Product Tips & Techniques

Tips and Tricks on how to get the most about Maple and MapleSim

Who else likes art?  I love art; doodling in my notebook between projects and classes is a great way to pass the time and keep my creativity sharp.  However, when I’m working in Maple Learn, I don’t need to get out my book; I can use the plot window as my canvas and get my drawing fix right then and there.

We’ve done a few blog posts on Maple Learn art, and we’re back at it again in even bigger and better ways.  Maple Learn’s recent update added some useful features that can be incorporated into art, including the ability to resize the plot window and animate using automatically-changing variables.

Even with all the previous posts, you may be thinking, “What’s all this?  How am I supposed to make art in a piece of math software?”  Well, there is a lot of beauty to mathematics.  Consider beautiful patterns and fractals, equations that produce surprisingly aesthetically interesting outputs, and the general use of mathematics to create technical art.  In Maple Learn, you don’t have to get that advanced (heck, unless you want to).  Art can be created by combining basic shapes and functions into any image you can imagine.  All of the images below were created in Maple Learn!

There are many ways you can harness artistic power in Maple Learn.  Here are the resources I recommend to get you started.

  1. I’ve recently made some YouTube videos (see the first one below) that provide a tutorial for Maple Learn art.  This series is less than 30 minutes in total, and covers - in three respective parts - the basics, some more advanced Learn techniques, and a full walkthrough of how I make my own art.
  2. Check out the Maple Learn document gallery art collection for some inspiration, the how-to documents for additional help, and the rest of the gallery to see even more Maple Learn in action!

Once you’re having fun and making art, consider submitting your art to the Maple Conference 2022 Maple Learn Art Showcase.  The due date for submission is October 14, 2022.  The Conference itself is on November 2-3, and is a free virtual event filled with presentations, discussions, and more.  Check it out!


I have polished up findings with custom components to share it here:

Optimized code generated with Maple’s codegen package cannot be used in the same way as it was possible with older versions of MapleSim’s Custom Component Template.

Intermediate variables `tx` (where x is an integer) of the optimized code are interpreted as physical parameters in the current template version and not as variables. This makes sense and is more consistent with MapleSim’s definition of variables and parameters, but leads to errors in MapleSim.

The attached model shows how optimized code can be generated for the current template and compares an older, still working (!) template with the new one.

The attached worksheet contains commands to programmatically generate optimized code for the current Custom Component Template.



New display of arbitrary constants and functions


When using computer algebra, first we want results. Right. And textbook-like typesetting was not fully developed 20+ years ago. So, in the name of getting those results, people somehow got used to the idea of "give up textbook-quality computer algebra display". But computers keep evolving, and nowadays textbook typesetting is fully developed, so we have better typesetting in place. For example, consider this differential equation:



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

Problem statement:
Determine the relativistic uniformly accelerated motion, i.e. the rectilinear motion for which the acceleration w in the proper reference frame (at each instant of time) remains constant.

As an application of the post presented by Dr Cheb Terrab in MaplePrimes on the principle of relativity ( found here ), we solve the problem stated on page 24 of Landau & Lifshitz book [1], which makes use of the relativistic invariant condition of the constancy of a four-scalar, viz., `w__μ`*w^mu where w^mu is the four-acceleration. This little problem exemplify beautifully how to use invariance in relativity. This is the so-called hyperbolic motion and we explain why at the end of this worksheet.


let's introduce the coordinate system, X = (x, y, z, tau)with tau = c*t 


Setup(coordinates = [X = (x, y, z, tau)])

[coordinatesystems = {X}]


%d_(s)^2 = g_[lineelement]

%d_(s)^2 = -Physics:-d_(x)^2-Physics:-d_(y)^2-Physics:-d_(z)^2+Physics:-d_(tau)^2





The four-velocity is defined by  u^mu = dx^mu/ds and dx^mu/ds = dx^mu/(c*sqrt(1-v^2/c^2)*dt) 

Define this quantity as a tensor.

Define(u[mu], quiet)

The four velocity can therefore be computing using

u[`~mu`] = d_(X[`~mu`])/%d_(s(tau))

u[`~mu`] = Physics:-d_(Physics:-SpaceTimeVector[`~mu`](X))/%d_(s(tau))



As to the interval d(s(tau)), it is easily obtained from (2) . See Equation (4.1.5)  here with d(diff(tau(x), x)) = d(s(tau)) for in the moving reference frame we have that d(diff(x, x)) = d(diff(y(x), x)) and d(diff(y(x), x)) = d(diff(z(x), x)) and d(diff(z(x), x)) = 0.

