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☖ Forums → Discussion Forum → Bit of maths help
Bit of maths help
2005-01-11, 3:46 PM #1
I just need to know the dy/dx of y=xe^(a(1-x)). I did it by expanding the exponential into two parts and just going from there, my friend used dy/dx = udv +vdu and got a whole different answer. I'm not sure what it should be, but the two answers I've tried I know are wrong.
<spe> maevie - proving dykes can't fly

<Dor> You're levelling up and gaining more polys!
2005-01-11, 3:49 PM #2
...
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2005-01-11, 3:55 PM #3
When you say you expanded the exponential into two parts, what exactly do you mean? You distributed the a(1-x) so you have y=x * e^(a-ax)? From there, you could rewrite the e^(a-ax) as a fraction, and use quotient rule from there, I think...

[Edit: Another way to do it is to move x onto the other side of the equation, so you have (y/x) = e^(a-ax), then rewrite it using ln and differentiate from there. You end up with something looking like (dy/dx)/y = ... at one point, but that's no problem since you have an expression for y already.]
2005-01-11, 4:09 PM #4
I get this. (will post work shortly; gotta scan)

dy/dx = (e^(a-ax)) + (x)(-ae^(a-ax))

cleaned up a bit:

dy/dx = (e^(a-ax))(1-ax)
May the mass times acceleration be with you.
2005-01-11, 4:35 PM #5
And the work. (Part of the image was cut off in scannning, so I reproduced it, crappily, from the original not-as-neat writing of the problem)

Did I do it right?
May the mass times acceleration be with you.
2005-01-11, 5:22 PM #6
I believe this is the answer:

dy/dx = (e^(a-ax)) - ax(e^(a-ax))

I could type of the steps but it would be terribly messy to
read...

Edit-

Slaw: It should be
u = x
v = e^(a-ax)

maevie:
http://en.wikipedia.org/wiki/Product_rule
2005-01-11, 7:16 PM #7
Um, no. Look again. All the stuff on the right, I was getting the derivative of e^(a-ax).

1 d2 + 2 d1 is the derivative of y in this case.
On the wikipedia site, they just used different names than I did. Below are the equivalents of what I used:
Code: 
Mine | Wiki
  1  =   u
  2  =   v
  d1 =   du/dx
  d2 =   dv/dx

And the corresponding values obviously are:
1 = u = x
2 = v = e^(a-ax)

We did get the same answer you know...
May the mass times acceleration be with you.
2005-01-12, 5:45 AM #8
Oops. Hah. I was in a rush. Sorry bout that.
2005-01-12, 6:01 AM #9
cheers guys. after going over it with my friend, we realised what each of us had done wrong, and now it seems we're all in agreement. I was going to post a pretty picture of the graph that this information has contributed to, but I can't be arsed.
<spe> maevie - proving dykes can't fly

<Dor> You're levelling up and gaining more polys!
2005-01-12, 6:23 AM #10
please do!
"The trouble with the world is that the stupid are cocksure and the intelligent are full of doubt. " - Bertrand Russell
The Triumph of Stupidity in Mortals and Others 1931-1935
2005-01-12, 6:43 AM #11
O_O

And I thought I had it bad...
2005-01-12, 7:08 AM #12
Oh, bloody hell.
Hey, Blue? I'm loving the things you do. From the very first time, the fight you fight for will always be mine.
2005-01-13, 11:34 AM #13
well I had to take a pic for the report, so now you can have it, the differential contributed to the Lyapunov exponent (bottom graph)
<spe> maevie - proving dykes can't fly

<Dor> You're levelling up and gaining more polys!
2005-01-13, 4:57 PM #14
:eek:

/as subject matter goes flying on by over head
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2005-01-13, 5:14 PM #15
Too . . . much . . . math . . . :eek:

/splodes
And when the moment is right, I'm gonna fly a kite.
2005-01-13, 5:34 PM #16
Quote:
Originally posted by maevie
well I had to take a pic for the report, so now you can have it, the differential contributed to the Lyapunov exponent (bottom graph)

You'll have to explain these us non math majors here.
Code to the left of him, code to the right of him, code in front of him compil'd and thundered. Programm'd at with shot and $SHELL. Boldly he typed and well. Into the jaws of C. Into the mouth of PERL. Debug'd the 0x258.

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