I need some help here - I don't get what's going on :)

Step 1:

Step 2:

What's happening between the two steps? It would be nice with an extra step or two, as this is very hard for me..

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- Dec 16th 2007, 03:35 PMRydbirkSome differential calculus..
I need some help here - I don't get what's going on :)

Step 1:

Step 2:

What's happening between the two steps? It would be nice with an extra step or two, as this is very hard for me.. - Dec 16th 2007, 04:51 PMTwistedOne151Partial fractions:
Rydbirk,

On the right hand side, the integration is simple, with A being the constant of integration, so I imagine you're asking about what they did on the left hand side.

First, they used the fact that to get

Next, you use the method of partial fractions:

for some A, B.

To find A and B, we multiply this by to get:

Equating the coefficients of powers of v, we get and

This gives and . Thus:

Now, note that , so ,

,

and

Then you use the fact that to get

which is the left hand side of their result.

--Kevin C. - Dec 16th 2007, 05:00 PMKrizalid
C'mon, do we really need partial fractions?

This is like integrating

The idea is the same. - Dec 17th 2007, 03:27 AMRydbirkHmm..
TwistedOne151:

Equating the coefficients of powers of v.. What do you mean by that? Can you reword it? :) - Dec 17th 2007, 03:57 AMtopsquark
:mad: First off, I would like to say that it would be much better to have posted this question in the original thread. If my explanation there was not good enough it would have been far more polite for you to have said so there, not to mention better for anyone else who had a similar question.

Now, as to your question:

If we have an expression, say

that is supposed to be true for all values of v, then we can say that A = 0 and B = 1. This is the only possible solution for A and B.

This technique generalizes. Suppose instead we have something like

that is true for all v. (Notice that this condition is critical!) Then we may say that

-Dan - Dec 17th 2007, 04:36 AMRydbirk
topsquark

I'm sorry, that was kind of rude of me. Your explanation was simply confusing to me :( I'll make a link to this thread in the first thread.

It's even more embarrising that your new explanation is so good :o Thanks, now I truly get it! - Dec 17th 2007, 04:39 AMtopsquark
- Dec 18th 2007, 12:05 PMRydbirkOh **** ..
http://www.mathhelpforum.com/math-he...-calculus.html

C=0,0067 and g = 9,82, and I guess A=0 (or what???)

What is wrong? This function is decreasing, but it's a velocity-function, it should be increasing!!

Help, it's for thursday and a big subject!! :( - Dec 18th 2007, 03:52 PMRydbirkPhew!
I found the problem:

My calculator gives me

instead of

which leads to the result

And that result I can use, it gives me the correct data. But why, what went wrong? Please help :) - Dec 18th 2007, 05:58 PMtopsquark
Okay, I found a bit of a problem in my solution. Let's go back to

I made a slight error in integrating this. It should be a Riemann integral with the intial condition v(0) = 0:

Giving:

In other words, the A vanishes.

So

Now, the problem I am having is that v is not defined at t = 0! But v(0) = 0 has already been applied to the Riemann integral. I have no explanation of why this is happening. (Doh)

-Dan - Dec 19th 2007, 03:35 AMRydbirkThanks Dan
That's one bit.

Have you looked at my reply just before yours? :)