Maple 2022 Questions and Posts

Equation for the area of a triangle given unknown ...

Asked by: Christopher2222 6035

June 02 2025

0 9

restart;
with(geometry);

triangle(ABC, [point(A, a, b), point(B, c, d), point(C, e, f)]);
AreCollinear: hint: could not determine if a*d-a*f-b*c+b*e+c*f-d*e is zero
Error, (in geometry:-triangle) not enough information: the three points might be AreCollinear

assume(a*d - a*f - b*c + b*e + c*f - d*e <> 0);

triangle(ABC, [point(A, a, b), point(B, c, d), point(C, e, f)]);
ABC

area(ABC);

What does the ~ after the point variables supposed to mean?

Average world temperature since 1850

by: Christopher2222 6035 Product: Maple 2022

May 23 2025

3 1

The workbook includes the excel data file but I also included the worksheet and data file separate for versions that can't load workbooks.

Average_World_Temperature_workbook-.maple

Average_World_Temperature_since_1850_worksheet.mw

HadCRUT5.zip

How do I refresh workbook to reflect changes in at...

Asked by: Son Ho 5

May 22 2025

1 0

I have just started to use Maple workbook (*.maple). In the workbook, there are 4 worksheets: "main.mw","calc1.mw","calc2.mw","calc3.mw". In "main.mw", I plot variables saved from "calc?.mw". This works. However, when I make changes in one of the worksheet "calc?.mw" and get the new expressions for the variables, they are not updated in "main.mw". I deleted the variables being changed but "main.mw" still run and show the old expressions of these variables.

How do I refresh so that "main.mw" will show the updated variables from "calc?.mw"?

Thanks.

Two font sizes in plot title...

Asked by: Christopher2222 6035

May 21 2025

3 8

Trying to produce two different font sizes within a plot title. I can achieve it with Typesetting but I'm getting brackets and fluff I don't want. Any other ideas?

For example

with(Typesetting):
plot(x^2, title = [cat(mtext("sort of", font = "TimesNewRoman", size = 30), mtext("works", size = 10))])

Why can't the calculated values be consistent ...

Asked by: GFY 75

May 19 2025

1 9

I’m trying to plot a 3D surface in Maple with variables l and m. I used numerical substitution to evaluate the function and the results are real and positive. However, when I plot the function over a range of l and m, the graph shows complex (imaginary) values instead.

This seems very strange to me and has been quite frustrating. I’ve tried many different approaches to resolve the issue, but nothing has worked so far.

Why is this happening? How can the function evaluate to real numbers with direct substitution, but show complex values during plotting?

Any suggestions or explanations would be greatly appreciated. Thank you!
gra423_Omega.mw

plot3d size issues...

Asked by: Christopher2222 6035

May 11 2025

0 6

I'm not sure if this is fixed in newer Maples or if there's a work around.

Whenever I size a 3d plot, that I'm trying to stretch out the width while keeping the height, never seems to work. For example using the size=[3000,800] produces a plot area that's bigger but NOT actually a plot stretched in the x axis. Going to size=[3000,3000] of course then makes the plot and the area bigger and so scales both x and y bigger. However I don't want the y axis scaled up - I'm trying to scale the plot up - not the area. And what happened to the window zoom, icon - we've lost zoom control to just magnify + and magnify - (at least in 2022) this seems like a regression.

Is this plot size a bug or just a plot command that fails to function like it should?

What is wrong with my logic in interpreting this s...

Asked by: zenterix 405

April 26 2025

0 1

I am very confused by the y-value of the rightmost point on the plot below.

I'd like to find the values of for which x^2/(10^(-8)-x) = 5*10^(-3) .

So I ask Maple to solve this equation.

evalf(solve(x^2/(10^(-8)-x) = 5*10^(-3)))

(1)

Do these solutions work?

eval(x^2/(10^(-8)-x), x = 1.0000*10^(-8)) =

eval(x^2/(10^(-8)-x), x = -0.5000010000e-2) =

Suppose I define the function

f := proc (x) options operator, arrow; x^2/(1/100000000-x) end proc =

Error, (in f) numeric exception: division by zero

=

Now, the function seems continuous between these two points

It is late, and perhaps I am just tired and not seeing things clearly. I expected the topmost point on the right to have a y-value of 0.00999998, ie almost 0.01.

I expected that the bottom leftmost point to be 0.0009999800001, ie almost 0.001, and it is.

And I thus expected to show that there must be some x for which we have f(x)=0.005, which if I am not mistaken is between the two numbers. After all, =

=

what am i missing here?

Download plotq.mw

How to solve this chemical equilibrium equation wi...

