hi. i am tottaly new to the maple and i have a problem.

consider function f with variables x & y which are not independent. x & y are functions of t and the relation is unknown.

for example i wanna to the below job:

define f as f=x^2+y^2

differentiate it with respect to t : diff(f,t) which should give me 2(dx/dt)x+2(dy/dt)y

i've googled it alot and i couldn't find anything usefull.

(the problem is how to set x as a function of t with unknown relation and use it in another function and then differentiate it with respect to t)

thanks alot :)

has anybody experienced issues with Maple connectiong to Solidworks?  There doen't seem to be much info ion trouble shooting a failed "OpenConnection" call.

thanks,

Bill

From a Maple Primes answer two years ago:

f(x,y) is the equation of a line through point [m,n]. The solve command finds values of a and b for which f(x,y) are lines through [m,n] and tangent to x^2 + y^2 = r^2.

f := proc (x, y) options operator, arrow; a*(x-m)+b*(y-n) end proc

solve([f(0, 0) = r, a^2+b^2 = 1], [a, b])

These commands are far from the conventional solution. Why do they provide the correct answers?

Hello,

I would like to plot gait diagrams (the lines you can see on the picture belowà from the solutions obtained with a NL oscillator (composed with 8 coupled odes). Here the result that I would like to obtain.

Initial plot:

Desired plot

I would like to obtain 4 lines corresponding to the 4 elliptic trajectories obtained with the NL oscillator. The four lines should be done like this. When the trajectory is above 0, the line should be colored in green. When the trajectory is below 0, the line should be colored in black. 

May you help me to define this kind of graph called gait diagrams from the solution of the NL oscillator ?

Here you can find my maple code:

K:=Matrix([<0, -1, 1, -1>,<-1, 0, -1, 1>,<-1, 1, 0,-1>,<1, -1, -1,0>]);

for i to 4
do
r[i]:=sqrt((u[i](t))^2+(v[i](t))^2):
omega[i]:=omega[sw]/(1+exp(b*v[i](t)))+omega[st]/(1+exp(-b*v[i](t))):
Equ[i]:=diff(u[i](t),t)=Au*(1-r[i]^2)*u[i](t)-omega[i]*v[i](t):
Eqv[i]:=diff(v[i](t),t)=Av*(1-r[i]^2)*v[i](t)+omega[i]*u[i](t)+MatrixVectorMultiply(K,<seq(v[i](t),i=1..4)>)[i]:
EqSys[i]:=[Equ[i],Eqv[i]]:
end do:

paramsCycle:=omega[st]=4*2*Pi,omega[sw]=2*Pi,Au=5,Av=50,b=100;
params:=paramsCycle;

Differential system 
sys:=map(op,eval([seq(EqSys[i],i=1..4)],[params]));
ic:=[u[1](0)=0, v[1](0)=0,u[2](0)=0, v[2](0)=-0.1,u[3](0)=0, v[3](0)=0.1,u[4](0)=0, v[4](0)=0.1];
Résolution1
res:=dsolve([sys[],ic[]],numeric):
Initial boundaries
tcalc:=4;
ic2:=[seq(u[i](0)=eval(u[i](t), res(tcalc)),i=1..4),seq(v[i](0)=eval(v[i](t), res(tcalc)),i=1..4)];
Résolution2
res:=dsolve([sys[],ic2[]],numeric):

tmax:= 40:
numpts:=100*tmax:
plots:-odeplot(res,[t,v[1](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[t,v[2](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[t,v[3](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[t,v[4](t)],0..tmax,thickness=2, view=[0..5, -1.5..1.5],numpoints = numpts);
plots:-odeplot(res,[seq([t,v[i](t)+i*5],i=1..4)],0..tmax,thickness=2,view=[0..5,0..25], numpoints = numpts);

Thanks a lot for your help

I'm not really an expert in maple. I'm stuck with a problem that could be related to the precision of the computational engine...

I defined a function which is relatively complicated as follows:

m := proc (x, l, T) options operator, arrow; (exp(l*(x-T)/(1-exp(-l*T)))*exp(l*T/(1-exp(-l*T)))*(1-exp(-l*T))+(2-exp(-l*T)-exp(l*T))*exp(l*(x-T)/(1-exp(-l*T))))/(exp(l*T)-1)-(l*(x-T)*exp(l*(x-T)/(1-exp(-l*T)))-2*exp(-l*T)+4-2*exp(l*T))/(exp(l*T)-1) end proc

whenever i simplify the equation or factor some of the terms, I get completely different results.

As an example, you can define this function as follows

m := proc (x, l, T) options operator, arrow; (exp(x*l/(1-exp(-l*T)))*(1-exp(-l*T))+(2-exp(-l*T)-exp(l*T))*exp(l*(x-T)/(1-exp(-l*T))))/(exp(l*T)-1)-(l*(x-T)*exp(l*(x-T)/(1-exp(-l*T)))-2*exp(-l*T)+4-2*exp(l*T))/(exp(l*T)-1) end proc

By drawing the function, you can see how things get messy.

plot(m(450, (1/250)*x, 250), x = 0 .. 100)

I know the problem occurs as a results of computation round-off. How could I ameliorate that? I have already increase the precision in the option menu, but that didn't help. By the way, how can I be sure that the anwers maple gives me is correct, except using my intuition?

Download MathematicalFunctionsAndFunctionAdvisor.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Hello,

I've tried to find the solution to my problem, but none of my attempts was succesful.

