## 7 Badges

18 years, 302 days

## Write x1(t) explicitly if x1...

Write x1(t) explicitly if x1 is a function of t.
```dsolve(diff(x1(t), t) = -2*a*x1(t)*(x1(t)-1)+(1-b)*x0);

/       a (2 x1(t) - 1)       \                             (1/2)
_C1 + arctan|-----------------------------| + t (a (-2 x0 + 2 x0 b - a))      =
|                        (1/2)|
\(a (-2 x0 + 2 x0 b - a))     /

/       /                           (1/2)
1  |       |  /                      2\
0, x1(t) = --- \a - tan\t \-2 a x0 + 2 a x0 b - a /
2 a

(1/2)\                          (1/2)\
/                      2\     | /                      2\     |
+ _C1 \-2 a x0 + 2 a x0 b - a /     / \-2 a x0 + 2 a x0 b - a /     /
```
Regards, --Jean-Marc

## It is not clear what you...

It is not clear what you actually tried, but if the function is complex valued on chosen interval, the plot will be empty. Regards, --Jean-Marc

## Here is what I get on my...

Here is what I get on my system: Regards, --Jean-Marc

## plots[pointplot3d] creates a 3-D point p...

I know plot3d(equation, range, range) gives you graph, but is there a way to plot a graph in only places where data sets are given instead of one smooth graph?
I believe what you are looking for is the function pointplot3d. See help("pointplot3d") for more info. HTH, --Jean-Marc

## Variable t5 is declared twice in the sec...

The variable t5 is declared twice in the second argument of solve. replace it by t3, so the last line reads:
```q := solve(eqns, {t0, t1, t2, t3, t4, t5, t6, t7});
```
Regards, --Jean-Marc

## Could you provide some...

```Could you provide some values/definitions for a5, n1, n2, n3?

Moreover,

> restart; with(plots);
> spacecurve({[x, y, z]}, t = 0 .. Pi, scaling = constrained,
color = blue, thickness = 3)

returns a blank 3d-plot on my system (Maple 12.02, Mac OS X).

Also, I cannot see the pictures you posted for I get the
following error message:

"Safari can’t open the page
“http://c:/Users/sarper/AppData/Local/Temp/moz-screenshot-1.jpg”
because it can’t find the server “c”."

Apparently the links you posted do not point to any Maplesolft's servers.

Regards,
--Jean-Marc
```

## :- is a shortcut for the use statement...

The symbol :- is a shortcut for the use statement. See help("use") for more info (or also in MaplePrimes). With Maple 12.02, if I enter :- in the search text box, I arrive to the help page for use. Regards, --Jean-Marc

## proc () plots:-setoptions(title = `My Pl...

Write plots:-setoptions as in
```proc () plots:-setoptions(title = `My Plot`) end proc
```
Regards, --Jean-Marc

## The correct Maple syntax for...

The correct Maple syntax for your equations is as follows (eq1 and eq2 are just symbolic names for easy reference to each equation):
```eq1:=(Psi(1, b)-Psi(1, a+b))/(Psi(0, b)-Psi(0, a+b))^2 = s2/s1^2;
eq2:=(Psi(2, b)-Psi(2, a+b))/(Psi(0, b)-Psi(0, a+b))^3 = s3/s1^3;
```
Regards, --Jean-Marc

## I cannot see the images. If...

I cannot see the images. If I try to force the display, I get the following error message, "Safari can’t open the page “http://c:/Users/Zach/AppData/Local/Temp/moz-screenshot-2.jpg” because it can’t find the server “c”." which seems to indicate that you have not uploaded the image on the MaplePrimes's server or pasted the wrong url. You should use the file manager, titled My files, located on the left hand side of the page to upload images and get the corresponding url on MaplePrimes. Regards, --Jean-Marc

## It looks like you really...

It looks like you really have only one equation of two variables: eq2 is a symbolic expression for q that can be replaced into eq1.
```restart;
eq1 := (q/a)^2+2*q1*f*cosh(-3*p*q2/(2*a))-1-q1^2*f^2 = 0;
eq2 := q = a*q1*f*sinh((3/2)*p*q2/a)*q2;
algsubs(eq2, eq1);
subs({a = 0.150e9, f = .25, p = 0.174e9, q = 0.199e9}, %);
solve(%, [q1, q2]);

2
q               /3 p q2\         2  2
eq1 := -- + 2 q1 f cosh|------| - 1 - q1  f  = 0
2              \ 2 a  /
a
/3 p q2\
eq2 := q = a q1 f sinh|------| q2
\ 2 a  /
2
2  2     2  2     /3 p q2\    2                  /3 p q2\
-q1  f  + q1  f  sinh|------|  q2  - 1 + 2 q1 f cosh|------| = 0
\ 2 a  /                       \ 2 a  /
2            2                     2   2
-0.0625 q1  + 0.0625 q1  sinh(1.740000000 q2)  q2  - 1

+ 0.50 q1 cosh(1.740000000 q2) = 0
[[              /         /  2           /      /        2          2
[[q1 = q1, q2 = \2. RootOf\_Z  + 4 q1 exp\RootOf\-7569 q1  (exp(_Z))

3                                      2
+ 30276 q1 (exp(_Z))  + 30276 q1 exp(_Z) - 121104 (exp(_Z))

2   2          4          2   2          2         2   2\\
+ 625 _Z  q1  (exp(_Z))  - 1250 _Z  q1  (exp(_Z))  + 625 _Z  q1 // + 4 q1

/   /      /        2          2                     3
\exp\RootOf\-7569 q1  (exp(_Z))  + 30276 q1 (exp(_Z))  + 30276 q1 exp(_Z)

2         2   2          4          2   2          2
- 121104 (exp(_Z))  + 625 _Z  q1  (exp(_Z))  - 1250 _Z  q1  (exp(_Z))

2   2\\\     /   2     \ /   /      /        2          2
+ 625 _Z  q1 ///^3 + \-q1  - 16/ \exp\RootOf\-7569 q1  (exp(_Z))

3                                      2
+ 30276 q1 (exp(_Z))  + 30276 q1 exp(_Z) - 121104 (exp(_Z))

2   2          4          2   2          2         2   2\\\
+ 625 _Z  q1  (exp(_Z))  - 1250 _Z  q1  (exp(_Z))  + 625 _Z  q1 ///^2,

\\//   /   /      /        2          2                     3
label = _L4// \q1 \exp\RootOf\-7569 q1  (exp(_Z))  + 30276 q1 (exp(_Z))

2         2   2          4
+ 30276 q1 exp(_Z) - 121104 (exp(_Z))  + 625 _Z  q1  (exp(_Z))

2   2          2         2   2\\\          \]
- 1250 _Z  q1  (exp(_Z))  + 625 _Z  q1 ///^2 - 1. q1/], [q1 = 4., q2 = 0.],

]
[q1 = -4., q2 = 1.805513019 I]]
```
Regards, --Jean-Marc

## The following website should help...

Assuming I have correctly understood your question, I think Alexander Bogomolny's web site should help you to answer it: "Inscribed and Central Angles in a Circle: What is this about? A Mathematical Droodle". (Note that the website contains a nice interactive JAVA applet and a link to some explanations about what's going on.) HTH, --Jean-Marc

## Premature evaluation...

You should search this web site for "premature evaluation" for similar questions to your arise quite often. One possible way to get the plot is as follows:
```> restart;
> t := proc (x) if 1 < x then x else x+1 end if end proc;
> plot(t, -1 .. 4);
```
Regards, --Jean-Marc

## 'e' is not exp()...

Regards, --Jean-Marc
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