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Calculus help!

Originally posted by bignickel
Thanks for your help 69dodge. I'm going thru it right now.

Got confused tho: I thought "u + du" was the dividend?
Click to expand...
Yes, sorry. I meant to say "divisor."

"v + dv" is correct.
 
One question 69dodge:

v + dv | u + du | u/v + du/v - udv/v^2
----------

I see how you got u/v and du/v, but where did - udv/v^2 come from?

u gets divided by v
du gets divided by v
where doe udv come from, and why does it get divided by v^2?
 
Originally posted by bignickel
I see how you got u/v and du/v, but where did - udv/v^2 come from?
Click to expand...
On the fifth line, the first term is (-u dv / v). Dividing that by v yields (-u dv / v<sup>2</sup>).
 
If w = at - (1/2 bt^2), find dw/dt. And explain each step.

Now, the book gives it's answer. But I can't figure it out.
a - bt.

Now, I can figure out how we lost the t in (at) and the t in the exponent of (bt^2). But where the heck did (1/2) go?
 
I think I got it.

Find the derivitive of both sides:
dw = adt - (1/6)b(t^3)dt or dt(a - (1/6)b(t^3)dt

Then divide both sides by dt to get dw/dt

dw/dt = a - (1/6)b(t^3)

I'm guessing since b is supposed to represent some variable then you can have (1/6) = b and thus getting:

dw/dt = a - b(t^3)


Edit: Doh! That isn't the answer in the book. It is impossible for me to be wrong so the book must be. ;)
 
The person who wrote the post above me is my evil stupider twin brother. This IS the real answer.

Ok we can rewrite it the equation as

w = t(a - .5bt)
The derivitive of that is:

dw = t(-.5b(dt)) + a(dt) - .5bt(dt) or dt(-.5bt + a -.5bt)
Divide both sides by dt:
dw/dt = a - bt

Remember the derivitve of (UV) is Udv + Vdu.
 
Well, I was kinda following along there, but:

where did the .5 go?


Also:
"dw = t(-.5b(dt)) + a(dt) - .5bt(dt) "

Where did the end part of '-5bt(dt)' come from?
 
bignickel said:

where did the .5 go?
Click to expand...
There is two -.5b(dt) and together they up to -b(dt).

Also:
"dw = t(-.5b(dt)) + a(dt) - .5bt(dt) "

Where did the end part of '-5bt(dt)' come from? [/B]
Click to expand...

I'm not sure what you mean.
 
epepke, very cool mnemonic

I love mnemonics , they have engrained such disparete elements like resistor color codes, the absorbsion on light in sea water, the clasification of star types, order of planets, math processes and more into this rapidly smoothing grey matter.
 
Ok we can rewrite it the equation as

w = t(a - .5bt)
The derivitive of that is:

dw = t(-.5b(dt)) + a(dt) - .5bt(dt)
Click to expand...

I got a bit lost here. You had t(a...) but when you went to write the derivative, the t only mulitplied -.5b(dt), but stopped multiplying a(dt). Plus, we have .5bt(dt) showing up out of nowhere.

I would think that the derivative of (w=t(a-.5bt) would be:
dw = t(-.5b(dt) + a(dt))

and that's it. But evedently, I'm wrong somehow.
 
bignickel said:

I would think that the derivative of (w=t(a-.5bt) would be:
dw = t(-.5b(dt) + a(dt))
Click to expand...

OK I think I understand your trouble.



The reason why you are wrong is that it is more complicated to find the derivitive of a product. You are just taking the derivitive of one of the products instead of the combined whole.

The products rule is the derivitive of UV = Udv + Udu

If we have U = t and V = (a - .5bt) then UV will equal t(a - .5bt) ,what we are trying to find the derivative of. To find the other side of the equation we just replace the variables of the equation of the products rule with their corresponding parts.

U = T
V = (a - .5bT)
du = dt
dv = -.5bdt

Replace the variables of (UV)' = Udv + Vdu and we get:
(t(a - .5bt))' = t(-.5bdt) +dt(a - .5bt)

Bringing out the dt and simplfeing:
(t(a - .5bt))' = dt(a - bt)

Of course:
(t(a - .5bt)) = W
(t(a - .5bt))' = dw

Rewriting it and dividing both sides by dt we get the answer

dw/dt = a - bt

I relize half of that probably wasn't important but I didn't want to leave out anything
 
bignickel said:
If w = at - (1/2 bt^2), find dw/dt. And explain each step.

Now, the book gives it's answer. But I can't figure it out.
a - bt.

Now, I can figure out how we lost the t in (at) and the t in the exponent of (bt^2). But where the heck did (1/2) go?
Click to expand...

The derivative of c*bt^2 with respect to t is 2cbt.

In your case, since c = 1/2, 1/2*2 = 1, so we are just left with bt.
 
Bonzo said:
This is how I figure it out when I forget the rule:

y = u/v

Take ln of both sides:

ln y = ln u - ln v

Differentiate:

y' / y = u' / u - v' / v where y' = dy/dx etc.

Multiply by y, substitute u/v for y and simplify.

To me, the manipulations are simpler using this method.
Click to expand...

Oh yeah? Well, this is the way *I* figure it out when I forget the rule.

I think of a compact manifold "M" with a nontrivial fibre bundle "B". I use the easily computed connection to define the gauge invariant derivative, and examine (almost any) non-homotopically equivalent path integrals - and voila, it just pops out... :)

On a hardly more serious note - I can never remember the trig rules, things like

sin(A+B)= blah blah

so I invariably re-derive them use exp(i*x)=cos(x)+i*sin(x). The same for the formulae like v=u+a*t for motion with constant acceleration - I always work them out by integrating v=dx/dt etc.
 
I must be seriously math impaired.

How did we get from:
"Replace the variables of (UV)' = Udv + Vdu and we get:
(t(a - .5bt))' = t(-.5bdt) +dt(a - .5bt)"

to

"Bringing out the dt and simplfeing:
(t(a - .5bt))' = dt(a - bt)"

?

What happened to t(-.5bdt) and -.5? Especially since -.5 +-.5 = 1?
 
We have the equation (t(a - .5bt)) = something, right?

Rewrite that as at - .5bt^2

The derivative of that is (a dt) - .5b(2t dt).

The factors .5 and 2 cancel each other out.
 
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