 Thus, remembering that the velocity is a function of the time and hence of tau, set

%d_(s(tau)) = d(tau)*sqrt(1-v(tau)^2/c^2)

%d_(s(tau)) = Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2)


subs(%d_(s(tau)) = Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2), u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/%d_(s(tau)))

u[`~mu`] = Physics:-d_(Physics:-SpaceTimeVector[`~mu`](X))/(Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2))


Rewriting the right-hand side in components,

lhs(u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))) = Library:-TensorComponents(rhs(u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))))

u[`~mu`] = [Physics:-d_(x)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics:-d_(y)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics:-d_(z)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), 1/(-(v(tau)^2-c^2)/c^2)^(1/2)]


Next we introduce explicitly the 3D velocity components while remembering that the moving reference frame travels along the positive x-axis


simplify(u[`~mu`] = [Physics[d_](x)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics[d_](y)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics[d_](z)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), 1/(-(v(tau)^2-c^2)/c^2)^(1/2)], {d_(x)/d_(tau) = v(tau)/c, d_(y)/d_(tau) = 0, d_(z)/d_(tau) = 0}, {d_(x), d_(y), d_(z)})

u[`~mu`] = [v(tau)/(c*((c^2-v(tau)^2)/c^2)^(1/2)), 0, 0, 1/(-(v(tau)^2-c^2)/c^2)^(1/2)]


Introduce now this explicit definition into the system

Define(u[`~mu`] = [v(tau)/(c*((c^2-v(tau)^2)/c^2)^(1/2)), 0, 0, 1/(-(v(tau)^2-c^2)/c^2)^(1/2)])

{Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], u[mu], w[`~mu`], w__o[`~mu`], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}



Computing the four-acceleration


This quantity is defined by the second derivative w^mu = d^2*x^mu/ds^2 and d^2*x^mu/ds^2 = du^mu/ds and du^mu/ds = du^mu/(c*sqrt(1-v^2/c^2)*dt)

Define this quantity as a tensor.

Define(w[mu], quiet)

Applying the definition just given,

w[`~mu`] = d_(u[`~mu`])/%d_(s(tau))

w[`~mu`] = Physics:-d_[nu](u[`~mu`], [X])*Physics:-d_(Physics:-SpaceTimeVector[`~nu`](X))/%d_(s(tau))


Substituting for d_(s(tau))from (1.2) above

subs(%d_(s(tau)) = Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2), w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/%d_(s(tau)))

w[`~mu`] = Physics:-d_[nu](u[`~mu`], [X])*Physics:-d_(Physics:-SpaceTimeVector[`~nu`](X))/(Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2))


Introducing now this definition (2.2)  into the system,

Define(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2)), quiet)

lhs(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))) = TensorArray(rhs(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))))

w[`~mu`] = Array(%id = 36893488148327765764)


Recalling that tau = c*t, we get


w[`~mu`] = Array(%id = 36893488148324030572)


Introducing anew this definition (2.4)  into the system,



In the proper referential, the velocity of the particle vanishes and the tridimensional acceleration is directed along the positive x-axis, denote its value by `#msub(mi("w"),mn("0"))`

Hence, proceeding to the relevant substitutions and introducing the corresponding definition into the system, the four-acceleration in the proper referential reads

  "Define(`w__o`[~mu]= subs(v(t)=`w__0`, v(t)=0,rhs(?)),quiet):"

w__o[`~mu`] = TensorArray(w__o[`~mu`])

w__o[`~mu`] = Array(%id = 36893488148076604940)



The differential equation solving the problem



Everything is now set up for us to establish the differential equation that will solve our problem. It is at this juncture that we make use of the invariant condition stated in the introduction.

The relativistic invariant condition of uniform acceleration must lie in the constancy of a 4-scalar coinciding with `w__μ`*w^mu  in the proper reference frame.

We simply write the stated invariance of the four scalar (d*u^mu*(1/(d*s)))^2 thus:

w[mu]^2 = w__o[mu]^2

w[mu]*w[`~mu`] = w__o[mu]*w__o[`~mu`]


TensorArray(w[mu]*w[`~mu`] = w__o[mu]*w__o[`~mu`])

(diff(v(t), t))^2*c^2/(v(t)^2-c^2)^3 = -w__0^2/c^4



This gives us a first order differential equation for the velocity.


Solving the differential equation for the velocity and computation of the distance travelled



Assuming the proper reference frame is starting from rest, with its origin at that instant coinciding with the origin of the fixed reference frame, and travelling along the positive x-axis, we get successively,


dsolve({(diff(v(t), t))^2*c^2/(v(t)^2-c^2)^3 = -w__0^2/c^4, v(0) = 0})

v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2), v(t) = -t*c*w__0/(t^2*w__0^2+c^2)^(1/2)



As just explained, the motion being along the positive x-axis, we take the first expression.