Asked by: zenterix 405

April 26 2025

0 2

I have a question about a calculation I was just trying to do. For some context, I am trying to calculate the pH of a solution of a weak acid in water. When the concentration of the weak acid is low enough, we need to consider the effect of ionization of the water itself (ie, autoprotolysis of water), since the order of magnitude of this ionization is the same as the order of magnitude of the ions due to the weak acid in this case of very low concentration.

There is a specific formula that can be used for this, which is shown below.

K_a is a constant that depends on the weak acid being considered, and K_w is a constant as well (for the autoprotolysis of water). I am interested in solving for the concentration of hydronium, given an initial concentration of the weak acid [HA]_initial.

I'd like to solve the equation for different values of [HA]_initial.

In the calculations below, I am trying to solve for in the expression

K__a = ([H__3*LinearAlgebra:-Transpose(O)])([H__3*LinearAlgebra:-Transpose(O)]-K__w/[H__3*LinearAlgebra:-Transpose(O)])/(HA__initial+[-H__3*LinearAlgebra:-Transpose(O)]+K__w/[H__3*LinearAlgebra:-Transpose(O)])

Below, I use as the concentration of hydronium and

evalf(solve((x^2-10^(-14))/(10^(-3)-x+10^(-14)/x) = 5*10^(-3), x))

(1)

If I use a manual simplification (noting that is extremely small compared to reasonable values of , then I am able to get real solutions.

evalf(solve(x^2/(10^(-3)-x) = 5*10^(-3), x))

(2)

Now, looking at the complex solutions, the imaginary parts are very small and the real parts of two of the three solutions match the real solutions above, so I guess perhaps I could have just ignored those complex parts.

On the other hand, if I try to do similar calculations but with , I get

evalf(solve((x^2-10^(-14))/(10^(-6)-x+10^(-14)/x) = 5*10^(-3), x))

(3)

evalf(solve(x^2/(10^(-6)-x) = 5*10^(-3), x))

(4)

So here, I don't see a complex solution that looks like the positive real solution above. Which means that I am not sure if the equation with the complex solutions (which is not simplified) is useful to me (I am doing various such calculations with different values of .

Download concentrations.mw

Plotting pH of a solution as a function of an init...

Asked by: zenterix 405

April 26 2025

0 4

Today I was working on some plots involving pH, which is defined as -log_10 ([hydronium]), that is, the negative of the base 10 logarithm of the concentration of hydronium in a solution.

I reached an expression for a variable x that is a function of an initial concentration C_i.

I wanted to plot the pH given by -log_10 (0.0001+x).

Note that x(0)=0, and so for this latter plot we should have the point (0, 4).

I am not able to see any part of the plot near (0,4), as can be seen below.

Download plotatzero.mw

In the context of nonlinear dynamics, how should o...

Asked by: GFY 75

April 23 2025

2 11

I aim to conduct a numerical frequency sweep analysis on a nonlinear, coupled two-degree-of-freedom vibration model and compare the results with the analytical solution. However, I am currently facing two main difficulties:

I am unsure how to determine whether the computed response has reached a steady state. If I simply use the maximum value to represent the steady-state amplitude, it can be misleading—since transient responses prior to reaching steady state may yield a higher peak, as seen in the uploaded code.
I do not know how to properly select the steady-state result from the previous frequency as the initial condition for the next frequency step.
2.mw

Approximation of ODEs with cubic splines

by: Kiefer Gerhard 20 Product: Maple 2022

April 06 2025

1 0

Approximation_of_ODEs_with_Cubic_Splines.mw

Show steps for viscous Burger's equation...

Asked by: 130

March 11 2025

0 2

Hi!

I am studying Burger's equation, and I would like to see the steps that Maple takes to solve this. "ShowSteps" doesn't seem to work.

Unfortunately, I am unable to share the worksheet I made.

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How to simplify the arguments of the exponential?...

Asked by: FZ 30

March 07 2025

0 4

How do we simplify the arguments of the exponential in (1)? Further how to express (1) into hyperbolic/trig functions?

>

(1)

>

(2)

>

(3)

>

(4)

>

(5)

>

Download argument.mw

How to perform simplifications to find unknown con...

Asked by: zenterix 405

March 03 2025

2 5

I am taking a course in quantum mechanics and trying to do the calculations in Maple.

In the worksheet below I try to solve a system of equations to find some constants.

It isn't very relevant that the domain is physics.

The issues I have are with manipulations of expressions (specifically, simplifications).

Levine - Ch. 2.4 - Particle in a Rectangular Well

Define

s__1 := sqrt(2*m*(V__0-E))/`&hbar;` = 2^(1/2)*(m*(V__0-E))^(1/2)/`&hbar;`

= -2^(1/2)*(m*(V__0-E))^(1/2)/`&hbar;`

(1)

and are wavefunctions.

=

I would also like to assume that and cannot both be zero simultaneously.

