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Surface btw spacecurves in 3d......

Asked by: MarcinM 30
October 05 2023
1  9

Hi,

Anyone knows how to plot a surface btw 2 spacecurves in 3d? For example a flat curve x^2+y^2=1 which maybe a upper half of above circle for y=0 to 1 and a spacecurve which is a function for example 2+x^2*y. So second curve should be above the first and limit a surface From the top. 

Can't find a solution in help and parametric curves does not solve a problem.

In fact it would be enough to have a vertical surface starting on a xy plane and limited by Givenchy function. As it is in line integral calculations.

Regards

Marcin

How Do I enter this into Maple?...

Asked by: EdisonSu223 5
linearalgebra homework do-my-homework-for-me
October 04 2023
1  1

How to manually build a loglog plot (without using...

Asked by: zenterix 405
plot plotting loglogplot
October 04 2023
0  1

Consider the following worksheet (perhaps it is better to download the worksheet and execute since the contents below aren't showing the commands used to plot the last two plots).

NULL

T = log(R)/(a+b*log(R))^2

plot(log(R)/(-1.16+.675*log(R))^2, R = 1000 .. 30000)

  

NULL

10^log[10](T) = log(R)/(a+b*log(R))^2

NULL

log[10](T) = log[10](log(R)/(a+b*log(R))^2)

NULL

w = log[10](z/(b*z+a)^2)

w = ln(z/(b*z+a)^2)/ln(10)

 (1)

NULL

plot(log[10](z/(-1.16+.675*z)^2), z = log(1000) .. log(30000))

  

NULL

``

NULL

plots:-loglogplot(log(R)/(-1.16+.675*log(R))^2, R = 1000 .. 30000)

  

NULL

 

My question is about making a loglog plot of the equation

for R between 1000 and 30000.

 

The second to last plot is of w as a function of z, as in the last equation below

and the plot command is (where I have subbed in a=-1.16 and b=0.675)

In the second to last plot, of course we have negative values of w. If we were to consider the underlying values of T, they would never be negative of course.

The last plot is the command

I think this last plot is what I want (though I am not sure because I am not totally sure what plots:-loglogplot is doing).

 

My question is how to obtain this loglog plot manually. That is, I want the axes to show values of R on the x axis and T on the y axis (just like a usual loglog plot shows).

In other words, how to go from the second to last plot to the same plot but showing the corresponding R and T values instead of z and w.

Download loglog.mw

display solution pdesolve...

Asked by: Zeineb 540
pde differential-equations
October 04 2023
0  1

Dear all

I would like to get the solution of a system : pde with boundary and initial condition. Everything well coded, but the code does not return the solution 

sol_heat.mw

Thanks for your help 

How to get Maple to output result of solve using b...

Asked by: zenterix 405
log
October 04 2023
2  4


Here is a small worksheet to illustrate my question

Consider the expression

 

solve(sqrt(log[10](R)/T) = a+b*log[10](R), T) = ln(R)*ln(10)/(ln(10)^2*a^2+2*ln(10)*ln(R)*a*b+ln(R)^2*b^2)NULL

NULL

Can we somehow tell Maple to keep the logarithms in base 10?

 

NULL


I'd like for the final expression to have only base 10 logarithms.

 

When I do it with pen and paper, I get

 

It is this last expression that I would like Maple to output.

Download log10.mw

The issue with mtaylor for a function of 2 variabl...

Asked by: Vortex 65
series mtaylor
October 04 2023
0  3

Hello.

I want to check my analytical calculations and to compare my results with the series expansion for x=0 and y=0 of the function

R0 := -(1-tanh((1/2)*l*sqrt(k*y^2+x^2)*d)^2)*x/(k*y^2+x^2)+2*tanh((1/2)*l*sqrt(k*y^2+x^2)*d)*x/(l*(k*y^2+x^2)^(3/2)*d)

by means of the command 

mtaylor(R0, [x, y], 5);

 

However, I have a message from Maple 17

Error, (in mtaylor) does not have a taylor expansion, try series()

I read here a little bit about some bugs of mtaylor function, but how to resolve this issue?

introduce the cosine...

