i need to calculate the radius of an arc. i have that the chord lenght is 96 inches and the distance between the chord and the arc is 6 inches perpendicular from the center of the chord. assuming that i am working with a circle.... how do i calculate the radius of the circle????

Let f:Rⁿ  -> R^m a linear function i need help about a proc about matrix of the linear map Df(x),

our input will be n,m and f

such that,f:R²->R³ ,f(x,y,z)->(x+y,y+2z,x+y+z) and i wish resulting output with derivative matrix with respect to the standard bases in R² and R³

it just won't work

simply hovering over tools crashes it

going into a menu crashes it

trying to execute a command crashes it

it just freezes up completely such that it won't even close down without using task manager

i know fully how to use maple, i've used it on my university network but using it on my home pc won't work for some reason, even though my computer is very modern and i am sure meets the specs.

is this a common thing?...

Hi,

i try to use your code (i mean other way )   that give to me but it doesnot work. it give x[c] and y[r] at label.

i have maple 14 .

could you please help what is the problem?

last answer is wrtten by

It can be done with the labels option (see ?plot/options

^Hi,

I have a problem with plot using maple. I want to plot x_c agaist y_r 

However, i have no idea how to make at label that   y_r  . In other word, i want to make r under y and ignore dash (-) that between y and r

and the same idea for x_c.

could you please help me in this plot label problem?

0 = 0, -(1/16)*cos(l1)*cos(l2)*cos(l3)*cos(l4)*(-Pi*x3+4*cos(Pi*t)) = 0, -(1/16)*cos(l1)*cos(l2)*cos(l3)*cos(l4)*(-Pi*x2+4*sin(Pi*t)) = 0, (1/16)*cos(l1)*cos(l2)*cos(l3)*cos(l4) = 0, 0 = 0, -(1/8)*cos(l1)*cos(l3)*cos(l4)*(-Pi*x3+4*cos(Pi*t)) = 0, -(1/8)*cos(l1)*cos(l2)*cos(l4)*(-Pi*x2+4*sin(Pi*t)) = 0, (1/8)*cos(l1)*cos(l2)*cos(l3) = 0

After I ran solve({eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8}, {l1, l2, l3, l4, t, x1, x2, x3}, AllSolutions); I got...

Here is my code where I get the problem of taking the coeff of theta dot . The output of matlab Cm shows the coeff in the output matlab ( Cmatrix , optimize) at t2727 I see coeff .Meaning when the loop ran coeff didn't execute.

> J[p1] := Matrix(3, 4, {(1, 1) = -sin(theta[1](t))*`Δx`[1]-cos(theta[1](t))*`Δz`[1], (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = cos(theta[1](t))*`Δx`[1]-sin(theta[1...

I keep getting this error although the code is running fine in maple 14 , Help is greatly appreciated Thanks alot

> restart; with(LinearAlgebra);
> A[0] := Vector(3, {(1) = 0, (2) = 0, (3) = -g});
> type(a[0], Vector);
                             false
> omega[0] := Vector(3, {(1) = 0, (2) = 0, (3) = 0});

Visualizing_the_SVD_.mw

Hola ev1:

I like to shorten the size of the command, through setoptions3d function. But it seems that it does not interact with ApplyLinearTransformPlot function.
How to get to label the vectors?

Gracias

I am working on hybrid dynamical systems an encounter the warning: "dependencies in discrete equations cannot be ordered". What does this mean and should I care about it?

> restart;

> with(plots);

BridgeHeight := 40;

BridgeFromCenter := 20;

BridgeToEdge := 35;

BridgeWidth := BridgeToEdge+BridgeFromCenter;

BridgeFromTheWater := 15;

BridgeLines := 10;

SpaceBetweenLines := 2*BridgeToEdge/BridgeLines;

LineStart := BridgeWidth;

CountLines := 0;

LineWidth := BridgeWidth;

while BridgeLines > CountLines do

CountLines := CountLines+1;

If i have the line:

S:=Sum( 1/16^k))*(4/(8*k+1)-2/ (8*k+4)-1/ (8*k+5)-1/ (8*k+6)), k=0..m);

and i want to establish the first m for which |Pi-Sm|<= 10^(-8)

when putting it into a while loop i get the error message that it cannot determine whether the expression is true or false:

while (Pi-S <= 10^(-8)) do

od;

etc,

what am i doing wrong here?

with(Optimization); nx := Maximize(abs(.5539395590*x^2+1.130864333*x+.9891410756-exp(x)), {-1. <= x, x <= 1.})

give [0.443368608590448687e-1, [x = 1.]]

but I know that it is not the maximum because:

abs(.5539395590*.56^2+1.130864333*.56+.9891410756-exp(.56))=0.045468048

with(Optimization); nx := maximize(abs((29.83428688*x-57.16914857)/(x-2.196093020)-exp(x)), x = 5.0 .. 6.0, location)

if we plot this function we see that the solution exists.

with(Optimization); nx := maximize(abs(.5547114632*x^2+1.130478381*x+.9883691714-exp(x)), x = -1 .. 1, location)

nx := 0.4568872387e-1, {[{x = 1.002012310}, 0.4568872387e-1]}

but 1.002 isn't in -1..1