As can be seen above, there are four unknown constants, and .

We can obtain values for these unknowns by imposing boundary conditions.

Some of the boundary conditions involve the first derivatives of the wavefunctions.

=

The boundary conditions are

`ψ__1`(0) = `ψ__2`(0)*`ψ__2`(l) and `ψ__2`(0)*`ψ__2`(l) = `ψ__3`(l)*(D(`ψ__1`))(0) and `ψ__3`(l)*(D(`ψ__1`))(0) = (D(`ψ__2`))(0)*(D(`ψ__2`))(l) and (D(`ψ__2`))(0)*(D(`ψ__2`))(l) = (D(`ψ__3`))(l)

My question is how to find the constants in Maple.

In this problem, and .

Solving the first boundary condition is easy.

$expr1 := solve(`ψ__1`(0) = `ψ__2`(0), {C})$ =

Next, I solve the third boundary condition and sub in the result from solving the first boundary condition. We get in terms of .

$expr3 := eval(solve(`ψ__1,d`(0) = `ψ__2,d`(0), {B}), expr1)$ = ${B = A*(m*(V__0-E))^(1/2)/(m*E)^(1/2)}$

Already here it is not clear to me why Maple does not cancel the 's.

Next, consider the second boundary condition

$expr2 := solve(`ψ__2`(l) = `ψ__3`(l), {G})$ = ${G = (A*cos(s*l)+B*sin(s*l))/exp(s__2*l)}$

We can obtain in terms of just by subbing in previously found relationships, as follows

${G = exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A*(sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2)+cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2))/E^(1/2)}$

(4)

Finally, consider the fourth boundary condition.

$solve(`ψ__2,d`(l) = `ψ__3,d`(l), {G})$

${G = (m*E)^(1/2)*(A*sin(2^(1/2)*(m*E)^(1/2)*l/`&hbar;`)-B*cos(2^(1/2)*(m*E)^(1/2)*l/`&hbar;`))/((m*(V__0-E))^(1/2)*exp(-2^(1/2)*(m*(V__0-E))^(1/2)*l/`&hbar;`))}$

(5)

$expr4 := simplify(subs(expr3, solve(`ψ__2,d`(l) = `ψ__3,d`(l), {G})))$

${G = (sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2)-cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2))*exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A/(V__0-E)^(1/2)}$

(6)

${exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A*(sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2)+cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2))/E^(1/2) = (sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2)-cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2))*exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A/(V__0-E)^(1/2)}$

(7)

At this point, we have this complicated expression.

The calculations above come from the book "Quantum Chemistry" by Levine. The variable "result" above should, after a few more steps of manipulating the expression, give us

(8)

I've done the calculation by hand. It is not difficult. The exponentials and the cancel on each side and we are left with an equation involving and which easily comes out to the desired equation above.

I would like to be able to do it here in Maple.

The first thing I would do at this point is to start eliminating some of the clutter. For example, I would like to replace with .

This is a first task (that I have asked about in a separate question).

Then, I would want Maple to do the cancellations for me, but I have already used "simplify".

Finally, I think I would like to collect terms involving and .

Here is what happens when I try to get Maple to solve the equations for me in one go

$solve({`ψ__1`(0) = `ψ__2`(0), `ψ__1,d`(0) = `ψ__2,d`(0), `ψ__2`(l) = `ψ__3`(l), `ψ__2,d`(l) = `ψ__3,d`(l)}, {A, B, C, G})$

(9)

Download solve_constants.mw

How to simplify factors that are square roots into...

Asked by: zenterix 405

March 03 2025

0 2

In a calculation, I obtained the following expression from Maple.

>	expr1:=Asqrt(-m(-V__0 + E))/sqrt(m*E)

A*(-m*(-V__0+E))^(1/2)/(m*E)^(1/2)

(1)

The first question I have is why the

Next, consider the expression

>	expr2:=exp(sqrt(2)sqrt(m)sqrt(-V__0 + E)lI/`&hbar;`)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(2)

>	simplify(expr2,[E-V__0=y])

(3)

>	simplify(expr2,[2*m=y])

exp(I*y^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(4)

Why doesn't the simplification below work?

>	simplify(expr2,[2m(E-V__0)=y])

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(5)

I also tried with other commands

>	eval(expr2,2*m=y)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(6)

>	eval(expr2,E-V__0=y)

(7)

>	eval(expr2,2m(E-V__0)=y)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(8)

>	algsubs(2*m=y,expr2)

exp(I*y^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(9)

>	algsubs(E-V__0=y,expr2)

(10)

>	algsubs(2m(E-V__0)=y,expr2)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(11)

In reality, I would like to the entirety of sqrt(2m(E-V__0)/hbar) a variable by the name of s.

Download simplify_roots.mw

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