Some differential calculus..

**Rydbirk** · December 16th 2007, 03:35 PM

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

Step 1:
$ \int \frac{dv}{Cv^2 - g} = \int~dt $

Step 2:
$\frac{1}{2\sqrt{Cg}}~ln \left ( \frac{v\sqrt{C} - \sqrt{g}}{v\sqrt{C} + \sqrt{g}} \right ) = t + A$

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

**TwistedOne151** · December 16th 2007, 04:51 PM

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 $Cv^2-g=(\sqrt{C}v)^2-(\sqrt{g})^2$ to get

$\int \frac{dv}{Cv^2 - g} = \int \frac{dv}{(\sqrt{C}v)^2-(\sqrt{g})^2}=\int \frac{dv}{(\sqrt{C}v-\sqrt{g})(\sqrt{C}v+\sqrt{g})}$

Next, you use the method of partial fractions:
$\frac{1}{(\sqrt{C}v-\sqrt{g})(\sqrt{C}v+\sqrt{g})}=\frac{A}{(\sqrt{C}v-\sqrt{g})}+\frac{B}{(\sqrt{C}v+\sqrt{g})}$ for some A, B.
To find A and B, we multiply this by $(\sqrt{C}v-\sqrt{g})(\sqrt{C}v+\sqrt{g})$ to get:
$1=(\sqrt{C}v+\sqrt{g})A+(\sqrt{C}v-\sqrt{g})B$
$1=(A+B)\sqrt{C}v+(A-B)\sqrt{g}$
Equating the coefficients of powers of v, we get $A+B=0$ and $(A-B)\sqrt{g}=1$
This gives $A=\frac{1}{2\sqrt{g}}$ and $B=-\frac{1}{2\sqrt{g}}$ . Thus:

$\int \frac{dv}{Cv^2 - g} = \frac{1}{2\sqrt{g}}\int \frac{dv}{\sqrt{C}v-\sqrt{g}} - \frac{1}{2\sqrt{g}}\int \frac{dv}{\sqrt{C}v+\sqrt{g}}$

Now, note that $\frac{d}{dx}\left ( \ln (ax+b) \right )=\frac{a}{ax+b}$ , so $\int \frac{dv}{\sqrt{C}v-\sqrt{g}}=\frac{1}{\sqrt{C}}\ln(\sqrt{C}v-\sqrt{g})$ ,
$\int \frac{dv}{\sqrt{C}v+\sqrt{g}}=\frac{1}{\sqrt{C}}\l n(\sqrt{C}v+\sqrt{g})$ ,
and
$\int \frac{dv}{Cv^2 - g} = \frac{1}{2\sqrt{Cg}}\left ( \ln(\sqrt{C}v-\sqrt{g}) - \ln(\sqrt{C}v+\sqrt{g}) \right )$
Then you use the fact that $\ln a - \ln b = \ln \frac{a}{b}$ to get $\int \frac{dv}{Cv^2 - g} = \frac{1}{2\sqrt{Cg}}\ln \left ( \frac{\sqrt{C}v-\sqrt{g}}{\sqrt{C}v+\sqrt{g}} \right )$
which is the left hand side of their result.

--Kevin C.

**Krizalid** · December 16th 2007, 05:00 PM

C'mon, do we really need partial fractions?

This is like integrating $\frac1{x^2-1}.$

$\frac{1} {{x^2 - 1}} = \frac{{(x + 1) - (x - 1)}} {{2(x + 1)(x - 1)}} = \frac{1} {2}\left[ {\frac{1} {{x - 1}} + \frac{1} {{x + 1}}} \right].$

The idea is the same.

**Rydbirk** · December 17th 2007, 03:27 AM

TwistedOne151:
Equating the coefficients of powers of v.. What do you mean by that? Can you reword it?

**topsquark** · December 17th 2007, 03:57 AM

Originally Posted by Rydbirk

TwistedOne151:
Equating the coefficients of powers of v.. What do you mean by that? Can you reword it?

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
$Av + B = 1$
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
$(A + B)v^2 - Bv + C = v^2 - 1$
that is true for all v. (Notice that this condition is critical!) Then we may say that
$A + B = 1$
$-B = 0$
$C = -1$

-Dan

**Rydbirk** · December 17th 2007, 04:36 AM

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

Thanks, now I truly get it!

**topsquark** · December 17th 2007, 04:39 AM

Originally Posted by Rydbirk

It's even more embarrising that your new explanation is so good

Thanks, now I truly get it!

Don't worry about it, we've all been there.

-Dan

**Rydbirk** · December 18th 2007, 12:05 PM

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!!

**Rydbirk** · December 18th 2007, 03:52 PM

I found the problem:

My calculator gives me

$ \frac{1}{2\sqrt{cg}}\ln\left(\frac{\sqrt g + v\sqrt c}{\sqrt g - v\sqrt c}\right)=a-t $

instead of

$ \frac{1}{2\sqrt{cg}}\ln\left(\frac{v\sqrt c - \sqrt g}{v\sqrt c + \sqrt g }\right)=t+a $

which leads to the result

$ v(t)=\frac{-tanh(\sqrt{g}(t-a)\sqrt{c})\sqrt{g}}{\sqrt{c}} $

And that result I can use, it gives me the correct data. But why, what went wrong? Please help

**topsquark** · December 18th 2007, 05:58 PM

Originally Posted by Rydbirk

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!!

Okay, I found a bit of a problem in my solution. Let's go back to
$\int \frac{dv}{Cv^2 - g} = \int~dt$

I made a slight error in integrating this. It should be a Riemann integral with the intial condition v(0) = 0:
$\int_0^v \frac{dv}{Cv^2 - g} = \int_0^t~dt$

Giving:
$\frac{1}{2 \sqrt{Cg}}~ln \left ( \frac{v \sqrt{C} - \sqrt{g}}{v \sqrt{C} + \sqrt{g}} \right ) = t$

In other words, the A vanishes.

So
$v(t) = -\sqrt{ \frac{g}{C}}~\frac{e^{2t\sqrt{Cg}} + 1}{e^{2t\sqrt{Cg}} - 1}$

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.

-Dan

**Rydbirk** · December 19th 2007, 03:35 AM

That's one bit.
Have you looked at my reply just before yours?

Thread: Some differential calculus..

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Some differential calculus..

Partial fractions:

Hmm..

Oh **** ..

Phew!

Thanks Dan

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