Is the smallest unit of time that Maple can use to report the computation time millisecond? When I use time(...) the output is a floating-point number with at most three digits after the floating point and since the unit in Maple help is said to be seconds, I'm thinking the smallest time unit it will report is millisecond. Is there any possibility to have more digits in the time reports, let's say up to 10^-5 seconds?

My interface has frozen, but above is a screen shot of what is by far the most unusual response from the CAS in the i guess 8 or so years ive been using it in total.

*updated situation its allowing me to interrupt evaluation

My "for loop" index is being treated as a letter and not a number. The 'h' in my for loop is is supposed to be the index for these tables but the result is tables that have indices of the letter 'h'.  

The indexing for the 'Z' part works Z[h][1] ,Z[h][2] and Z[h][3] have numbers instead of the letter 'h' but the x1[h], y1[h] and Q[h][1],Q[h][2] and Q[h][3] are all 'h' indexed with no number as an index when I view this table in the variable viewer.

for h to 50 do :

Z[h][1] := s -> evalf(subs(x = x1[h](s), y = y1[h](s), Q[h][1])) ;

Z[h][2] := s -> evalf(subs(x = x1[h](s), y = y1[h](s), Q[h][2]));

Z[h][3] := s -> evalf(subs(x = x1[h](s), y = y1[h](s), Q[h][3])); 

end do:

the result is just using the last value of index 'h' and puting it in the first index of 'Q'. There is nothing defined with index 51.

Z[2][3](4*(1/50));
                            Q[51][3]
Z[9][3](4*(1/50));
                            Q[51][3]
 

if I hard code the index numbers then it works correctly but I do not want to hard code all 50 of these.  

Why is the for loop index variable 'h' treated as a letter and not evaluated to be a number in the indexing?

thanks . . . 

Although a very general question... does the rope break or not?

A 100 kg weight is hung from the center of a 1/4" polypropelene rope making an angle of 10 degrees at both ends with the horizontal.  According to some characteristics, the breaking strength of the rope is 1125 lbs or 5kN.  The safe load (safety factor of 12) drops the weight to  93.8 lbs or 0.417kN. Find the tension in the rope?  Is it strong enough?

Fairly easy to solve in Maple - just two equations two unknowns making t1 tension on one side and t2 tension on the other side

eq1 := t1*cos(80*Pi*(1/180))+t2*cos(80*Pi*(1/180)) = 100*9.8 # ΣFy=mg

eq2 := t1*sin(80*Pi*(1/180))-t2*sin(80*Pi*(1/180)) = 0 #ΣFx=0 hence basically t1=t2

solve([eq1,eq2])

                 {t1 = 2821.797537, t2 = 2821.797537}

Therefore the tension on the rope t1=t2 is 2821 N  (looks like way beyond the safety factor) so this means the rope will not break, correct?  or is it that the total tension becomes t1+t2=5642 N putting us over the 5kN and the rope breaks?

Hi!
I'm very new to Maple, so I apologize ahead if this is a stupid question.

It seems that my "Determinant" function is not working properly, as shown in the picture.
I simply defined the matrix M and called for the determinant, as instructed by the help system in Maple.

Does anyone know what's happening?

Thank you very much.

Hi

I need some code in maple thats help me to verify my hand notes:

Find the set of points z  ( z complex number ) such that the three points A=1, B=z, and C=1+z^2  will be  aligned.

Many thanks

B(t)B' '(t)A(t) - A(t)(B'(t))^2 -A' '(t)(B(t))^2 + A'(t)B(t)B'(t) - A(t)=0

hi

im beginner in maple, i have a parametric matrix that i want to muliply it in a vector of differential equations like below.

i have functions like thees;

u(x,y,z)=u₀(x,y)+z*u₁(x,y)

v(x,y,z)=v₀(x,y)+z*v₁(x,y)

matrix:  A=[a,b;c,d]

vector : b = [du/dx, dv/dy]

now i want to do like this:

C=A.B

but i dont know how to do this.

I am new to maple. how do i segregate different term in long PDE in maple

The development of the calculation of moments using force vectors is clearly observed by taking a point and also a line. Different exercises are solved with the help of Maple syntax. We can also visualize the vector behavior in the different configurations of the position vector. Applications designed exclusively for engineering students. In Spanish.

Moment_of_a_force_using_vectors_updated.mw

Lenin Araujo Castillo

Ambassador of Maple

The MAPLE worksheet associated with the link below attempts to generate a sequence of signals to be symmetric about t=0.  It worked for the 1^st plot, but not for the 3^rd plot.  I believe the problem resides with the Heaviside function.  I need to somehow create a function that is the mirror image of the Heaviside function.  Does anyone know how to do that?

untitled4.mw

As can be seen in the example i have shown below, i have iterated the "collect" function a number of times as required for the number of distinct terms. Can someone please show me how i would specify that i would like to iterate a particular function 'N' times in the same manner i have done only 3 times in the example, ie suppose instead of the 3 variables i wish to collect for here (a[0],a[1],a[2]) i now must do so for another definition of F that has 'N' variables (a[0],a[1],a[2],....a[N]) 

Download Asking_Again_About_iterations.mw

Currently working on Fourier series. Having trouble plotting the graph of the Fourier series and would love to add and use a package such as the one found here: https://www.maplesoft.com/applications/view.aspx?SID=4857 . I'm having trouble installing and using the package, though. Could anyone be so kind as to assist me in adding and using packages not already found in Maple?

I have the following integral, which is solved. int(14*t*exp(-t/3),t=0..infinity); It equals 126. I now need to find 90%, of the integral. This is just math and the answer is 113.40. The challenge is I need to prove this using maple. So I need to show how to get the upper limit to 11.67, which yeilds the answer of 113.40(rounded).  I just started pluging in numbers for the upper limit till I got my answer, but clearly, this is now what my instructor wants. I need the process for finding the upper limit. Any help would be appreciated.