I understand your method but why am I getting another solution using the transferfunction. Could someone solve the equation using the transferfunction?

I should have posted this in the student math advice section. Excuses.

EDIT: I found a way to inser text. But isn't there a way to set the input of text standard to "text" en de input of math standard to "maple input". In the classic worksheet you can by pressing the backspace easily switch between text and math.

It's not exact what I've had in mind. The problem is that I want to plot the function: huur:=t->400*(1.01)^(t); with t a positive integer. When I plot this function (even when I'm assuming that t is a posint) I get a continu function and that's of course wrong. I didn't found a better way then making use of the ceil function to get a correct plot. I defined the function: huur:=t->400*(1.01)^ceil(t); This gives me the correct discontinu plot. My question was if there isn't a better method to plot the original function without redifining it using the ceil function.

EDIT: Has to be the ceil function instead of the floor function.

Always the same trouble posting some part of my worksheet here. Took a picture: http://img221.imageshack.us/img221/1630/probxv9.jpg By the way the command Curl is form the VectorCalculus bib.

Problem already solved. Sorry for the trouble.

Hi, I don't really get the explanations in here and I don't want to open up a new topic. So here is my question. How can I insert x < = a in a piecewise statement?

Can someone tell me whats my problem here? >Iz := -1/2*a^4*Pi*rho*2^(1/2)+1/2*b^4*Pi*rho*2^(1/2) >Mass := -a^2*Pi*rho*2^(1/2)+b^2*Pi*rho*2^(1/2) >simplify(Iz,{M=Mass}); -1/2*a^4*Pi*rho*2^(1/2)+1/2*b^4*Pi*rho*2^(1/2) <\prev> It seems like Maple doesn't reconize it.

Edit: Title must be: trouble understanding Maple commando. My excuses. I've copied the full commando but parts are missing in the post. Weird.. Tried it again an the same problem. Before the parameters a function is missing. It's not so crucial for understanding my problem.

Excuse me...

Edit: cirkely:=1*sin(theta)+1; It's an intersection between a regular ellips with a=1 and b=2 and a circle with midpoint(0,1)

Thank you. I already have a new question but because it's form another type I'm going to open a new topic.

Ok I've got what you mean. But what does the commando eval mean? I only saw evalf in class to round off. So there is no way to solve this integral with the commando Doubleint and some extra specifications like assume that ... . You must do it in two steps?

Is there really no way to solve it with double integrals? Can not really folow the way that alec does. I have a double integral of the form r * dr * dtheta. I need to double integrate over r going form R*cos(alpha)/cos(theta) to R and theta going from -alpha to alpha. The result as giving in my handbook must be: R^2*alpha-R^2/2*sin(2*alpha)