Thomas Dean

197 Reputation

7 Badges

15 years, 306 days

MaplePrimes Activity

These are answers submitted by Thomas Dean

Apply sort.


Help -> Maple Help -> solve



Aplply isolve() to your inital equation.

This will give the variables a,m,n,x in terms of _Z1 and _Z2, both integers.

You can subst integer for say Z1 and then solve for Z2, etc .

int(cos(3+sin(x)), x = 0 .. 5,numeric=true)

Look at help Statistics[Sample]

eq := -Pi/2 - arctan(25*x) - Pi = -(2*Pi)/3;

plot(lhs(eq)); ## shows the lhs is negative.

limit(lhs(eq), x = -infinity) ## shows the limit to be -Pi.

and -Pi < -2*Pi/3;



Tom Dean

eqs:=solve({fS,fV,fC,fR},{x1,x2,x3,x4},explicit); # No RootOfs


The problem, as stated involved floating point values and associated errors.

In calculating the target position, an approximation was used.  The values obtained gave different slopes for the bearing lines.

Correcting the calculation of the slopes of the bearing line resulted in a better solution, to several decimal places.

Sorry for the noise.

Tom Dean

help pdsolve[boundaryconditions] has an example that, on the surface, looks sililar.



S := {W,X,Y};

S minus {X};


Tom Dean


## Note:  Here, you are assigning a value to q.
q := -k*(T2-T1)/t;
S := {T1 = 550*Unit('K'), T2 = 50*Unit(Unit('K')), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), t = 2*Unit('cm')};
eval(q, S);

## Note:  At this point, q still has the value assigned above.
##        Maple simplifies the input and the result is t = t!
t := -k*(T2-T1)/q;

S := {T1 = 550*Unit('K'), T2 = 50*Unit(Unit('K')), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), q = 477.5*Unit('kW'/'m'^2)};
eval(t, S);

## Note:  If you avoid using variables that already have values
##        the result is different.  I used zz rather than q to make
##        the name more obvoius.
##        use variable names that make sense in the application.

t := -k*(T2-T1)/zz;

S := {T1 = 550*Unit('K'), T2 = 50*Unit(Unit('K')), k = 19.1*Unit('W')/(Unit('m')*Unit('K')), zz = 477.5*Unit('kW'/'m'^2)};
eval(t, S);

## the same applies to the remainder of your code.

q := 'q';  ## clear the previous value assigned to q.

Tom Dean

1 2 Page 1 of 2