Not sure exactly how i could achieve this but:

how do i determine the value of k for which the graphs p(x) = x^2+2x+3 and q(x) = k+5x-7x^2 enclose an area of exactly 36?

I have to do it in maple and using i guess area under the curve.

Thanks

I need to show that the least square solution x that I obtained as x=pseudoinverse(A).b is the solution of Transpose(A).A.x=Transpose(A).b with the smallest norm. I've obtained the norm for the RHS of this expression as well as the norm of x but I'm unsure of how to conclude that this is the lowest possible norm using these values

The issue I am currently having is that, while analyticity (and physics) indicates a certain function must have roots, fsolve is having trouble finding them. In fact, I have even found roots manually in a certain region myself, simply inputting into the function various values until I found them. However, fsolve does not seem to want to find these roots, and I believe it is a numerics issue: when I changed the digits around, for extremely low values of Digits, it would find a root (even though it was incorrect). Further, this problem arose elsewhere in the domain of interest for other values of Digits (in particular, for Digits:=5, fsolve failed in a region it had not failed before).

The region of interest is the "peak" of the output of poleR(M0, 0.935, mK), which should be somewhere around M0 = 0.95 or so. However, because fsolve cannot find the roots, the plot cannot be made.

Anyone have any ideas as to why fsolve cannot find the roots? I was also experiencing issues with some of these functions having multiple roots, which itself is weird as well (note that I am working over the complex plane).

Attached is the document.

pole-dragging-mapleprimes.mw

Any ideas?

Hi everybody!

It's nice to join in this forum.

I'm trying to get the analytic solution of the Bernouilli-Euler beam equation, with the next boundary conditions:

w(x,t) = displacements.

w(0,t) = 0   -> It's a cantilever beam. At the x=0 it's clamped.

diff(w(x,t),x) = 0.   -> the gyro in the clamp is zero.

E*I*diff(w(L,t),x,x) = 0  -> the moment at the end of the beam (x=L) is zero.

E*I*diff(w(L,t),x,x,x) = 0  -> the shear at the end of the beam (x=L) is zero too.

I'm not able to introduce the second and the third derivatives as boundary conditions in the pdsolve equation. I post the hole code:

restart;
ode := I*E*(diff(w(x, t), x, x, x, x))+m*(diff(w(x, t), t, t)) = 0;

s := pdsolve(ode, w(x, t));

ode1 := op([2, 1, 1], s);

ode2 := op([2, 1, 2], s);

f1 := op(4, rhs(ode1));

f2 := op(2, rhs(ode2));

sol1 := dsolve(ode1, f1);

sol2 := dsolve(ode2, f2);

sol := rhs(sol1)*rhs(sol2);

conds := [w(0, t) = 0, (D[1](w))(0, t) = 0, eval(I*E*(D[1, 1](w))(x, t), x = L) = 0, eval(I*E*(D[1, 1, 1](w))(x, t), x = L) = 0];

pde := [ode, conds];

pdsolve(pde, w(u, t));

And I get this error:

"Error, (in PDEtools/pdsolve) invalid input: `pdsolve/sys` expects its 1st argument, SYS, to be of type {list({`<>`, `=`, algebraic}), set({`<>`, `=`, algebraic})}, but received [I*E*(diff(diff(diff(diff(w(x, t), x), x), x), x))+m*(diff(diff(w(x, t), t), t)) = 0, [w(0, t) = 0, (D[1](w))(0, t) = 0, I*E*(D[1, 1](w))(L, t) = 0, I*E*(D[1, 1, 1](w))(L, t) = 0]]"

It's seems I'm introducing the Boundary conditions of the second and third derivatives in a wrong way, but I can't discover how to do it.

Thanks very much in advance to everybody!!

Ger89

P.D. - I have use this "tutorial" to write the code ( http://homepages.math.uic.edu/~jan/mcs494f02/Lec34/pde.html ). Thanks very much again. 

I am trying to write a question which asks the student to convert a random decimal number into hexadecimal. I can't get MapleTA to recognize that a correct result is input. I think the problem is that I am not correctly dealing with the hexadecimal number which is represented as a symbol from convert.  

