Can someone explain how to do this?

Prove that if we describe the circle of center (a; b) and radius r using
the parameters (a; b; k), with k = a2+b2-r2, rather than the more natural parameters
(a; b; r), then the error function H(a; b; k) = E(a, b,rad(a^2 + b^2 -k) is quadratic in a; b
and k. What does this imply about the number of critical points?

Hi there,

I typed exactly the same thing from my lecture notes to reproduce some graphs in Maple Classical Worksheet

however result in 'empty plots'

Is it that I have to load some packages using 'with( XXX )'?
 

Can someone help me with that?

Thanks

Download 7845_Coursework Maple.mws
View file details

Hi, assume  i have some data and want to fit a model function to it (Maple12). The help pages show a way:

with(Statistics):
X := Vector([1, 2, 3, 4, 5, 6], datatype=float):
Y := Vector([2, 3, 4.8, 10.2, 15.6, 30.9], datatype=float):

f:=a+b*exp(c*t);

Fit(f, X, Y, t);

It works as expected. Now assume that the model function is not explicitly given,  it may be just the solution of a  differential equation:

f:=diff(y(t),t)+u*y(t)=v; y(0)=0

when i run a program that returns a expression, example :

f=3%1 + 2 cos(phi2 + phi3)
%1 := sin(phi3 + phi4 + phi2)

How to return f =3sin(phi3 + phi4 + phi2)+2 cos(phi2 + phi3)

I'm using MAPLE TOOLBOX FOR MATLAB, run in Matlab windown.

K heres my problem.

Say i have a function, for example y=(1-x) and i want to find out which values of x will give me a real answer (i.e. no complex numbers!)

Preferrable i would like the output to be something along the lines of... "y is real for a<x<b..."

Hello,

Is there a way to merge two lists?

For example, to merge:
a=[[1,2],[3,4]]
b=[[5,6],[7,8],[9,10]]

and get:
[[1,2],[3,4],[5,6],[7,8],[9,10]]

Thanks

I need your help and guidance to resolve
this problem.

Simplify expression under summation. How could simplify the expression inside the sum, as much as possible. I have tried with:

simplify, expand, Normalize, Power...

But do not get the results I want.

Hi all,

I have a system of differential equations:

Eq1:=diff(R(t), t) = -0.22*W(t)-3000+0.3*R(t)+3488*exp(0.295e-3*t)+.744*(diff(W(t), t))

Eq2:=diff(W(t), t) = 0.304*W(t)-0.1916e-1*R(t)-2981*exp(0.295e-3*t)+2180-.36*H(t)+.6*(diff(R(t), t))

To solve, I could plug Eq1 into Eq2. However, maple will also solve this system without plugging Eq1 into Eq2. But, I see the answers are different. So, which method is correct?

I get the proper graph with this function however I need to zoom on the intersection to show that the limit does not exist.

implicitplot3d((x^2-y^2)/(x^2+y^2), x = -.5 .. .5, y = -.5 .. .5, z = -2 .. 2, grid = [30, 30, 30])

 When I enter this function, I only get a blank plot window.  When I drag and drop it in the plot window I get an error.

implicitplot3d(z = xy^2/(x^2+y^4), x = -4 .. 4, y = -4 .. 4, z = -.5 .. .5, grid = [30, 30, 30])

  Any help is greatly appreciated,

 Jim Z

Hey!

So I am trying to write a nifty little program to show that for any x^n Leibiz method of dividing the area under the curve into rectangles of base r,r^2,r^4... so on, can this be done?

Thanks!

#96

r=625       theta=pi/2

I found the 4th root of 625....it = 5.

Now, I'm supposed to use DeMoivre's Theorem. Then divide pi/2 by 4 and add k/4 times 2pi with k going between 0 and 3.

The 2nd angle should be 5pi/8, third should be 9pi/8, and the last should be 13pi/8.

((Show me how please. Thanks!))

#30

llvll=3              u=<4,-4>

-Find the magnitude of the given vector and divide each of the components by that magnitude. That will produce a unit vector that is the same direction of the given vector. Since it wants the magnitude to be equal to 3, multiply each component by 3. Make sure the radicans are out of the denominator. The answer should be a vector in component form.

#48

llvll=3            u=3i+4j

I am trying to find the stationary ponts of f(x,y) 5*x^3-12*x-3*x*y+6*y^2-4;

so far, I have got

A:=diff(f,x)=0

B:=diff(f,y)=0

e1:=RootOf(A,x)

e2:=RootOg(B,x)

G:=solve([A,B],[x,y])

do I now need to use allvalues?

if so, I get

V:=allvalues(G). This gives me two values of x and y.  (one with a positive root and one with a negative root)

ie. v=[[x=c - sqrtb,y=4c - 4sqrtb],[x=c+sqrtb,y=4c+4sqrtb]]

r:=diff(f,x$2) =30x

s:=diff(f,x,y) = -3