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Can't calculate partial derivative for a large equasion

Posted 8 years ago

Good day, community.

I'm new to wolfram alpha and it's principles of understanding things.

I need to take a partial derivative (and then solve that equasion relatively to that variable, but that's another question) for an equasion that big, it doesn't fit into "computing bar". I tried to use "data input" field, but WA doesn't seem to understand what I want.

I, of course, can split that equasion into two and then differentiate it, but as was mentioned earlier, I need to equate the result of partial derivativation to zero and Solve the equasion relatively to variable that the equasion was differentiated by. (In other words, I need to find an optimal value of variable).

Could you kindly help me with my problem?

P.S. Here is the equasion: 1.28((3.84F+8.96U+5600)((an/2)x+5)+(7.382B+10.6U+2976)(0.63(x^(-0.5))+0.37x)(3a(n^2)-12an+32))+3.1693.53500((0.7p)/((U^2)B))((n-2)a(n+n^2)S^2+2a(-n+n^2)S^2)(0.33x+0.67x^(-0.5))+3.1693.53500((0.77p(n^4)(S^2))/((U^2)F2)))(0.5anx+5). It has to be differentiated by x. Result needs to be equated to zero and Solved relatively to x.

POSTED BY: Anton Yakunin
3 Replies
Posted 8 years ago

As you suggest, your problem neatly divides into two parts and you can take the partial of each of those.

Adding those two results gives

Solve[1.28 (1.92 a n (1458.33+F+2.33333 U)+(32+3 a (n-4) n)(2976+7.382 B+10.6 U)(0.37-0.315/x^1.5)) +
 (0.652797 a n^5 p S^2)/(F^2 U^2)+(1.1869 p (2 a (n^2-n) S^2+a (n-2)(n+n^2) S^2)(0.33-0.335/x^1.5))/(B U^2)==0,x]

and that is small enough to fit into WolframAlpha.

Unfortunately the powers of x in the denominator seem to be too much for either WolframAlpha or Mathematica to quickly solve.

Since x seems to only appear in two places as 1/x^1.5 I try a substitution

Solve[1.28 (1.92 a n (1458.33+F+2.33333 U)+(32+3 a (n-4) n)(2976+7.382 B+10.6 U)(0.37-0.315 q)) +
 (0.652797 a n^5 p S^2)/(F^2 U^2)+(1.1869 p (2 a (n^2-n) S^2+a (n-2)(n+n^2) S^2)(0.33-0.335 q))/(B U^2)==0,q]

Unfortunately I have not been able to find a way to coax WolframAlpha into solving that, but I suspect it should be able to do that with the appropriate posing of the question.

Fortunately Mathematica is less reluctant.

In[9]:= Simplify[Solve[1.28 (1.92 a n (1458.33+F+2.33333 U)+(32+3 a (n-4) n) (2976+7.382 B+10.6 U)
   (0.37-0.315 q))+(0.652797 a n^5 p S^2)/(F^2 U^2)+(1.1869 p (2 a (n^2-n) S^2+a (n-2) (n+n^2) S^2)
   (0.33 - 0.335 q))/(B U^2) == 0, q]]

Out[9]= {{q -> (-((0.652797 a n^5 p S^2)/(F^2 U^2)) - (0.391677 a n (-4.+1. n+n^2) p S^2)/(B U^2) - 
     2.4576 a n (1458.33+F+2.33333 U)-0.4736 (32.+3. a (-4.+n) n) (2976.+7.382 B+10.6 U))/
     (-((0.397612 a n (-4.+1. n+n^2) p S^2)/(B U^2))-0.4032 (32.+3. a (-4.+n) n) (2976.+7.382 B+10.6 U))}}

I think with some fiddling you should be able to get WolframAlpha to give you the same result.

A couple of tricks that can sometimes help you if you at the limit of what WolframAlpha will accept:

WolframAlpha usually interprets x2 or c2 as x squared or c squared but does not interpret (x+y)2 as (x+y) squared. That can sometimes save you a few characters if the length of your input is slightly too long.

WolframAlpha seems to sometimes do better when variables are named x or sometimes y or z while coefficients are named single lowercase letters from the beginning of the alphabet, but watch out for it sometimes interpreting e as Euler's constant, although it seems to usually ask if that is correct.

POSTED BY: Bill Simpson
Posted 8 years ago

I tried another substitution before I saw the answer and the results are the same in the end. Thank you for your help. Really appreciate that.

POSTED BY: Anton Yakunin
Posted 8 years ago

It's very hard to read your expression, but it looks like it has the form

A x + B x^(-1/2), where (A,B) are independent of x.

If so, you want to solve 2 A - B x^(-3/2) = 0 and the answer is x = (2 A/B)^(-2/3).

If I'm wrong about the form, please substitute numerical example values for everything in your equation except x and re-post it.

p.s. I'm not a Wolfram|Alpha user, but typing "solve 2 A - B x^(-3/2) = 0 for x" worked fine. It you want to do things like collecting all the terms in your expression into A and B, that is easily done in Mathematica, but perhaps not in Wolfram|Alpha. Perhaps you should pick up a trial or home copy of Mathematica for that.

POSTED BY: Alan Lewis
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