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I need to plot a path with curvilinear coordinates. Knowing the curvature theta and length s of  parts that compose it, I found the equations of x and y as a function of the parameter s that goes from zero to the total length.
I make an example below:

for 0 <s <2,     x = s     and

                     y = 0 (straight line)

 

for 2 <s <2 + Pi,        x = 2 + 2 * sin ((s-2) / 2)      and

                              y = 2 * (1-cos ((s-2) / 2)

Now I want to find the corresponding plot of x (s), y (s) (without showing s as coordinate) but I do not know what commands I can use. Can you suggest me something? Thank you

hi.

how i can solve two equation with respect to parameter sigma1

SOLL.mw

restart; pprime11 := -16395.36603*q1+5.811117425*q1*sigma1^2+3526.724044*p1-1.250000000*p1*sigma1^2+4.999870968*10^11*p1^3+4.999870970*10^11*p1*q1^2+7.967307034*10^14*p1^2*q1+4.999870966*10^12*sigma1*p1^2*q1-2.655769012*10^14*q1^3+4.999870968*10^12*sigma1*q1^3-17633.62022*q1*sigma1+6.250000000*q1*sigma1^3

qprime11 := 2.655769012*10^14*p1^3-7.967307034*10^14*p1*q1^2+4.999870970*10^11*p1^2*q1-4.999870968*10^12*sigma1*p1^3-4.999870966*10^12*p1*sigma1*q1^2+3526.724044*q1-1.250000000*q1*sigma1^2+16395.36603*p1-5.811117425*p1*sigma1^2+4.999870968*10^11*q1^3+17633.62022*p1*sigma1-6.250000000*p1*sigma1^3:

-50 < sigma1 and sigma1 < 50:

sigma1 <> 53.11665685, -53.11665685:

SOLL := solve({pprime11, qprime11}, real)

Warning,  computation interrupted

 

``

 

Download SOLL.mw

I want to ask., I put delta as my constant in maple program and I want the answer are in delta as well., but the thing is., when running., it let delta=0, delta=-1, and delta=delta.,
the condition is we cannot let delta=1 or delta=0 because it is just same for s5 and s7.,.(delta is refer to the s8). How can I get answer as delta? with the condition? here I attach my maple programme..

 

> derivation := proc (A, n)
local i, j, k, t, s5, s7, s8, m, D,
sols5, sols7, sols8, eqns5, eqns7, eqns8,
BChange5, BChange7, BChange8; eqns5 := {}; eqns7 := {}; eqns8 := {};
D := matrix(n, n);
BChange5 := matrix(n, n); BChange7 := matrix(n, n); BChange8 := matrix(n, n);
for i to n do for j to n do for m to n do
s5 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+A[i, k, m]*D[k, j], k = 1 .. n));
s7 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+0*A[i, k, m]*D[k, j], k = 1 .. n));
s8 := sum(0*A[i, j, k]*D[m, k], k = 1 .. n)-(sum(A[k, j, m]*D[k, i]+delta*A[i, k, m]*D[k, j], k = 1 .. n));
eqns5 := `union`(eqns5, {s5}); eqns7 := `union`(eqns7, {s7}); eqns8 := `union`(eqns8, {s8})
end do end do end do;
sols5 := [solve(eqns5)]; sols7 := [solve(eqns7)]; sols8 := [solve(eqns8)];
t := nops(sols5); t := nops(sols7); t := nops(sols8);
for i to t do for j to n do for k to n do
BChange5[k, j] := subs(sols5[i], D[k, j]);
BChange7[k, j] := subs(sols7[i], D[k, j]);
BChange8[k, j] := subs(sols8[i], D[k, j])
end do end do;
print("eqns&Assign;", eqns5); print("sols:=", sols5); print("BChange5:=", BChange5);
print("eqns&Assign;", eqns7); print("sols:=", sols7); print("BChange8:=", BChange7);
print("eqns&Assign;", eqns8); print("sols:=", sols8); print("BChange8:=", BChange8)
end do end proc;

