rlopez

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14 years, 172 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

Again, please avoid using the old student package. It hasn't been actively supported for more than 15 years.

The following command will immediately evaluate to the correct answer.

int(1/(1+x+y+z)^3,[z=0..1-x-y, y=0..1-x, x=0..1])

Also relevant are the following two task templates:

Tools/Tasks/Browse/Calculus-Multivariate/Integration/MultipleIntegration/Cartesian 3-D

Tools/Tasks/Browse/Calculus-Multivariate/Integration/Visualizing Regions of Integration/Cartesian 3-D

Please stop using the unsuppprted and outdated student package. It's not been maintained for more than 15 years.

Maple does not recognize "inf" as the symbol for "infinity." Type out the full word or, in 2D math, use the appropriate symbol from the Common Symbols palette.

If you want to integrate over the first quadrant portion of the interior of a circle of radius "a", do not then integrate over the whole of the first quadrant.

The following command will immediately return the value of the integral.

int(1/sqrt(a^2-x^2-y^2),[y=0..sqrt(a^2-x^2),x=0..a])

There are two task-templates that are of interest here.

Tools/Tasks/Browse/Calculus-Multivariate/Integration/MultipleIntegration/Cartesian 2-D

Tools/Tasks/Browse/Calculus-Multivariate/Integration/Visualizing Regions of Integration/Cartesian 2-D

If you turn the help page into a worksheet (click the icon in the toolbar), you then can make changes in style. Unfortunately, you have to do this for each help page as you bring it up.

I believe the built-in command you want to use is "infnorm" in the numapprox package.

I tried it on the test example in this thread and it produces the correct result.

The thing to keep in mind when using spherical coordinates in the VectorCalculus packages is that the names of the variables are irrelevant. It's their position in the triple spherical[u,v,w] that determines the definitions of the variables. The first name in the triple is the radial variable; the second, the angle down from the z-axis; and the third, the angle around the z-azis.

Calling the middle variable "theta" as you have done with the command SetCoordinates('spherical'[r, theta, phi]), changes the name of the "dropping down" angle to theta, but doesn't make theta match the very first image in your post.

One of the benefits of using the Student VectorCalculus package is that for Cartesian, polar, cylindrical, and spherical systems, there are default coordinate variable names recognized. For spherical, it's [r, phi, theta], with phi, being the middle name, the angle down from the z-axis, as in the diagram in your post. Your figure follows the definition of spherical coordinates found in most math books. Physics and engineering texts tend to interchange the names of the two angles. Thus, they make theta the angle down from the z-axis, and phi the angle around the z-axis.

So, the engineer of physicist using the VectorCalculus package would define a vector field in spherical coordinates with the syntax

VectorVield(<f(r,theta,phi),g(r,theta,phi),h(r,theta,phi)>,coords=[r,theta,phi])

This overwrites the defaults in the Student package and precludes the need for SetCoordinates in the VectorCalculus package.

The default symbol for the imaginary unit (sqrt(-1)) is the letter I, in upper case. In the Common Symbols palette you see the three characters i, j, and I, all of which point to the default imaginary unit name, I. If you enter either the i or the j and hit the Enter key, the echo will be I.

So, in your trial worksheet, you typed the letter i, which is not interpreted as the i in the Common Symbols palette. It's just the letter i.

If the interface command is used to change from I to i as sqrt(-1), then a typed i would point to sqrt(-1).

In other words, what the symbols in the paletts do is establish a "dictionary" in which the palette symbol points to some underlying thing in Maple. If you were to type many of those same symbols, the typed character would not necessarily point to the appropriate underlying Maple item.

Finally, try hovering over the i, j, and I in the Common Symbols palette. You should see that all three point to the same thing, namely, the default imaginary unit, I.

with(Student:-MultivariateCalculus):

L:=Line([x1,y1,z1],[x2,y2,z2]);

simplify(Projection([x3,y3,z3],L));

In fact, these two calculations can be done in a syntax-free way via the Context Panel once the Student:-Multivariate package is loaded (which can also be done syntax-free via the Tools/Load Package menu).

Of course, the detailed solutions provided by the other contributors give insight into the math that the solution by built-in functions perhaps hides. But I always find it helpful to obtain an answer first, with a minimum of "fuss" before I dig in to see how the math works.

