fandomsecrets | [ SECRET POST #2317 ]

⌈ Secret Post #2317 ⌋

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Notes:

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Flat | Top-Level Comments Only

Hi! I'm logged in now. Take a look at this page (http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/usubdirectory/) and start reading the section from the words "Now the method of u-substitution will be illustrated on this same example. Begin with"

The sentence "Now "pretend" that the differentiation notation" (the 4th sentence down from "Now the method of u-substitution...") is the part you're getting stuck at? Hm. Maybe if I wrote your example this way it'll help you understand.

Let Y = 1+X
dY/dx = 1
(dY/dx)*(dx) = (1)*(dx)
(dY)*(dx/dx) = (1)*(dx)
dY = 1 dx (because dx/dx = 1 )

Do you see how I got from the 2nd equation (dY/dx = 1) to the last equation (dY = 1 dx)?

Now instead of Y, we say U

Let U = 1+X
dU/dx = 1
(dU/dx)*(dx) = (1)*(dx)
(dU)*(dx/dx) = (1)*(dx)
dU = 1 dx

Are you okay with this so far?

Question, how did you get (dx) to (dx/dx)? Wouldn't (dx/dx)=1 and =/= (dx)?

The first (dx) you are talking about, which equation and which side of the equation are you talking about?

If you're talking about the (dx) I multiplied in, the *(dx), I multipled (dx) to the left side and the right side of the equation. It wasn't a random addition.

Let U = 1+X
dU/dx = 1
(dU/dx)*(dx) = (1)*(dx)
(dU)*(dx/dx) = (1)*(dx)
dU = 1 dx

That one. Is it because you had (^dU/_{dX_{)(dX), and you pulled the dx from ^dU/_{dX_over?}}}

Wow, htmfail, sorry. :(

Yep! Do you remember the associative property of multiplication? It's the one that says a multiplication equation has the same product no matter what order its factors are in? A*(B*C)=(A*C)*B=B*(C*A)=etc.?

Division is essentially multiplication written in the form of (1/something). Let's take a look at just the left side of equation 3, (dU/dx)*(dx)

(dU/dx)*(dx)
=(dU)*(1/dx)*(dx)
=(dU)*(dx)*(1/dx)
=(dU)*(dx/dx)

:D?

I won't lie, I had forgotten that. |D

Yes, that makes sense now, thank you~! And I understood everything in your post with that clarification. :3

You're welcome! I see dreemyweird and another nonnie have been helping you out too. Is there anything else about u-substitution you want clarified?

(Since you're working with definite integrals where x has upper and lower bound limits, don't forget to change the limits to upper and lower bounds of U when you're integrating f(u) instead of f(x) ;P)

So, I change the bounds by plugging the upper and lower bounds into the U-formula, right?

Yes. Sometimes equations try to trick you by having the upper and lower bounds go from positive to negative (like 2 to -8) in terms of X but negative to positive (like -3 to 10) in terms of U. Lable your variables and don't let that confuse you!

Kay.

Um. Where do I plug in the bounds again? ^^;

The U = Xsomething equation.

U=Xsomething
U'=X'something
U" = X"something
Uupper = Xuppersomethingsomething
Ulower = somethingXlowerXlower*somethingoranother

It doesn't matter how you lable your upper and lower bound equations as long as you're consistant. It may be one more step, but I re-write the upper and lower bound equations in terms of U and x before substituting the numerical values. For example, if the equation is U = X^2 + 3X - 1 and the upper limit of X is 9 and the lower limit of X is (-1), I'd write something like:
U = X^2 + 3X - 1, X" = 9, X' = (-1)

U" = X"^2 + 3X" - 1
= (9)^2 + 3(9) - 1
=81+27-1
=107

U' = X'^2 + 3X' - 1
=(-1)^2 + 3(-1) - 1
=1 - 3 - 1
=(-3)
Notice how I re-wrote the U = xsomething equation in terms of the upper and lower bounds of U and x instead of writing U (upper/lower bound) = eqn2 even though I knew which X value went with bound? It's one more step but it helps other people to follow what you're doing. It's also easier to find mistakes when it's easy to tell what you're trying to do by looking at the equations.

I'm not completely sure I understand, but I think I'll be able to get it if I go through some problems I know the answer to, and see if I get the right answers, thank you so much!

Uh, I just realized I should have used subscripts to denote upper and lower bounds. What if I write it like this?

X_upper = 9 (((X_upper is a variation of the variable X I'm using to denotes the upper bound of X. Don't use superscripts like ' or " - which I did, which is BAD! - or any superscripted numbers. Use a subscript number or letter instead!)))

U = X² + 3X - 1
(((Rewrite using U_upper and X_upper so I'm clear which number I want to substitue in and other people can see why I used the numbers I did)))
U_upper = (X_upper)² + 3*(X_upper) - 1
U_upper = (9)^2 + 3*(9) - 1
U_upper = 81 + 27 - 1 = 107

Therefore the upper bound of U is 107.

Better?

Edited (html fail -_-) 2013-05-08 01:13 (UTC)

Yes! I thought you were taking the derivatives of x, and I was slightly confused. ^^;

Yeah, sorry, that was completely my fault. If only I can erase my previous comment and substitute it with the clearer version. Brain fart, my bad!

If you've any more questions about u-substitution or other aspects of calculs, please don't feel too shy to ask! I'm sure someone will answer your questions. If not, I'll take a look - if you don't mind some waiting! (I have to be asleep in 30 minutes and I probably won't be back until after 0100 UTC May 9, 2013.) Though I will be check back on this thread for the next 3 days or so. :)

I think I'm good on this thread, but I will post in other GC comments in the future! You all explained things a lot more clearly then my teachers did. (To be fair, they have so many students, and so much material to get through.)

Too many students at different knowledge levels is an issue that teachers can't solve easily. Sometimes it isn't really the teacher being at fault because they are so used to students not understanding a topic they're talking about NOW, they forget it might be a much more basic concept that's causing students problems. You can try fixing the 11th wheel in the machine as much as you want, but the machine won't work if the 1st wheel is missing! (I don't know if this analogy works, haha...)

Good luck!

[ SECRET POST #2317 ]

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