> AS1 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 2) = 1]);
> derivation(AS1, 2);

> AS2 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1]);
> derivation(AS2, 2);

> AS3 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (2, 1, 2) = 1]);
> derivation(AS3, 2);

> AS4 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (2, 2, 2) = 1]);
> derivation(AS4, 2);

> AS5 := array(sparse, 1 .. 2, 1 .. 2, 1 .. 2, [(1, 1, 1) = 1, (1, 2, 2) = 1, (2, 1, 2) = 1]);
> derivation(AS5, 2);

> AS1 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (3, 1, 2) = 1]);
> derivation(AS1, 3);

> AS2 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (3, 1, 2) = alpha]);
> derivation(AS2, 3);

> AS3 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (1, 2, 3) = 1, (2, 1, 3) = 1]);
> derivation(AS3, 3);

> AS4 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (2, 3, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS4, 3);

> AS5 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 3, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS5, 3);

> AS6 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(3, 1, 2) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS6, 3);

> AS7 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 2, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS7, 3);

> AS8 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 3, 3) = 1]);
> derivation(AS8, 3);

> AS9 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS9, 3);

> AS10 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS10, 3);

> AS11 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 3, 2) = 1, (2, 3, 2) = 1, (3, 1, 2) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS11, 3);

> AS12 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (1, 3, 1) = 1, (2, 3, 2) = 1, (3, 1, 1) = 1, (3, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS12, 3);

> AS13 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS13, 3);

> AS14 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS14, 3);

> AS15 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 2, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS15, 3);

> AS16 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(2, 1, 1) = 1, (2, 2, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS16, 3);

> AS17 := array(sparse, 1 .. 3, 1 .. 3, 1 .. 3, [(1, 1, 2) = 1, (3, 3, 3) = 1]);
> derivation(AS17, 3);
>

Dear all

I have the following equaion

Eq := diff(phi(x, k), x, x)+(k^2+2*sech(x))*phi(x, k) = 0;
          
The solution is given by 

phi := (I*k-tanh(x))*exp(I*k*x)/(I*k-1);

My question : At what value of k is there a bound state and in this case can we give a simple form of the solution phi(x,k)

 

With best regards

 

Hello everyone,

I am trying to solve numerically int( f(t,z) , t=0..T ) = 0 , in z for a cumbersome f.

I tried z1=fsolve( int( f(t,z) , t=0..T ) = 0 , z). But then I tried int( f(t,z1) , t=0..T ) and the result is clearly not zero nor anything small.

It looks like Maple evaluates analytically the integral, and does it wrong (check this for more details) so fsolve uses the wrong equations.

Anyone knows how I can force Maple to evaluate numerically the integral at each step of the fsolve function?

Thank you!

I want to solve one equation with one variable and the variable is also in definite integral delimiter. When trying fsolve, I get the error:

"Error, (in fsolve) Can't handle expressions with typed procedures"

code

Here is worksheet.mw

How can I obtain solution with method other from simple manual testing Te values?

I have a problem with finding a interval where parametr "a" is changing.
My funcion:

                 exp(x2(a-x)) for xcR.
I did this, and I don't know what's next.

g:=x->exp(x2(a-x));
dg:=diff(g(x),x);
solve(dg=0,x);
                      0,2/3a
g(0);
g(2/3a);

I am solving a complicated ODE and I would like to know if there is a way for Maple to output the ODEs without doing any numerical substitutions for known parameters. Say one of my parameters, call it P, is initialized (there are many more but Ill just simplify and consider one here) to a value of 10. In order to chekc that I have coded the ODEs correctly it would help me if Maple does not substitue with the numerical value 10 for P when displayng teh ODEs, but rather keeps P, as a parameter. Is there a way to achieve this?

