## 1762 Reputation

12 years, 49 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.

## Define the center as a "point"...

with(geometry):

point(cent,2,3);

circle(C,[cent,5]);

This will define a circle named C, center at (2,3), and radius 5.

## Tools/Options?...

Because the Help icon is mentioned, I infer that the question is about the Maple icons shown in a launched copy of Maple.

If that inference is correct, then note that the Tools/Options dialog contains a setting for large icons under the Interface tab.

## The secret is the Equate command...

The old linalg package required the use of the evalm command. The new LinearAlgebra package does not. The old linalg package use the "matrix" command but the new LinearAlgebra package uses the "Matrix" command. The noncommutative multiplication operator in LinearAlgebra is the period, not &*. So, it would be helpful to work within the confines of the newer LinearAlgebra package.

The simplest way to equate structures elementwise is to use the Equate command.

A complete solution to the stated problem is given in the attached worksheet.Matrix_calculation.mw

## Use a sequence not a loop...

First, when you use a[i] as a variable (i.e., a Maple name), the element a[i] is actually a member of a table whose name is also "a". So, when you assign each equation to the same name "a" each time the loop cycles, you end up with just one equation, namely, the last one. And you have really clobbered your table.

So, rather than name each equation and then later feed all the names to another command, generate a sequence of unnamed equations. For example, write eqns := seq(a[i]+a[i+1]=i^2, i=0..99). Thus, you can send these equations to the solve command with the syntax solve({eqns}), However, you will have 100 equations in 101 unknowns (a[0], a[1], ..., a[100]), so Maple will pick the indeterminate variable. Either supply one more equation or provide the set of names you want solved for.

Perhaps you could use solve({eqns},{seq(a[k],k=1..100)}). This will return a[1],...,a[100] in terms of a[0], provided solve can obtain the solutions. If you have to resort to a numeric solution via the fsolve command, then there can be no indeterminates, and you definitely would have to provide that one more equation to have the same number of equations as variables.

## Built-in Tools...

Carl Love has provided a complete solution based on first principles. There are three built-in tools that would have drawn the region of integration and provided an appropriate integral for the volume inside the sphere but outside the cylinder.

The VolumeOfRevolution tutor in Student Calculus1 will implement the method of shells for any vertical axis of rotation.

There are two relevant Task Templates in Tools/Tasks/Browse: Calculus-Multivariate/Integration/Visualizing Regions of Integration/3D, one for integration in cylindrical coordinates, and one for integration in spherical coordinates.

See the attached worksheet for details.Tools_for_Volume.mw

## Use the CenterOfMass command...

The CenterOfMass command in the Student MultivariateCalculus package will provide the CM for both 2D and 3D regions that can be swept by a single mulltiple integral. Give the density function and the ranges of integration as equations of the form x=a..b, y=c..d, etc, and Maple returns either the unevaluated integrals, of the values for the coordinates of the CM.

## Use dchange...

The dchange command in the PDEtools package will also do it.

PDEtools:-dchange(_Z1=n,symbolic,[n])

## Need "surd"...

q:= (x^(5/3)-5*x^(2/3)) ;

Q:=convert(q, surd);

plot(Q,x=-1..1);

Roots are principal roots, and can be complex numbers. To obtain the real root in these cases, change the expression to a surd (an old-fashioned word for radical).

## Syntax for Int command...

RJL Maplesoft

The OP mentions "...I can't get the integral to work with the dy." This leads me to suspect that the OP, apparently new to this forum and probably to Maple, does not realize that the Int command does not require a user to input the differential dy. The syntax for an unevaluated integral is Int(f(y),y=a..b). Maple really does not have a concept of a "differential".

If the integral defining BesselJ(0,x) is implemented as an inert integral by means of the Int command, then it can be evaluated by applying the value command.

## Yes, it takes some retraining....

RJL Maplesoft

I agree that without a global switch to see which text-field equations are "live" it can be tedious to hover over each one to check.

Prior to Maple 2016, one would exit typeset math by pressing Function Key F5. Now, you have to train yourself to press Shft-F5 to make the typeset math inert, then press F5 again to exit the typeset math field.

I have asked GUI to fix this so that Shift-F5 both makes the typeset math inert, and simultaneously exits the field. We'll have to see if the next version of Maple makes this more convenient.

## Two issues...

RJL Maplesoft

Looks to me like there are two problems in the last line where delta is computed. The inner sum needs its range inside the parenttheses, and the second sum needs a range, not just the equation j=i. The first problem is easily corrected, but the second one appears to be structural. You might need to rethink what you are trying to calculate in delta.

## Try the MultivariateCalculus Package...

rlopez

There are tools for this in the Student MultivariateCalculus package.

with(Student:-MultivariateCalculus):

L:=Line([1,2,3],[1,-1,1]);

GetRepresentation(L,form=combined_vector);

GetPlot(L);

Now all of these operations are also possible via the Context Menu system after the package has been installed. Also, the form of the line is not restricted to the one chosen here. See the help page for GetRepresentation.

## Try the intersectplot command...

@Annonymouse  Add in a graph of the intersection drawn by the intersectplot command from the plots package.

## One line missing...

lopez@maplesoft.com

There is no value declared for x[0]. Add such a line to the first batch of initializations and the code runs.

## Root-Locus...

If w is real and z complex, this is the root-locus problem of control theory. As noted by other contributors, there is no way to solve a sixth-degree polynomial equation in closed form, so obtaining exact (analytic) solutions for z=z(w) is a challenge. Depending on what you need z(w) for, some of the following devices might be of help.

Although Maple's plots package contains a rootlocus command, it numerically solves and graphs solutions for 1+k*h(z), where k is a real parameter (called a "gain" in the language of controls).

If z is real, then a graph of z=z(w) can be obtained by solving for w=w(z), then graphing via plot([w(z),z,z=a..b]).

If z is complex, solve for w=w(z) and replace z with x+I*y. Let Wr be the real part and Wi be the imaginary part. (This takes evalc(Re(... and evalc(Im(... but for polynomials, Maple can do this.) Now, if w is real, then Wi is zero, so an implicitplot of Wi=0 is a graph of the root locus. Each point on the resulting graph is a z=x+I*y value in the xy-plane for which some w satisfies the original equation P=0. The trick now is to obtain that w-value for each point on the curve.

For example, pick an x-value and count how many y-values will correspond. Then set x in Wr equal to that value and fsolve for y. Make a procedure of this process and you will be able to see the values of w for each z on the root locus. In control theory, a value of z indicates a possible state of a system. The idea is to determine the gain that will put the system into that state, so knowing the gain as a function of the state is a useful thing, something not ordinarily discussed in elementary texts in the subject.

As an isolated mathematical problem, I have always found this topic interesting. Just what is the trajectory in the z-plane of the solutions of P=0 as w varies, even if w isn't real. It's a hard problem to solve directly, so looking at it "backwards" certainly leads to interesting options.

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