rlopez

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16 years, 153 days

Dr. Robert J. Lopez, Emeritus Professor of Mathematics at the Rose-Hulman Institute of Technology in Terre Haute, Indiana, USA, is an award winning educator in mathematics and is the author of several books including Advanced Engineering Mathematics (Addison-Wesley 2001). For over two decades, Dr. Lopez has also been a visionary figure in the introduction of Maplesoft technology into undergraduate education. Dr. Lopez earned his Ph.D. in mathematics from Purdue University, his MS from the University of Missouri - Rolla, and his BA from Marist College. He has held academic appointments at Rose-Hulman (1985-2003), Memorial University of Newfoundland (1973-1985), and the University of Nebraska - Lincoln (1970-1973). His publication and research history includes manuscripts and papers in a variety of pure and applied mathematics topics. He has received numerous awards for outstanding scholarship and teaching.

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These are answers submitted by rlopez

At the left edge of the lower toolbar for a graph, there's a drop-down box. It shows the word "Plot" by default, but the box also contains the option "Drawing." So, the Drawing toolbar still exists in Maple 2021.

For graphs drawn with the Plot Builder, the Drawing toobar is the default toolbar because all the Plot options are provided by the Plot Builder.

Also, after an initial scare that the animation toolbar no longer was available in Maple 2021, further experiments (Windows 10) verify that this toolbar does appear for animations created with the Plot Builder and with the plots:-animate command.

 

Under the radicals, the RootOf expressions require an explicit solution of a fifth-degree polynomial. In the 1800s, such formulas were proven not to exist. What you want is a mathematical impossibility. I believe that is the "something small" that has escaped you.

Let's simplify the issue by considering polar coordinates.

Points in polar coordinates are represented in the VC packages as "vectors," sums of components times fake unit vectors in a rectangular version of the polar plane. Thus, the point where r=a, theta=b has a polar representation in terms of unbarred basis vectors e_r and e_theta. This allows the polar point to be represented in this rectangular polar plane as a "position vector" in that plane, a vector from the "origin" of that plane to the "point" (r,t). This is what is meant by the "vector" r*e_r+t*e_t, where these basis vectors are the unbarred basis vectors.

That's why the VC packages also have barred basis vectors as the moving, point-wise determined basis vectors needed for vector fields. When changing coordinates in a vector field, the basis vectors also have to change. But this is not the case for "vectors" that represent points and use the unbarred basis vectors.

MapToBasis(Vector(<r,t>,polar),cartesian[x,y]) produces the Cartesian vector (and hence, the Cartesian point (r cos(t), r sin(t))) expressed with the unbarred basis vectors e_x and e_y that can be interpreted as i and j.

Thus, the help page is correct when it says just components are converted because the argument to PositionVector is never a VectorField, just a Vector that represents the points along a curve or surface. 

I do not see how Equation A becomes Equation B. Eqn A has c^2 on the right, but it becomes c^4 in Equation B. Could that be why the Equation Manipulator gave a messy result and not Equation B?

Here's the device I use. Not super convenient, but it works.

Into a GUI table place the input that creates the equation you want displayed. Execute the input. Then, use the table properties dialog to hide the input. (It's a checkbox near the bottom.) You can also hide the bounding lines of the table if you wish.

Re-execution of the worksheet re-executes code hidden in a table cell, so the equation label is re-constructed.

Execute the command 

?DetermineRadiusOfConvergence

The help page for this Task Template will be returned. You can use it to determine the radius of convergence of any power series. Convergence at an endpoint always has to tested separately.

The Task Template itself can be accessed directly from Tools/Tasks/Browse/Calculus-Integral/Series/Radius of Convergence.

If you really need f to be a function, then don't use the syntax f(x):=x^2. That does not create a Maple function.

Use instead, f := x-> x^2

If you don't need f to be a function, then just assign to the name "f" via the syntax f := x^2. In this case, f is not a function, and the input f(t) will return some nonsensical stuff, and not t^2.

implicitplot(eq,x=-4..4,y=-4..4)

What is puzzling, however, is "...recieve in response x^4=4." Is that actually the response or is that just a typo in the post?

After changing pi to Pi, the simplification succeeds if accompanied with the assumption that rho__m is real.

simplify(expression) assuming rho__m::real

Replace the function being integrated over the hemisphere with the generic f(x,y,z) and add the option "inert" to the integral so the unevaluated integral is returned. It will contain f(1,s,t), indicating that x, y, z, were respectively replaced with 1, s, and t, not with the correct x(phi,theta), y(phi,theta), and z(phi). The error is the incorrect handling of the function being integrated over the surface.

f := piecewise(t<0,0,t=0,1,t>0,0);

plot(f,t=-1..1,discont=[showremovable=0],color=red, thickness=3, symbol=solidcircle, symbolsize=15)

The essence of this approach is the use of the showremovable option. The other options to plot simply enhance the graph.

q := a^2 + 2*a*b + b^2 + c^2;
Student:-Precalculus:-CompleteSquare(q);

The attached worksheet shows a stepwise solution that a student would be able to implement by hand. It parallels what I learned more than 50 years ago when I first met this material.

By the way, the standard way to show the existence of a scalar potential is to show that the curl of the vector field is the zero vector. That's what all the OP's derivatives are - the components of the curl.

Scalar_Potential.mw

If you mean that you didn't get a 3D graph at the end, well, you did. It has constrained scaling, a vertical of 800, and a base 20x20. So the image is pretty squished. Remove the constrained scaling and you will see a surface.

The February 2013 Maple Reporter contained the article that this worksheet generated. A worksheet was used to generate a pdf that was then disseminated as the Reporter article. The worksheet was then stored in the Maple Application Center without any adjustments needed to make the worksheet run correctly.

I dug out the original worksheet and noted that in Maple 2020, several changes are needed for the worksheet to run properly now. Those adjustments are in the attached version of the worksheet.

Basically, the issues are that several of the display commands were never executed, so when the worksheet is re-executed stepwise with the single exclamation or entirely with the triple exclamation, these display statements won't execute. Then, The procedure in Method 3 was set in non-executable math. Also, the rules for declaring locals and procedure parameters appear to have changed in the last 7 years.

R-90_Animated_Trace_of_a_Curve_Drawn_by_Radius_Vector.mw

(The R-90 designation means this was the 90th article in a series that ran to a little over 100.)

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