Hey everybody,

I'm doing a project for a biomechanics class, and I've been stuck on trying to integrate this one equation! To say the least, it's been driving me crazy. My brother helped me to get to an answer, although we are both unsure if it is correct, if someone could so kindly check it out and let me know if you get the same results I would GREATLY appreciate it. Thanks so much.

Here is the question:

http://image.pbase.com/u42/mtech/medium/27611600.Dscf0019.jpg [Broken]

And this is what I got as my solution:

http://genji.image.pbase.com/u42/mtech/medium/27611611.Dscf0020.jpg

You will probably have to copy and paste those links if they don't work by clicking.

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## Answers and Replies

I can't get those links to work, tried copy/paste too.

Thanks Braumin, original links are fixed, if you right click and "copy shortcut" and then paste, it should work

Um..I still can't get the link. Its forbidden for some reason.

- Harsh

The equation is

$$\int_0^h \,dy = -\int_{\sqrt{2gD}}^0 \frac{v}{g + \frac{KA}{2m}v^2} \, dv$$

$$y = \frac{m}{KA} \ln{(g + \frac{KAgD}{m})}$$

Which, at least, Mathematica doesn't agree with.

$$\sqrt{\frac{2m}{A g K}} \arctan{\left(\sqrt{\frac{ADgK}{gm}\right)}$$

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ahrkron
Staff Emeritus
Gold Member
Not to nitpick, but the RHS should be h instead of y.

Hm...

I think the reason mathematica isn't agreeing is because you have the V in the numerator different from the v in the denominator. This would explain the arctan, as the "dx/(1+u^2)" structure can be seen in his problem, if the v(V?)'s are different. I <i>almost</i> got what you got, except for one tiny thing. it's -m/KA[ <b>ln|g|</b> - all that other stuff] (because there is a zero in the integrand, when you plug in zero for g, you'll get ln|g + 0|... do you see that?). But besides that, excellent work.

Ah, the joys of handwriting. Sorry about that.

Mathematica now gives:

$$\frac{m\,\left( -\log (g\,m) + \log (g\,\left( A\,D\,K + m \right) ) \right) }{A\,K}$$

I just wanted to see if Mathematica's TeX output worked, actually...

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Hmm... I did it by hand (subsitution, t = g + 1/2 kA/m v^2) and got this:

$$y = -\frac{m}{kA}ln(1 + \frac{kAD}{m})$$

Which is the same as mathematica returned, but with an extra minus sign. Is it an error on my behalf?

Wow, yeah, I've been getting a few different answers from people. A few others have gotten what my brother and I ended up with. But I went to school today and plugged it into mathematica and came up with the same as cookiemonsters second solution. Oh and ahrkon, you are right, I made a mistake while copying it out. Thanks for the correction.

I'd guess Mathematica's right. I've never seen it wrong before.

Did you remember the minus sign when differentiating t?

philosophking said:
I think the reason mathematica isn't agreeing is because you have the V in the numerator different from the v in the denominator. This would explain the arctan, as the "dx/(1+u^2)" structure can be seen in his problem, if the v(V?)'s are different. I <i>almost</i> got what you got, except for one tiny thing. it's -m/KA[ <b>ln|g|</b> - all that other stuff] (because there is a zero in the integrand, when you plug in zero for g, you'll get ln|g + 0|... do you see that?). But besides that, excellent work.

Hmm, yeah, that makes sense. Thanks philosophking :)

I'd guess Mathematica's right. I've never seen it wrong before.

Did you remember the minus sign when differentiating t?

Yeah, I did. I've never used mathematica before, and I actually got my friend to enter it in today between classes. I'm pretty much between the mathematica answer and my original answer. If mathematica is as reliable as you say it is, I will go with it. I've tried using both answers in my modeling project and they both produce reasonable figures.

Oh, by the way, I was told that when mathematica puts out "log" it means "ln"...is this correct?

Yes, log = ln. Get used to seeing it that way. It happens more and more often the higer the level classes you're taking and it's standard for mathematical references like CRC or Mathematica. I'm not sure if Maple does it, but I imagine it does as well.

philosophking was mentioning something about that negative. Maybe you should read his post?

Yep, I already replied to his post above. I took a look at it, and I do believe he is correct about the negative, at least it makes sense to me. Although, for now I will model using the mathematica equation and get it looked over by my prof.

Thanks again everyone, much appreciated

Cheers

hehe...

I'm glad I could be of help!