brian bovril

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15 years, 44 days

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These are replies submitted by brian bovril

@mmcdara The other way to get a is more direct, albeit taking longer. I initially did this but thought I could get 'a' from the physical properties of the material.

DIV := diff(y(x), x, x) = a*sqrt(1 + diff(y(x), x)^2):
RV  := y(0)=2.6, y(3)=2.1:
sol := rhs(dsolve({DIV, RV}, y(x)));
L   := Int(sqrt(1+diff(sol, x)^2), x=0..3);
a   := select(is, [solve(value(L)=3.6)], positive)[];


Just used this method for the triangular sail. For L=4.33m, got a= -1.015016, but with your method +1.015016

Anyway, the sag was ~1.4m for this case

@brian bovril C is a scaling factor

@mmcdara Whats big c? "for instance, let C > 0, and denote..."

@vv Thats got it!  thanks vv.......

@Carl Love I could do ST[2] and get the result I wanted. I did give an upvote if it's any consolation.

@tomleslie I like this one liner, but can the input be stripped off its apostrophes?
2 instead of



@mmcdara Thanks.

edited: using numeral 2 rather than name ‘2’ using Carl Loves Table:

rhs(ST[ListTools:-Search(2, lhs~(ST))]):-mu; #doesnt work.

@Carl Love My actual function is a product of two Gaussians. Thus no closed-form solution afaik. For the sake of brevity, I supplied the above expression. To avoid this palava, it's a pity sum doesn't have an inbuilt step parameter like seq. Anyways the good news is, by cutting down the number of terms by 10, my calculation time has gone down by a factor of 20.  Thanks for your efforts.

@tomleslieChanging to this convention worked! inadvertent bug discovery...

@mmcdara To answer your question, separation of U and V are necessary because X__2 and X__4 are variables of integration.  Your method works in isolation. But in my actual Record Procedure, I need the expression for u to be as a function of x only (the numerical integration I perform later protests if I have X__2 and X__4 in the expression ).


nok := iquo(ulim-llim+1, step);

U:=unapply(sum(u(step*i, X__2),i=1..nok),X__2);


doesnt quite work, but there may be a workaround...


@vv I like the simplicity of your procedure, but thanks one and all

@Carl Love just what I wanted, Thankyou


cheers for part a)

For part b) assuming (gamma = beta) using this method does not seem to produce a solution in closed form (without integrals).

Odd for an assignment question.

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