Asked by: jalal 435
embedded-components trigonometry
October 04 2023
0  0

Hi, I'm looking to create a discovery activity to introduce the cosine . Any ideas for using Maple components with a slider to vary the position of a point on a line while displaying distances and their ratios? Thank you

Doc2.pdf

principal measurement of an angle...

Asked by: jalal 435
radians mod
October 04 2023
0  1

Hi,

How can one obtain the principal measurement of an angle (which belongs to the interval ]-π;π]?

Thank you

QuestionMesPrincipale.mw

How does Maple arrive at an implicit solution to i...

Asked by: Joseph Poveromo 25
October 04 2023
0  3
Equation:=int(y'(x)*w(x)) = [int([y(x)^(1/2)]*w(x))]^(-2/3) When plugging this Equation into dsolve, Maple provides the following implicit solution: Solution:= [3*y(x)^(4/3)]/4 + int[(2*y(x)^(5/6)/[3*int([(y(x)^(1/2))*w(x)]/[3*int[(y(x)^(1/2))*w(x)]^(5/3) When going to odeadvisor, the suggestion was to first convert to the form y = G(x,y'(x) and then utilize the method of 'patterns', which I could not apply to this equation. If anyone can fill in the steps between 'Equation' and 'Solution', it would be greatly, appreciated. P.S.Sadly, I am unable to attach the actual Maplesoft worksheet.

Error (in fprintf)...

Asked by: Carolyn Davis 15
loop homework printf
October 03 2023
0  1

Can anyone assist with this error please?

> 

#Clear memory and load package.
restart;  
with(LinearAlgebra):

#Initialise variables,matrices and vectors.
b:=<<18>,<-2>>:
c:=<<1>,<1>>:
i:=0:
P:=<1.08,1.37,1.56,1.61>;
t:=<5,10,15,20>;
tol:=1e-6:

#Initialise Gauss-Newton matrices.
n:=Dimension(t):
f:=Matrix(n,1):
J:=Matrix(n,2):

#Display initial parameter values.
printf("Gauss-Newton Method\n");
printf("-------------------\n");
printf("Before iterations,A = %f and B = %f\n",b(1),b(2));

#Perform the Gauss-Newton method.
while max(abs(evalf(c)))>tol do
  i:=i+1;
  for r from 1 to n do
     f(r,1):=evalf((In(b(1)+t[r])+b(2))-P[r]);
     J(r,1):=evalf(1/b(1)+t[r]);
     J(r,2):=evalf(1);
end do;
c:=Multiply(MatrixInverse(Multiply(Transpose(J),J)),Multiply(Transpose(J),f));
b:=b-c;
printf("After iterations %d, A = %f and B = %f\n",i,b(1),b(2));
end do:

Vector(4, {(1) = 1.08, (2) = 1.37, (3) = 1.56, (4) = 1.61})

  

Vector[column](%id = 36893488152076174028)

  

Gauss-Newton Method
-------------------
Before iterations,A = 18.000000 and B = -2.000000
After iterations 1, A =

 

Error, (in fprintf) number expected for floating point format

  
> 

NULL

Download Asst_3_Q4b.mw

Error - final value in loop must be numeric or cha...

Asked by: Carolyn Davis 15
loop homework
October 03 2023
0  1

The x region is z<x<1 and y region is 0<y<pi//3, I think there is an error in the j values in the loop

Can anyone assist please?

> 

#Clear memory.
restart;

#Initialise variables and arrays.
h:=0.1;
k:=Pi/30;
c:=1:
iter:=0:
tol:=1e-5;
w:=1.5;
u:=Array(0..10,0..10):
U:=Array(0..10,0..10):