In the VectorCalculus package (Student or otherwise) there are two relevant commands, namely, PositionVector, and PlotPositionVector. The first is used to define a curve (or surface) parametrically; and the second, to graph the curve (or surface) along with any number of different vector fields. Here, the vector field to include would be tangents. Of course, this still requires two commands, and a look at the help pages for getting the syntax right, but I find this pair of commands to be both powerful and useful.

Given that the OP asks for derivatives with respect to b and varphi, let's assume that these are the independent variables, and that x and r are just parameters. And for simplicity, let's call varphi just phi. Then the analysis can be simplified to w=w(b,phi) with the constraint f(b,phi)=0.

I believe that the constraint implies there can be only one independent variable. So, if a derivative with respect to b is expected, then the functional dependencies are w(b, phi(b)) =  W1(b) and the derivative with respect to b would be obtained via

diff(w,b)+diff(w,phi)*implicitdiff(f,phi,b) (using Maple's implicitdiff command).

When I calculate d_W1/db this way, and replace phi with phi(b,x,r) from the constrant, I obtain the same result as Rouben did by eliminating phi immediately.

If a derivative with respect to phi is expected, the the functional dependencies are w(b(phi),phi) = W2(phi) and the derivative with respect to phi would be obtained via

diff(w,phi)+diff(w,b)*implicitdiff(f,phi,b).

When I calculate d_W2/d_phi by this recipe, and replace phi with it's equivalent, I do not get the second result that Rouben obtained. I suspect that w(x,b,r,phi(x,b,r)) does not lead to the correct derivative with respect to phi, given that r is assumed to be a parameter and that b is the other independent variable.

I looked at this problem because it appeared to be one of the more challenging ones of the type that appear in Advanced Calculus textbooks where the intricacies of the chain rule for partial derivatives are expounded. (The issue of determining the independent variables seems to be important here.) If my interpretation is flawed, I would be pleased to be enlightened.

 

 

Maple's pdsolve command can only solve such a BVP with just one edge carrying a nonhomogeneous BC. So, break this problem into a sum of problems, each with just one edge carrying a nonhomogeneous BC. Then add the resulting solutions. Let q be Laplace's equation. Since the left edge already has a homogeneous BC, there are just three subproblems to solve. The first and third of the following return infinite series; the second one returns a closed-form expression.

pdsolve({q, u(0, y) = 0, u(Pi, y) = sinh(Pi)*cos(y), u(x, 0) = 0, u(x, Pi) = 0})

pdsolve({q, u(0, y) = 0, u(Pi, y) = 0, u(x, 0) = sin(x), u(x, Pi) = 0})

pdsolve({q, u(0, y) = 0, u(Pi, y) = 0, u(x, 0) = sin(0), u(x, Pi) = -sinh(x)})

 

 

 

The given equation is quadratic in x, so for k not 7, there are two distinct solutions. The first one Maple returned is equivalent to the one you declared to be "correct."

Under the radical, bring the minus sign into the parentheses and split the radical into sqrt(k)*sqrt(7-k). The sqrt(k) out front combines with the k in the denominator to give sqrt(k) in the denominator. Then the sqrt(7-k) in the numerator combines with the sqrt(k) in the denominator to give the form declared to be "correct."

 

Of course, some of these manipulations require that k be positive. That's why Maple returned the form it did, a form that would be valid for any k, even a complex number.

Have you tried the Import Data Assistant available via the Tools menu?

The easiest way to graph vectors is via the PlotVectors command in either of the VectorCalculus packages.

VectorCalculus:-PlotVector([`<1, 0, 0)>,<0, 1, 0>,<0, 0, 1>], color = [black, red, green]);

Notice how this command maps colors onto the list of vectors.

 

It looks like you are calculating arc length in polar coordinates, but without knowing rho(theta), we can't experiment with the integral. Apparently, LL is discontinuous. Try either an indefinite integral, or make assumptions on phi to obtain correct expressions between the discontinuities.

I've seen something like this come up when constructing involutes of a plane curve. (The unwinding of a string wrapped around the curve requires computation of arc length. Some curves need a strictly increasing arc-length function; for others, a discontinuous one!)

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