (v+at)dt= vt+(1/2)at^2 ... Also in the second part, v turns into (v+at) because of dt correct? That it has to assume the same value but it must include time, right?

Since velocity is the derivative (rate of change) of position, v(t) = x'(t), position would be the integral of velocity: x(t) = Int[ v(t) dt ]. The above is actually incorrect as stated, but it would make sense if the v's do not refer to the same thing--in particular, if the velocity itself is changing at a constant rate, then v(t) = v(0) + at, where v(0) is the initial velocity, and so x(t) = Int[ v(t) dt ] = Int[ (v(0)+at) dt ] = v(0)t+at²/2+x(0), where x(0) is the initial position.

[SCRawl]

Oy, then I guess I mixed the terms up. I didn't quite well understand it in class and as I said I never have taken calculus before.

My high school physics class had no calculus in it whatsover. Our AP Physics class was B which didn't have any either.

Quick tutorial.

A derivative of an equation is taking the slope of the tangent. In other words, for, say, y=x^2 (accept this for the moment) the derivative of y with respect to x (denoted as dy/dx) is 2x. (This is also called y'=2x, with the ' or "prime" denoting the first derivative of y with respect to x.) If you take a look at any point on that curve (for example, where x=0) then the slope of the tangent can be determined by taking the derivative and plugging in the value for x. That isn't the most common use for the derivative, and it's easy to lose sight of the fact that that's what it is: simply the slope of the tangent.

An integral of an equation takes the area under a curve, which is effectively the antiderivative. Again, if we take that y=x^2, then the antiderivative is (x^3)/3 + c (where "c" is a constant, which becomes clear once you learn how to differentiate). To obtain the area under the curve you have to take two points and resolve this expression.

I'm sure that others here can explain this better than I can. Alternatively, you can google about a million tutorials about how to learn what you need. I found one here that looks pretty good, but he's hardly the only one.

[Dangermouse]

SCRawl pretty much answered it for you. Heres a little bit more detail.

These equations are the equations of kinematics. One major assumption for this derivation is that accelaration must be constant.

1. Start with acceleration as the derivative of velocity with respect to time
a = dv / dt.

2. Multiple both sides by dt.
a dt = dv

2. Integrate both sides
a t +C = v

C is a constant and will be the initial velocity Vo.

at +Vo = v
or
v=Vo + at

3. Note that v is really dx / dt (change of position with respect to time)

v=dx/dt=Vo+at

Integrate again with respect to t
x = Vot+ at^2 * 1/2 +B

So the 1/2 comes from integrating the at. The B is the initial position. The final equation should be x = Vot+at^2*1/2 + Xo. Its a double integration problem.

Hereas how your teacher did it (I am guessing):

v*dt
v=final velocity= initial velocity + at
v = vo+at, Sub in above
vdt=(vo+at)dt
v=vo+at
v=dx/dt
dx/dt=vo+at

Integrate both sides
x=vo t + at^2*1/2+xo

I am not sure why your teacher did not include the constant xo (unless they are assuming starting from a zero position.

[SCRawl]

Ghetto edit: the page I provided has a number of "cheat sheets", which is not really a how-to. Kuroneko clearly has a better handle on this stuff (and other stuff, I've observed) than I do, and his first-principles explanation is about as dumbed-down as it gets. I mean that in a good way.

[Soontir C'boath]

Wow. All of you were very, very helpful. I understand clearly now but I want to clarify this.

v=dx/dt=Vo+at

Integrate again with respect to t
x = Vot+ at^2 * 1/2 +B

From what I see here, this should look like,

x = Int[(Vo+at)dt]+B

Correct?

Would I then say that since the coefficient of Vo is one and in which when t is integrated with it, it would only equal to one too because of it?
I mean, it's the reason why it is just Vot?

[Dangermouse]

Wow. All of you were very, very helpful. I understand clearly now but I want to clarify this.

v=dx/dt=Vo+at

Integrate again with respect to t
x = Vot+ at^2 * 1/2 +B

From what I see here, this should look like,

x = Int[(Vo+at)dt]+B

Correct?

Yes, this is right. The b comes from the integration as well.

Would I then say that since the coefficient of Vo is one and in which when t is integrated with it, it would only equal to one too because of it?

Sort of. Vo is a constant, not a variable. Anytime you integrate a constant you get the variable you are integrating with respect to. Think of it like this:

Int[dt] = t

5*Int[dt] = 5t

Int[5*dt] = 5t

You can always move a constant to outside the integration operation. While the coefficient of Vo is one, it is not affected by the integration because Vo has no dependence on t. Your initial velocity will never change with time; it is just a number. So, when you integrate Vo dt, you will get Vo t..