[v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2), v(t) = -t*c*w__0/(t^2*w__0^2+c^2)^(1/2)][1]

v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)


This can be rewritten thus

v(t) = w__0*t/sqrt(1+w__0^2*t^2/c^2)

v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2)


It is interesting to note that the ultimate speed reached is the speed of light, as it should be.

`assuming`([limit(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = infinity)], [w__0 > 0, c > 0])

limit(v(t), t = infinity) = c



The space travelled is simply

x(t) = Int(rhs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2)), t = 0 .. t)

x(t) = Int(w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = 0 .. t)


`assuming`([value(x(t) = Int(w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = 0 .. t))], [c > 0])

x(t) = c*((t^2*w__0^2+c^2)^(1/2)-c)/w__0


expand(x(t) = c*((t^2*w__0^2+c^2)^(1/2)-c)/w__0)

x(t) = c*(t^2*w__0^2+c^2)^(1/2)/w__0-c^2/w__0


This can be rewritten in the form

x(t) = c^2*(sqrt(1+w__0^2*t^2/c^2)-1)/w__0

x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0



The classical limit corresponds to an infinite velocity of light; this entails an instantaneous propagation of the interactions, as is conjectured in Newtonian mechanics.
The asymptotic development gives,

lhs(x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0) = asympt(rhs(x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0), c, 4)

x(t) = (1/2)*w__0*t^2+O(1/c^2)


As for the velocity, we get

lhs(v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)) = asympt(rhs(v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)), c, 2)

v(t) = t*w__0+O(1/c^2)


Thus, the classical laws are recovered.


Proper time



This quantity is given by "t'= ∫ dt sqrt(1-(v^(2))/(c^(2)))" the integral being  taken between the initial and final improper instants of time

Here the initial instant is the origin and we denote the final instant of time t.


`#mrow(mi("t"),mo("′"))` = Int(sqrt(1-rhs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2))^2/c^2), t = 0 .. t)

`#mrow(mi("t"),mo("′"))` = Int((1-w__0^2*t^2/((1+w__0^2*t^2/c^2)*c^2))^(1/2), t = 0 .. t)


Finally the proper time reads

`assuming`([value(`#mrow(mi("t"),mo("′"))` = Int((1-w__0^2*t^2/((1+w__0^2*t^2/c^2)*c^2))^(1/2), t = 0 .. t))], [w__0 > 0, c > 0, t > 0])

`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0


When proc (t) options operator, arrow; infinity end proc, the proper time grows much more slowly than t according to the law

`assuming`([lhs(`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0) = asympt(rhs(`#mrow(mi("t"),mo("′"))` = arcsinh(t*w__0/c)*c/w__0), t, 1)], [w__0 > 0, c > 0])

`#mrow(mi("t"),mo("′"))` = (ln(2*w__0/c)+ln(t))*c/w__0+O(1/t^2)


combine(`#mrow(mi("t"),mo("′"))` = (ln(2*w__0/c)+ln(t))*c/w__0+O(1/t^2), ln, symbolic)

`#mrow(mi("t"),mo("′"))` = ln(2*t*w__0/c)*c/w__0+O(1/t^2)



Evolution of the four-acceleration of the moving frame as observed from the fixed reference frame



To obtain the four-acceleration as a function of time, simply substitute for the 3-velocity (4.3)  in the 4-acceleration (2.4)

" simplify(subs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2),?),symbolic)"

w[`~mu`] = Array(%id = 36893488148142539108)


" w[t->infinity]^(  mu)=map(limit,rhs(?),t=infinity) assuming `w__0`>0,c>0"

`#msubsup(mi("w"),mrow(mi("t"),mo("→"),mo("∞")),mrow(mo("⁢"),mo("⁢"),mi("μ",fontstyle = "normal")))` = Array(%id = 36893488148142506460)


We observe that the non-vanishing components of the four-acceleration of the accelerating reference frame get infinite while those components in the moving reference frame keep their constant values . (2.5)


Evolution of the three-acceleration as observed from the fixed reference frame



This quantity is obtained simply by differentiating the velocity v(t)given by  with respect to the time t.


simplify(diff(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t), size)

diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)


Here also, it is interesting to note that the three-acceleration tends to zero. This fact was somewhat unexpected.

map(limit, diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2), t = infinity)

limit(diff(v(t), t), t = infinity) = 0



At the beginning of the motion, the acceleration should be w__0, as Newton's mechanics applies then


`assuming`([lhs(diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)) = series(rhs(diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)), t = 0, 2)], [c > 0])

diff(v(t), t) = series(w__0+O(t^2),t,2)



Justification of the name hyperbolic motion



Recall the expressions for x and diff(t(x), x)and obtain a parametric description of a curve, with diff(t(x), x)as parameter. This curve will turn out to be a hyperbola.

subs(x(t) = x, x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0)

x = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0