I have written the following coade in Maple:
r := 50;
l1 := 0.2742e-10;
s := I*w;
l := (-1.342110665*10^22*c^2*(Pi^4)-4.225000000*10^25*c^2*(Pi^2)+2.316990000*10^11*c1*(Pi^2)-1)/(-1.342110665*10^22*c^2*c1*(Pi^4)-7.140250000*10^43*c^2*c1*r^2*(Pi^4)+1.957856550*10^33*c^2*(Pi^4)+9.789282750*10^32*c*c1*(Pi^4)-1.690000*10^22*c*(Pi^2)-4.22500*10^21*c1*(Pi^2));
z1 := (c*l1*s^2+1)/(c*s);
z2 := l*s/(c1*l*s^2+1);
h := (z1+2*z2)*((z1+2*r)*(z1+3*z2)/(2*r)-2*z2)/z2-(1/2)*z2*(z1+2*r)*r;
f := h*(z1+3*z2)/z2-(z1+2*r)(2*r)*(z1+3*z2)+2*z2;
gain := 2*z2/f;
a := abs(gain);
d := diff(a, w);
s := subs(w = 2*pi*0.325e11, d)
Now, I have a function named "s" which I want to set to zero, and calculate the relationship between variables c & c1 in order to achieve this. How should it be done?
Thanks.

I have the following function

where A,B,Ψ, K1,K2,K3,α,β are all constants.

How to find the value of m for which the above expression is 0 or approximate to 0 for different values fo the constants.

e.g., Fixing all the parameters except A, I want to find the values of m for different values of A. How to do that in maple?

 

I hoped that Maple would return the value of 1 in all commands (see below). However, introducing a scaling parameter, sigma, yields the unevaluated expression. Why? I still think it should evaluate to the value of 1.

 

kind regards,

Harry (not a mathematician, but a psychologist)

 

 

 

integral.mw

how can i improve doing the graph with various parameter

do anyone have abother numerical method in maple rather than rk45 felhberg

like keller box/homotopy/ or anything

i have attchassignment.mwsassignment.pdf

restart

with(plots)

%?

Eq1 := diff(f(eta), `$`(eta, 3))+(diff(f(eta), `$`(eta, 2)))*f(eta)-(diff(f(eta), eta))^2+4 = 0

diff(diff(diff(f(eta), eta), eta), eta)+(diff(diff(f(eta), eta), eta))*f(eta)-(diff(f(eta), eta))^2+4 = 0

(1)

%?

Eq2 := diff(theta(eta), `$`(eta, 2))+Pr*(diff(theta(eta), eta))*f(eta) = 0

diff(diff(theta(eta), eta), eta)+Pr*(diff(theta(eta), eta))*f(eta) = 0

(2)

%?

VPr := [0.1e-1, 0.2e-1, 0.3e-1]

etainf := 27

bcs := (D(f))(0) = 0, f(0) = 0, (D(theta))(0) = -1, (D(f))(etainf) = 2, theta(etainf) = 0

(D(f))(0) = 0, f(0) = 0, (D(theta))(0) = -1, (D(f))(27) = 2, theta(27) = 0

(3)

dsys := {Eq1, Eq2, bcs}

for i to 3 do Pr := VPr[i]; dsol[i] := dsolve(dsys, numeric); print(Pr); print(dsol[i](0)) end do

0.1e-1

[eta = 0., f(eta) = HFloat(0.0), diff(f(eta), eta) = HFloat(0.0), diff(diff(f(eta), eta), eta) = HFloat(3.4862842650940435), theta(eta) = HFloat(9.305856096452466), diff(theta(eta), eta) = HFloat(-0.9999999999999998)]

0.2e-1

[eta = 0., f(eta) = HFloat(0.0), diff(f(eta), eta) = HFloat(0.0), diff(diff(f(eta), eta), eta) = HFloat(3.4862842653392216), theta(eta) = HFloat(6.7064688361073745), diff(theta(eta), eta) = HFloat(-1.0)]