On the other hand, a t will also have a coefficient of one. However,
Int[t dt ] will become t^2 / 2 since t is the variable you are integrating with respect to.

[Kuroneko]

From what I see here, this should look like,

x = Int[(Vo+at)dt]+B

Correct?
Would I then say that since the coefficient of Vo is one and in which when t is integrated with it, it would only equal to one too because of it?
I mean, it's the reason why it is just Vot?

Yes, but they are equal. In general, Int[ (f(t)+g(t))dt ] = Int[ f(t)dt ] + Int[ g(t)dt ] and Int[ c*h(t)dt ] = c*Int[ h(t)dt ] for any constant c, so one can split things up with impunity: Int[(Vo+at)dt] = Vo*Int[1 dt] + a*Int[t dt] = Vo*t + at²/2 + B. The latter integral has already been shown; as for the former, ask yourself: what kind of function of t has a constant rate of change? The answer is, somewhat obviously, a line--in this case, with slope 1, i.e., t+C, for any constant C. In practice, it is not necessary to introduce constants for every term, as the sum of any number of arbitrary constant is still an arbitrary constant. As in:
Vo*Int[1 dt] + a*Int[t dt] = Vo[t+C] + a[t²/2+D] = Vot + at²/2 + (VoC+aD),
and one may simply rename (VoC+aD) to be "B", since all of the terms involved in it are constants. There is therefore no harm in having a single constant term from the start when dealing with a sum of multiple integrals. When you gain a bit of experience, it will not be necessary to split up the integral in the first place, and evaluating such simple integrals will become fairly obvious just by inspection.

[Soontir C'boath]

That really cleared things up. Thanks for all the help.

It feels damn good to understand!

[Academia Nut]

Yeah, once you understand more about calculus you really start to see a lot of relationships between physics formulas that you just memorize at first (or at least that's how it works at my high school). Suddenly you realize that all those time, distance, velocity, and acceleration formulas all tie into one another quite nicely. Even better, when you take the derivative of kinetic energy (E=.5m*v^2) you end up with momentum (p=mv). It's kind of weird when something clicks like that, although if it happens during English class you might annoy your teacher .

I can't wait for university to start so I can take even more physics and calculus.

[sketerpot]

Yeah, once you understand more about calculus you really start to see a lot of relationships between physics formulas that you just memorize at first (or at least that's how it works at my high school).

My favorite is Einstein's formula for kinetic energy. If you find the Taylor series expansion, the first term is the classic energy = 1/2mv^2 that you learn in Physics.

By the way, Soontir C'boath: if you just understand what integration is, you can figure out what's happening when somebody else does it. But if you ever have to use it yourself in this class, you're screwed unless you get studying.

[Dangermouse]

My favorite is Einstein's formula for kinetic energy. If you find the Taylor series expansion, the first term is the classic energy = 1/2mv^2 that you learn in Physics.

Definitely. Learning how to use and apply Taylor series is one of the most useful things I learned in school. Its one of the more useful techniques, especially when you want to
linearize and simplify expressions. I think I was using a taylor series expansion nearly every day in acoustics.

[Soontir C'boath]

By the way, Soontir C'boath: if you just understand what integration is, you can figure out what's happening when somebody else does it. But if you ever have to use it yourself in this class, you're screwed unless you get studying.

I definitely will as I am locked in my physics class as of today, I believe.

[sketerpot]

By the way, Soontir C'boath: if you just understand what integration is, you can figure out what's happening when somebody else does it. But if you ever have to use it yourself in this class, you're screwed unless you get studying.

I definitely will as I am locked in my physics class as of today, I believe.

If you use it responsibly to avoid being totally screwed by your lack of calculus (and you plan to rectify this lack of calculus later), I recommend getting a computer algebra system and learning how to use it. For a computer, Maxima is a decent free one, and straightforward to use. Some calculators have a CAS built in, like the TI-89 and TI-92. That way you can focus your early studies on understanding what's going on and fill in your grasp of the mechanics of integration later.

[Kuroneko]

I would recommend against gaining dependence on such a thing; it is not honest. If nothing else, it won't save you on the test either. Get a calculus textbook quickly, consider it one of your classes, and study. If, after some effort, you still can't get through something, well, there is a number of competent people willing to help here, as long as you show evidence of your own effort. I doubt that the moderators here will mind an occasional question (perhaps it would be best to keep them in a single thread, however).