In the algorithm section I use:

$dec=range(100,200,5);
$hhh=maple("convert($dec,hex)");

The question was:

Convert the decimal bumber $dec into Hexadecimal.   

In the text input section I use:

is(($RESPONSE)-($hhh) = 0);

in the grading section of a Maple graded part of the question. I have tried converting the $RESPONSE to decimal and comparing with $dec.

I allow the students 4 attempts at each question, so I don't want to do a multiple choice question.

Hi

What can i do to prevemt maple from resizing my drawings and inserting empty space beneath drawings?

It happens when i export to pdf or print the document out.

Is there any other way i can export to a readable format without this happening?

Hello

I want to fill an area between multiple numerical curves. I have shown my curves in the below picture. Could you please help?

Thank you.

I've got a vector x=[0.36,1.3279,1.6882] (1*3) obtained as x:=pseudoInverse(A)*b where A=[<2,4,1>|<1,-1,1>|<3,3,2>] where <2,4,1>,<1,-1,1>,<3,3,2> are the columns of A and b is [8,5,4](1*3). Now, when I find the rowspace of A using RowSpace(A) I get the row vectors <1,0,1> and <0,1,1>, neither of which are equivalent to x. How do I arrive at my result that x is in the rowspace?

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

Hi,

I'm pretty new into MAPLE andI'm trying to get into it.

I have these four equations:

eq1:=1.6*10^(-7)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)-1.6*10^(-14)*R^2*cos(t)^2+4.2*10^(-14)*R^2-1.3+2.1*10^(-9)*R*cos(t)=0;

eq2 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.9*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0;

eq3 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.2*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0;

eq4 :=2.1*10^(-9)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)+1.6*10^(-14)*R^2*cos(t)^2+2.6*10^(-14)*R^2-1.3+1.6*10^(-7)*R*cos(t)=0;

i want to solve the 2nd order ODE y''=0.8y'arctan(x)+xy    -2<x<1

the maple code used is:

>maplesol := dsolve({ (D(D(y)))(x) =0.8*arctan(x)*(D(y))(x)+x*y(x),y(-2)=1.3,D(y)(-2)=0.7});

and maple only gives out maplesol:=

I think its something to do with the arctan(x) because when i remove it, i get an answer.

I am a middle school teacher trying to program an app where a student will type in a linear equation in any form. They are to solve this for the variable Y and graph the line. I have most of this up and running, but I would like to include a step by step help system, where after entering the equation, the students can click on a button and it will display the 1st step to solve the equation. I am using the built-in function “LinearSolveSteps” in the Students [Basics] package. Every repeated click of the button would reveal the next step. I have this working to a point, but I don't seem to be able to add the next step into the math container without deleting the contents that are already in there. I have tried rebuilding the string by using the "+" but the + shows up in the equation and it is not on a new line. Any ideas would be appreciated. Thank you.

It seems a frequent issue that exported 3d plots are not shown as wished. I experience the same problem. Although I exported in the .eps format into a .tex latex-file the resulting .pdf-file shows a somewhat pixelated image of my 3d plot as if it was created in "Paint". Is there a solution for this in Maple13?

Take a 3d plot of some uneven surface in the xyz space and you want to have the surface colored according to local z-coordinates (e.g. a valley is blue, a peak is red with all rainbow colors inbetween). For such a color-coding one can in principal select the plot option "Color->Z (Hue)". What do you do if the valley and the peak are still more or less green colored? How can you force the valley and peak to have different colors?

I have this surface which I display with the following commands:

     Belt:=plots[surfdata](Surface,color=C,labels=["x","y","z"]):

     display({...,Belt});

It all works, and the surface has nice default rainbow colors, but when I choose Color->Z (Hue) everything turns to green. Also when I manually write in the first of the above lines "color=["Blue","Red"]" the surface turns all red. The ranges of the x and y axis are larger than the z axis by a factor of about 50. Might this be the problem? How can I adjust the color ranges for my small z axis?

Hello

I wqant to minimize a function that has som parameters (here number of parameters are two). how can i do that?

I have attache a picture from my target function. Could you please help me?

Tahnk you.