0.3e-1

[eta = 0., f(eta) = HFloat(0.0), diff(f(eta), eta) = HFloat(0.0), diff(diff(f(eta), eta), eta) = HFloat(3.4862842648459234), theta(eta) = HFloat(5.552583608770695), diff(theta(eta), eta) = HFloat(-0.9999999999999998)]

(4)

SDf1 := odeplot(dsol[1], [eta, f(eta)], 0 .. etainf, color = green, axes = box); SDf2 := odeplot(dsol[2], [eta, f(eta)], 0 .. etainf, color = red); SDf3 := odeplot(dsol[3], [eta, f(eta)], 0 .. etainf, color = blue)

display([SDf1, SDf2, SDf3], labels = ["&eta;", "f (&eta;)"], labeldirections = [horizontal, vertical], labelfont = [italic, 16, bold], axes = boxed, axesfont = [times, 14], thickness = 3)
%?

 

%?

SDfd1 := odeplot(dsol[1], [eta, diff(f(eta), eta)], 0 .. etainf, color = green, axes = box); SDfd2 := odeplot(dsol[2], [eta, diff(f(eta), eta)], 0 .. etainf, color = red); SDfd3 := odeplot(dsol[3], [eta, diff(f(eta), eta)], 0 .. etainf, color = blue)

%?

display([SDfd1, SDfd2, SDfd3], labels = ["&eta;", "f  ' (&eta;)"], labeldirections = [horizontal, vertical], labelfont = [italic, 16, bold], axes = boxed, axesfont = [times, 14], thickness = 3)

 

%?

`S&theta;1` := odeplot(dsol[1], [eta, theta(eta)], 0 .. etainf, color = green, axes = box); `S&theta;2` := odeplot(dsol[2], [eta, theta(eta)], 0 .. etainf, color = red); `S&theta;3` := odeplot(dsol[3], [eta, theta(eta)], 0 .. etainf, color = black)

display([`S&theta;1`, `S&theta;2`, `S&theta;3`], labels = ["&eta;", "&theta; (&eta;)"], labeldirections = [horizontal, vertical], labelfont = [italic, 16, bold], axes = boxed, axesfont = [times, 14], thickness = 3)

 

%?

`S&theta;d1` := odeplot(dsol[1], [eta, diff(theta(eta), eta)], 0 .. etainf, color = green, axes = box); `S&theta;d2` := odeplot(dsol[2], [eta, diff(theta(eta), eta)], 0 .. etainf, color = red); `S&theta;d3` := odeplot(dsol[3], [eta, diff(theta(eta), eta)], 0 .. etainf, color = black)

display([`S&theta;d1`, `S&theta;d2`, `S&theta;d3`], labels = ["&eta;", "&theta; '(&eta;)"], labeldirections = [horizontal, vertical], labelfont = [italic, 16, bold], axes = boxed, axesfont = [times, 14], thickness = 3)

 

%?

 

 

Download assignment.mws

 

hi .how i can clarify expression in maple that in running dont asked me again..

for example attached file

2)how i can deleted one parameter of 
Memory in maple or
Restart it??

hi.i encounter error in pdsole equations with unknown parameter(N)??

please help me for solve it....thanks alotmaple_prime.mw

Dear Maple users

Physical experiment: I dropped a ball with low mass from a height of approximately 7 meters and wanted to test if the air resistance was proportional to the square of the velocity. I filmed the fall and used the program Logger Pro to collect data: a number of datapoints (time,height) was collected. I copy/pasted the datapoints into MS Excel, from where I could import data into Maple via Tools > Assistants > Import Data ... Then I wanted to make a fit with the theoretical solution, given by a function having just one parameter: the Drag coefficient. Unfortunately I received an error "complex values encountered" (see below). I can solve the problem manual by making a number of guesses for the drag coefficient, until the theoretical curve approximates the data points well. I wanted to make Maple do the fitting job, though. I will appreciate if someone could give an idea how to fit the data properly.

NB! Mass m and g is defined above in the Maple document. The Statistics and plots package is called too.

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