I have a function which is one-to-one in a particular domain which I am interested in. I would like to get the inverse function of it only in this domain. Here is my function and plot for xp=0..10000:

x := xp-> (-1)*720.5668720*sinh(0.2043018094e-3*xp-0.8532729286)+84952.59423+4.014460003*10^5*arcsinh(0.1219272144e-1*sinh(0.2043018094e-3*xp-0.8532729286)-0.2032498888)

I would appreciate any help,

Iza

I am tryng to write and solve within Maple the equations of movement for a single body subject to a gravitatinal field, F_=-GMm/r2

But I get an error message when trying to define the Angular Momentum which doesn't make sense for me.

Thank you for any help on this topíc.

restart; with(Physics[Vectors]); conventions, Setup(mathematicalnotation = true);
conventions, [mathematicalnotation = true]
r_ := rho*_rho;
r_ := rho _rho
_rho(t);
_rho(t)
rho(t);
rho(t)
v_ := diff(rho(t)*_rho(t), t);
/ d \ / d \
v_ := |--- rho(t)| _rho(t) + rho(t) |--- phi(t)| _phi(t)
\ dt / \ dt /
a_ := diff(%, t);
/ 2 \
| d | / d \ / d \
a_ := |---- rho(t)| _rho(t) + 2 |--- rho(t)| |--- phi(t)| _phi(t)
| 2 | \ dt / \ dt /
\ dt /

/ 2 \ 2
| d | / d \
+ rho(t) |---- phi(t)| _phi(t) - rho(t) |--- phi(t)| _rho(t)
| 2 | \ dt /
\ dt /

eq[1] := -G*M*_rho(t)/r^2-a_ = 0;
/ / 2 \ 2\
| G M | d | / d \ |
eq[1] := _rho(t) |- --- - |---- rho(t)| + rho(t) |--- phi(t)| |
| 2 | 2 | \ dt / |
\ r \ dt / /

/ / 2 \
| / d \ / d \ | d |
+ _phi(t) |-2 |--- rho(t)| |--- phi(t)| - rho(t) |---- phi(t)|
| \ dt / \ dt / | 2 |
\ \ dt /

\
|
| = 0
|
/
Eq[1, 2] := seq(Component(lhs(eq[1]), n) = 0, n = 1 .. 2);
/ 2 \ 2
G M | d | / d \
Eq[1, 2] := - --- - |---- rho(t)| + rho(t) |--- phi(t)| = 0,
2 | 2 | \ dt /
r \ dt /

/ 2 \
/ d \ / d \ | d |
-2 |--- rho(t)| |--- phi(t)| - rho(t) |---- phi(t)| = 0
\ dt / \ dt / | 2 |
\ dt /
NULL;

L_ := `&x`(r_, m*v_);
Error, (in Physics:-Vectors:-&x) found the unit vector _rho present also as a function _rho(t); either one form or the other - not both - can be present in an algebraic expression

HI, I am trying to solve two PDEs but in boundry conditions there is arising an error plz help.
Nazi.mw

Hi,

I recently noticed that Maple 2015 has become irresponsive on mac (macbook pro) with the latest java release (Java 8 update 66). Did anywone experience the same problem ?

Hi,

I have a first order differential eq. for some variable say $r(x)$, where $x$ is the independent variable.

After solving this differential equation numerically, I want to use its solution in other expression for $r(x)$ and plot the expession with $x$.

Please let me know how to do it.

Thanks in advance.

I'm trying to use the CodeGeneration package to generate code for a series expansion. I'd like to wrap it in a function that specifies the arguments, so that the code generation package can generate a complete function definition along with definitions for all the temporary variables.

with(CodeGeneration):
f := proc(r): x->r end proc:
to_translate := f(convert(series(sin(x),x,20),polynom));

to_translate := proc (x) options operator, arrow; x-(1/6)*x^3+(1/120)*x^5-(1/5040)*x^7+(1/362880)*x^9-(1/39916800)*x^11+(1/6227020800)*x^13-(1/1307674368000)*x^15+(1/355687428096000)*x^17-(1/121645100408832000)*x^19 end proc

CodeGeneration['C']( to_translate );

Warning, procedure/module options ignored
double to_translate (double x)
{
  return(r);
}

Instead of using the value for 'r' passed in, CodeGenerate is producing a function which returns the bare symbol 'r' which is treated as a double. It shouldn't be an issue with lazy evalution because 'to_translate' is evaluated on the statement before the call to CodeGeneration, and to_translate has the full expression that I want to generate code for. How do I get CodeGeneration to produce the intented result?

psi = x^(4/5)*f(x, eta)/(1+x)^(1/20)

eta = y/(x^(1/5)*(1+x)^(1/20))

how can i (diff(psi, x))

how can i (diff(psi, y))

Has anyone solved this problem from an older Putnam paper?

An ellipse sitting in the first quadrant with its major axis parallel to the x axis is tangent to the positive x and y axes.

It slides clockwise within the first quadrant while maintaining tangency to both positive axes until its major axis is parallel to the y axis.

Prove that the locus of its centre is the arc of a circle.

I have crudely animated this motion by sliding the axes around the stationary ellipse. Is there a more elegant animation which slides the ellipse against stationary axes?

I have functions like

g := (y, t)->sin(y(t))

diff(g(y, t), t) # ok

How to determine the derivative below with maple :

diff(g(y, t), y(t))

Thanks