## 261 Reputation

17 years, 334 days

## Something has got wrong...

Something has got wrong getting my Maple commando's in here. I've you type oppp:=Doubleint(integrand(pool),r=R*cos(alpha)/cos(theta)..R,theta=-alpha..alpha); You will see where is my problem. Same thing happens with this formula. opp:=Doubleint(sqrt((R-x)^2+(-y)^2),y=-sqrt(R^2-x^2)..sqrt(R^2-x^2),x=-R..R); I think it has something to do with the parameter R. Previous excersisses without worked perfectly.

## cirkelx:=theta->1*cos(theta);...

>cirkelx:=theta->1*cos(theta); cirkelx := theta -> cos(theta) > cirkely:=theta->1+1*sin(theta); cirkely := theta -> 1 + sin(theta) > ellipsx:=theta->2*cos(theta); ellipsx := theta -> 2 cos(theta) > ellipsy:=theta->1*sin(theta); ellipsy := theta -> sin(theta) > solve({cirkelx=ellipsx,cirkely=ellipsy},{theta}); These are parameterfunctions of a circle with midpoint (0,1) and of an ellips with a=2 and b=1. Why wouldn't Maple solve this? I checked it out with a plot and they have two intersections.

## Because my topic is still...

Because my topic is still open I will post a new question here. Is there a way to round off to three digits after the decimal points? I already know the command evalf to round off to three significant digits and convert( .. , float) to display all the digits after the decimal point.

## Today I was again trying to...

Today I was again trying to combine multiple commando's but I just would not not work. For example: solve(afg1=0,t); solve(afg1=1,t); I would combine them like: solve([afg1=0 and afg1=1],t); Can somebody correct me.. (again :) ). I have the same prob getting the following in one command: substituting two different values of a variable in a functions to get two different solutions. (could somebody also tell me what happens when I use following command as advised by alec solve({K,y=6})?)

## Here it is. Two easy...

Here it is. Two easy questions I think but it could make my worksheet easier!
```Oefening 6
Implicit Equation
> restart:
> K:=x^2-x*y+y^2=27;

2          2
K := x  - x y + y  = 27

First deriviative.
> afg1:=implicitdiff(K,y(x),x);

2 x - y
afg1 := -------
x - 2 y

When is the tangent horizontal or vertical?
> solve(afg1=0,[x,y]);

[[x = y/2, y = y]]

> solve(1/afg1=0,[x,y]);

[[x = 2 y, y = y]]

Which points are these?.
> subs(x=y/2,K);

2
3 y
---- = 27
4

> solve(%);

6, -6

>
Now is my question if there is an easier way to substitute this points. I have calculated the
2 values for y but could not substitute them in K because I had already substituded x = y/2
in it. It would result in an error. Is there an easy way to reset K or do I have to restart
all over? A second questions is if I could put the 2 solve commando's from the first
deriviative in one command. Something like solve(afg1=0 and 1/afg1=0,[x,y]). Thank you :)

```
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