## 1080 Reputation

17 years, 342 days

## One pleasant discovery...

I am happy to report that MAPLE11 seems to compile whole lot faster than MAPLE10 !

## Help for 'while'...

Here's something I think TPTB at Maplesoft should be aware of. In MAPLE10, when you type ?while you get an explanation of 'while' statement used in conjunction with the 'for' statement - you don't get an explanation of the 'while' loop. For that explanation you have to go to the books. Perhaps this is corrected in MAPLE11.

## tan(a+b)...

People on this forum have been unbelievably helpful. I am trying to write some worksheets to help flatten the learning curve for folks who are new to MAPLE. Trouble is, being not far from the newbie stage myself, I may very well be making significant mistakes about the capabilities of MAPLE and thus teaching people cumbersome and inefficient ways of doing things. With that in mind, if anyone has the time to critique the following, I would be most appreciative.

## Brain teaser # 1...

Maple
A few years ago I saw somewhere in a math forum a brain-teaser type problem which I'm about to present. I wasn't able to solve it at the time. I haven't been able to find the original forum, so I don't know whether or not my answer is right. I do know that I could never have attacked the problem without MAPLE. We have a cube of edge a and a drill bit of diameter b. We drill one hole from the center of the front face to the center of the back face, and another from the center of the right face to the center of the left. What is the total volume of material removed? I get (Pi*(b^2)*a)/2 - ((2/3)*b^3) Am I right?

## Why do certain definite integrals make M...

I wonder if someone out there would help me understand the inner workings of MAPLE as indicated by the following: MAPLE has no trouble with int(sqrt(1-x^2)-x1,x=x1..sqrt(1-x1^2)); nor with this f1:=x->sqrt(1-x^2)-x1:int(f1(x),x=x1..sqrt(1-x1^2)); this one works too int(sqrt(r^2-x^2)-x1,x=x1..sqrt(r^2-x1^2)) assuming r>0,x1>0,r>x1; but for some reason not this one f2:=x->sqrt(r^2-x^2)-x1:int(f2(x),x=x1..sqrt(r^2-x1^2)) assuming r>0,x1>0,r>x1; Curiously, this works gg:=x->sqrt(r^2-x^2)-x1:int(gg(x),x):z:=unapply(%,x):z(sqrt(r^2-x1^2))-z(x1)
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