hello, i'm using maple, in order to draw a graphic in wich, different functions are plotted, 
these functions came from one with 4 variables, wich represents losses in a phenomenon (extraction) depending on diffenrent variables, so my objective is to varry only one parameter and then draw its curve, and this for each parameter ; i had 4 functions.

with(plots):

f1:=plot((n-1)/(-1+n*10^(6.77*((n-1)/n))), n=1.0537..1.2688):

f2:=plot(0.2365/(-1+1.2365*10^(0.0155*O)), O=76.42..99.92):...

I want to create a package.  I want to be able to load it using the usual with(Accellib); construct.

how can i do it?

hi

please help me to solve the following system of ODE

diff(u(eta), eta, eta)+.5699919300/(0.9930000000e-3+0.3883623000e-1*phi(eta)+.5301627000*phi(eta)^2)+1/(eta-1)+(0.3883623000e-1+1.060325400*phi(eta))*(diff(phi(eta), eta))*(diff(u(eta), eta))/(0.9930000000e-3+0.3883623000e-1*phi(eta)+.5301627000*phi(eta)^2) = 0;

diff(T(eta), eta, eta)+(5.05659*((-1.1752324*10^6*phi(eta)+4.1744724*10^6)*u(eta)/(int((1-eta)*(-1.1752324*10^6*phi(eta)+4.1744724*10^6)*u(eta), eta = 0 .. .5))+7.47*(diff(phi(eta), eta))/(1+7.47*phi[0])-(.1977617327*(.597+4.45959*phi(eta)))*(diff(T(eta), eta))/(1-eta)))/(.597+4.45959*phi(eta))=0;

diff(phi(eta), eta)-5.000000000*phi(eta)*(diff(T(eta), eta)) = 0;

with following boundary condition

u(0)=0,u(0.5)=0,phi(0)=1,T(0)=0,D(T)(0)=0

thanks for your attention in advance

Hi,

I want to check this inequality (check below) for the set of parameters that holds, and I also want to restrict my variable to be real and non-negative (i.e k>=0), I am getting the usual message that "Solutions may have being lost". Can somebody please help me on how i can handle this ? Thanks 

alpha*beta*(1-alpha)*k^(alpha-1)/(1+beta)-1-(2*alpha-1)*alpha*(1-alpha)*beta*k^(2*(alpha-1))/(1+beta)+alpha^2*k^(alpha-1)

I've been playing around with power towers and thought my latest might be interesting enough to post.

I used Wolfram Alpha on the big numbers (n=5..50). Some corrected.

Hi,

I obtain  curve pH = f(Vb) with following code

restart;with(plots);

 Vtot := V0+Vb+Vind;h := 10^(-ph);C0 := 10^(-2);Cind := .83;Vind := 0.1e-1;V0 := 10;Ka1 := 10^(-4.8);Ka2 := 10^(-9);Cb := 0.1e-1;Kaind := 10^(-8);

eq := h+Cb*Vb/Vtot = 10^(-14)/h+C0*V0/(Vtot*(1+h/Ka1))+Cind*Vind/(Vtot*(1+h/Kaind));

 implicitplot(eq, Vb = 0 .. 40, gridrefine = 3, ph...

What do you think about spliting of single worksheet where code and results are placed into two separate worksheets where code and results of computation are placed separately?

Thank you.

Hello,

I installed Maple 17 on Slackware Linux 14.0 today and everything seems to work except that when I want to search for something through Maple Help my mouse doesn't seem to activate the text search field. Is this happening with anyone else or is this just because I am running under Slackware?

Thanks,

Jovan

Hi,

I obtain  curve pH = f(Vb) with following code but now, i would have tangent for two points near Veq. I don't know the procedure with implicitplot.

I think the best way is to attribute the points to a function but dn't know how to do

restart;with(plots);

 Vtot := V0+Vb+Vind;h := 10^(-ph);C0 := 10^(-2);Cind := .83;Vind := 0.1e-1;V0 := 10;Ka1 := 10^(-4.8);Ka2 := 10^(-9);Cb := 0.1e-1;Kaind := 10^(-8);

eq := h+Cb*Vb/Vtot = 10^(-14...

i have 6 non linear diff equation. for the quad copter i want to get its state space representation and its done in a master  but they didnot say how they did it ???

soo any one can help me?

Dear guys!

I have two expressions in terms of one variable. For example A(t) and B(t). Now, I want to plot A in terms of B. What should I do?

I cannot obtain this relation analytically because both expressions in terms of "t" are very complicated. But I can plot both of them. Now, I need to plot A in terms of B.

Thanks a lot.

> test1r := dsolve((Diff(f(x,y), t, t))*x+(Diff(f(x,y), t))*y = 0, f);

Error, (in dsolve) expected an ordinary differential equation. Received: 0 = 0

another question is

i do not know whether can 

 dsolve((Diff(f(x,y), (x,y), (x,y)))*x+(Diff(f(x,y), (x,y)))*y = 0, f);

Hello,

I want to report a strange behaviour when working with bras and kets from the physics package. Working in a discrete basis, Ket(Phi,m) and Ket(Phi,n) should be orthogonal as long as m neq n. At least this is said so in the Physics-Example document.

However when I enter Bra(Phi,0).Ket(Phi,4) Maple returns 1! Just with 4 - all other integer values n > 1 give zero, as it should be! Maybe I am completely misunderstanding something here or this result is...

I consider a PDE system (?pdsolve,numeric)

PDE := {diff(u(x,t),t)=-1/20*diff(v(x,t),x,x),
diff(v(x,t),t)=diff(u(x,t),x,x)}:

IBC := {u(x,0)=sin(Pi*x), u(0,t)=0, u(1,t)=0,
v(x,0)=1-x, v(0,t)=1, v(1,t)=0}:

pds := pdsolve(PDE, IBC, numeric, spacestep=1/50);

Suppose, I need to use the data diff(u(x,t),x,x)  or diff(v(x,t),x,x) at x=0.