#Calculate boundary u values.
for i from 0 to 10 do

   u[i,Pi/3]:=0;
   u[i,0]:=0;
end do:

for j from 0 to Pi/3 do
   u[0,j]:=0;   
   u[10,j]:=sin(3*j*k)/2;
end do:
#Perform the Gauss-Seidel method.
while c>tol do
   iter:=iter+1;
   if iter>99 then
      printf("\nGauss-Seidel method did not converge,change w.\n");
   break;
end if:
for i from 1 to 9 do
  for j from 0 to Pi/3 do
     U[i,j]:=u[i,j];
     u[i,j]:=(10*Pi^2*u[i-1,j]+9*u[i,j-1]+9*u[i,j+1]+2*Pi^2*u[i+1,j])/20;
     u[i,j]:=w*u[i,j]+(1-w)*U[i,j];
  end do:
end do:
c:=max(abs(u[1..9,1..9]-U[1..9,1..9]));
printf("After iteration %2d,the maximum change in u is %7.1e\n",iter,c);
end do:

#Display the final u values if method converged.
if iter<100 then
    printf("\nGauss-Seidel Method\n");
    printf("-------------------\n");
    for j from 10 by -1 to 0 do
      for i from 0 to 10 do
        printf("%6.4f",u[i,j]);
    end do:
    printf("\n");
end do:
end if:

.1

  

(1/30)*Pi

  

0.1e-4

  

1.5

  

Error, bad index into Array

  

Error, final value in for loop must be numeric or character

  

Error, final value in for loop must be numeric or character

  


Gauss-Seidel Method
-------------------
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000
0.00000.00000.00000.00000.00000.00000.00000.00000.00000.00000.0000

  
> 

NULL

Download Asst_3_Q3c.mw

Error, (in DEtools/phaseportrait) vars must be dec...

Asked by: Missis 15
ode syntax differential-equations
October 03 2023
2  4

Hello. I have a problem with the phaseportrait and I hope could help me with this:

can you share a documet ...

Asked by: Andr387m 5
group
October 03 2023
1  1

can you do groupwork on a single documet 

Array error out of range...

Asked by: Carolyn Davis 15
array syntax homework
October 02 2023
0  1

Can anyone assist with the error please?
 

> 

#Clear memory.
restart;

#Initialise variables and arrays.
h:=0.1;
n:=10;
k:=0.1;  
m:=10;
t:=Array(0..m):
x:=Array(0..n):
u:=Array(0..n,0..m):

#Initialise the x array and the initial u(x,0) boundary.
for i from 0 to n do
    x[i]:=i*h;
    u[i,0]:=exp(x[i]);
end do:

#Initialise the t array and the u(x,t) side boundaries.
for j from 0 to m do
   t[j]:=j*k;
   u[0,j]:=0;
   u[n,j]:=t[j];
end do:

#Use the 2D CTCS explicit wave method.
for i from 1 to n-1 do
   u[i,1]:=(9*u[i-1,0]+14*u[i,0]+9*u[i+1,0])/32+[i]/10;
end do:
 
for j from 1 to m-1 do
  for i from 0 to n-1 do
   u[i,j+1]:=(9*u[i-1,j]+14**u[i,j]+9*u[i+1,j])/16-u[i,j-1];
  end do;
end do:

#Display the u(x,t)values.
printf("2D CTCS Explicit Wave Method\n");
printf("----------------------------\n");
printf("x\t\t t\t\t u\n");
for i from 0 to n do
   printf("% f\t% f\t% f\n",x[i],t[m],u[i,m]);
end do;

.1

  

10

  

.1

  

10

  

Error, Array index out of range

  

2D CTCS Explicit Wave Method
----------------------------
x                 t                 u
 0.000000         1.000000         0.000000
 0.100000         1.000000         0.000000
 0.200000         1.000000         0.000000
 0.300000         1.000000         0.000000
 0.400000         1.000000         0.000000
 0.500000         1.000000         0.000000
 0.600000         1.000000         0.000000
 0.700000         1.000000         0.000000
 0.800000         1.000000         0.000000
 0.900000         1.000000         0.000000
 1.000000         1.000000         1.000000

  
> 

NULL


 

Download Asst_3_Q2b.mw

How to use simplify command?...

Asked by: raj2018 55
solve simplification
October 02 2023
2  3

HI,
I want to simplify an algebraic experssion. How do I do this with Maple?
A1.mw

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