Question

(Analytical Mechanics-Calculus&ODE Focused) I'm having trouble understanding part of my book, will post below shortly, but I just don't understand how what it's saying is right, involving substitutions and a derivative.

The Physics is irrelevant, my bone to pick is with the math statements.

https://s3.amazonaws.com/pushbullet-uploads/ujvx2jieYTs-jwsWhCEHrn6EIMQmUIKzqswvIPV4Ksos/IMG_20150120_232225.jpg

Answer 1

"So far, we have discussed the terminal speed of a projectile (moving vertically) but we must now discuss how the projectile approaches that speed. This is determined by the equation of motion (2.25) which we can rewrite as

\[m \dot{v}_{y}=-b(v_y -v_{ter})\]

(Remember that v_ter = mg/b.) This differential equation can be solved in several ways. Perhaps the simplest is to note that it is almost the same as Equation (2.15) for the horizontal motion, except that on the right we now have (v_y - v_ter) instead of v_x. The solution for the horizontal case was the exponential function (2.20). The trick to solving our new vertical equation (2.29) is to introduce the new variable u = (v_y - v_ter), which satisfies mu(dot) = ibu (because v_ter is constant). Since this is exactly the same as equation (2.15) for the horizontal motion, the solution for u is the same exponential , u = Ae^(-t/tau). [Remember that the constant k in (2.20) became k = 1/tau.] Therefore.

\[v_{y}-v_{ter}=Ae^{-t/\tau}.\]
When t=0, v_y = v_y0, so A = v_y0 - v_ter and our final solution for v_y as a function of t is

\[v_{y}(t)=v_{ter}+(v_{y0}-v_{ter})e^{-t/\tau}\]\[=v_{y0}e^{-t/\tau}+v_{ter}(1-e^{-t/\tau}).\]

Answer 2

My question here isn't in the Physics, it's in the implied math, which, in three different ways, makes absolutely zero sense to me. First thing is the substitution, u. I'll elaborate on why I don't get that in one sec.

Answer 3

Welp, there goes everything I wrote.

What I don't understand is how the statement \[\dot{u}=-bu\]
is true for \[u = (v_y - v_{ter})\]
Where v_ter is a constant. If I differentiate that whole expression with respect to t, I'll get \[\dot{u}=\frac{dv_{y}}{dt}\]Correct? How the heck is the derivative of u in this instance equal to mu times a constant?

Answer 4

errata: supposed to be u, not mu, in that paragraph, and the i in the expression should be a negative sign.

Answer 5

https://s3.amazonaws.com/pushbullet-uploads/ujvx2jieYTs-jwsWhCEHrn6EIMQmUIKzqswvIPV4Ksos/IMG_20150120_232225.jpg

Answer 6

Also free of errors and better to read IMO, sadly, than my failed attempt at a LaTeX writeup, lol.

Answer 7

I don't understand how it is hard to get?

Answer 8

The derivative of u, from what I'm reading, does not equal u times a constant.

Answer 9

\[m \dot{v}_{y}=-b(v_y -v_{ter})\]

plugin \(u = v_y -v_{ter}\) and \(\dot{v}_{y}=\dot{u}\), you get :

\[m \dot{u}=-b(u)\]

Answer 10

You guys are actually doing the differentiation in your mind, not just moving around the symbols, right?

\[u = (v_y - v_{ter}), \ \ \ \dot{u}=\frac{d}{dt}(u) = \frac{d}{dt}(v_y - v_{ter})\]Whether m is there or not, taking the derivative of some function, minus a constant, is going to just yield a one-term answer. The original term mu has two terms, v sub y and v sub ter, taking the derivative of v sub y minus v sub ter will make v sub ter disappear since it's a constant, and you will be left with \[\frac{d}{dx}(function - constant) = \frac{d}{dx}(function)\]
The original substitution had two terms, the derivative of u would yield only one, how can the derivative of u possibly be u times some constant if it doesn't even resemble u?

Answer 11

(I'm not talking about exponential functions or anything else, my point is, the second term would disappear; if this statement were true, (It is, I just don't get stuff ever, apparently), it would look like this:

Answer 12

The derivative of \(u\) doesn't equal \(u\) times a constant. They're plugging it back into equation 2.29 as @rational mentioned.

Answer 13

\[m \dot{u}=-b(v_{y}-v_{ter})\]

Answer 14

Alright, that's what confused me, I guess....doesn't the statement \[m \dot{u} = -bu\]........That looks 100% like taking the derivative of something yields the undifferentiated term, times some constant. I don't understand how this statement, as written, isn't implying that.

Answer 15

If I write \[\frac{d}{dx}(f(x)) = c \cdot f(x)\]

......Isn't that me saying that if I take the derivative of f(x), I'll get the original function, times a constant?

Answer 16

thats what you get when something is resisting the velocity :
\[v' = -kv\]

Answer 17

As you get faster and approach terminal velocity, the air resistance increases and slowly balances out with gravity

Answer 18

"The derivative of u doesn't equal u times a constant. They're plugging it back into equation 2.29 as @rational mentioned."

Yeah, but in order for the statement to be mathematically true and not just a random rearrangement of symbols, I have to be able to take the derivative of that quantity, multiply it by m, and get the un-derived original expression for it, times some negative constant, correct? Is that true, or not?

Answer 19

PS, thank you guys. I don't understand how I manage to learn everything pretty much retricebackwards somehow from everybody else does, but thanks for trying to help.

Answer 20

Equation 2.29: \(mv_y' = -b\left(v_y - v_{ter}\right)\)
The substitution they use: \(\color{aquamarine}{u = v_y - v_{ter}}\)
Take the derivative of the substitution term: \(\color{green}{u' = v_y'}\)
Substitute back into 2.29:
\(m\color{aquamarine}{u'} = -b\color{red}{u}\)

Answer 21

Oh crap flipped the colors, I meant

Equation 2.29: \(mv_y' = -b\left(v_y - v_{ter}\right)\)
The substitution they use: \(\color{red}{u = v_y - v_{ter}}\)
Take the derivative of the substitution term: \(\color{aquamarine}{u' = v_y'}\)
Substitute back into 2.29:
\(m\color{aquamarine}{u'} = -b\color{red}{u}\)

Answer 22

\[
(e^{kt}-C)' = ke^{kt} = k([e^{kt}-C]+C)
\]So \[
\dot v = k(v+C)
\]

Answer 23

Alright, I got it, at least from the way DisplayError said it.

Wio, I get the literal Physics behind it, but I didn't understand the math until someone illustrated what was going on in a different way.

Answer 24

Thanks, anyways, guys, I was looking at that for far too long trying to figure out what the hell I couldn't get.

Answer 25

That was the first thing I didn't get, in any case. The second and third thing, still taking a look at them after learning that to see if I get them.

Answer 26

Alright, got it, figured out the stuff with the exponential terms and stuff that happens later on. Thanks.