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xddxogm3
08-30-2005, 04:17 PM
I have a question.
I'm doing a cube.
(a+b)^3
the book says it should be the following.
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

but one of my b variable is delta

(1) unsure how to make delta on the post.
so we will use this to resemble delta :eek:

so the actual problem is this.

(x+:eek:x)^3

when cubing this, would i square and cube delta?

x^3+3:eek:x^3+3x^3:eek:^2+:eek:^3x^3

valis
08-30-2005, 04:27 PM
yes you do, dx is simply (x + a bit), treat it just like a normal variable

Zach L.
08-30-2005, 06:07 PM
It is a normal variable. What you end up finding out, though, is that the higher order terms go much more quickly to zero than the lower order terms, making the higher order terms negligible.

That is, (dx)^2 goes much more quickly to 0 than dx.

Rashakil Fol
08-30-2005, 09:14 PM
Erm, there's a huge difference between 'dx' and 'delta x', so it's best not to confuse the two. The latter is usually used as an actual quantity, the difference between two distinct x values. The former is a differential, a different beast that when treated like an actual variable can cause incorrectness, unless you happen to correctly treat it as an actual value.

(x+:eek:x)^3

when cubing this, would i square and cube delta?

x^3+3:eek:x^3+3x^3:eek:^2+:eek:^3x^3

Unless the smilies went out of control, that looks horribly wrong.

You don't square and cube delta alone -- delta is just part of the variable name "deltax" or "Δx" (I wrote two versions because maybe the Unicode doesn't work for everybody.) The delta character is part of the variable name the same way that you can use multiple letters in variable names in C. In other words,

(x + deltax)^3 = (x^2 + x*deltax + deltax^2)(x + deltax) =
x^3 + 3*(x^2)*deltax + 3*x*((deltax)^2) + deltax^3

And let's see if the Unicode works:
(x + Δx)^3 = (x^2 + x*Δx + Δx^2)(x + Δx) =
x^3 + 3*(x^2)*Δx + 3*x*((Δx)^2) + Δx^3.

So, any time you see "Δx" or "Δy" etc, it always means the same thing as "(Δx)" or "(Δy)". So (Δx)^3 is just a slightly more verbose way of saying Δx^3 -- they mean the same thing, since Δx is a single variable.

Generally speaking, delta represents "change in", so one might want to read Δx as "change in x." Though "delta x" works too. They're the same amount of syllables :-)

Zach L.
08-31-2005, 06:02 AM
Eh... most of the books I've encountered lately used 'little delta x' as the differential; that's what I was speaking of. But yes, you are correct, there is a difference between the differential and the 'change in'. My mistake.

xddxogm3
09-05-2005, 08:39 PM
so if you have delta and x combined as in deltax, it can not be seperated for further calculations? that makes since. I just wanted to make sure i couldn't take the x away from the delta when performing these calculations. so if i was factoring i would have to factor out a deltax and not just the x in the deltax.

Thantos
09-05-2005, 09:11 PM
When in doubt use one of the most powerful tools avialable to you: Let

As in: Let y = Δx
And then redo the equation.

xddxogm3
09-06-2005, 08:54 PM
I have a problem that i'm trying to work.

lim [2(x+Δx) - 2x]/Δx = (2x+2Δx-2x)/Δx = 2Δx/Δx
Δx->0

The book answer to the limit is 2, but i get 0 due to i go from

2Δx/Δx = Δx

why is this incorrect?

substituting y does the same thing for me.
y = Δx
lim [2(x+y) - 2x]/y = (2x+2y-2x)/y = 2y/y = 2y/y = y
y->0

Dweia
09-06-2005, 09:05 PM
(2*3)/3 = 2.

(2*y)/y = 2

(2*Δx)/Δx = 2

xddxogm3
09-06-2005, 09:11 PM
:o ok, i'm blind.
opps.
is there a list of operations you can do with Δx
similar to an order of precedence/operations list