Does anyone know how to use Maple well? I'm having some trouble plotting Riemann Sums.

When I try to plot
"my_plot1 := rightbox(11+6*x^4,x=5..9,4);"
all I get is "my_plot1 := PLOT(`...`)"

What am I doing wrong?

8/30/2007 9:05:53 PM

http://ellerbruch.nmu.edu/Maple/Leftrightsum.html

that might help, not sure though.

8/30/2007 10:08:59 PM

If you assign a plot to a variable name like my_plot1, I've never seen Maple actually display the plot. Just copy the plot info to a new line and hit enter, then you will see it.

If the plot info is the correct answer, the eGrader thing will count it correct when it is assigned to my_plot1.

This is just from my experience, btw.

This is the Calc 2 Maple HW 1 isn't it?

8/31/2007 12:42:45 AM

I don't have Maple in front of me, but perhaps you need to add "plot(my_plot1);" after the assignment statement.

8/31/2007 9:53:04 AM

^^ yep

8/31/2007 12:53:10 PM

or you could just get the maple labs from previous years since they are the exact same every year. My freshman year my friend sent me an email that had every single maple from calc 1 to calc 3. I never did one Maple lab on my own....mostly because nobody in real life uses maple after calc classes

8/31/2007 3:40:59 PM

It begins

8/31/2007 6:30:31 PM

^^ well im in grad school and i use both maple and matlab

what do you say to that son

9/3/2007 7:50:12 PM

oh yeah, while I'm in grad school and I don't use either of em, so there.

9/3/2007 9:22:34 PM

Quote :

"mostly because nobody in real life uses maple after calc classes"

Not true, I used Maple for all sorts of complex differentiation and integration throughout my academic career. I didn't use it for any of the super detailed stuff they make you learn, but it was a life saver when integrating an entire page of crap.

9/4/2007 9:41:47 AM

yep, I used maple quite a bit in grad school and now I use it from time to time at work.

9/5/2007 8:50:22 AM

But by the time you're in grad school you will have probably forgotten most of that old calc maple you learned, unless you keep your assignments...

I could see myself using matlab a lot in the future but not maple

9/5/2007 9:39:14 PM

ok question for maple then - if you have 2 vectors, convert them to unit vectors and find the cross product of them (because they are supposed to be orthogonal) - they ask for 2 unit vectors - i found the first one and thought that the second was the negative of the first, but I was obviously wrong - what would the second unit vector be?

[Edited on September 5, 2007 at 11:01 PM. Reason : ]

9/5/2007 10:59:55 PM

gram-schmidt

take the first vector minus the projection of the first onto the second then normalize

9/6/2007 8:49:49 AM

^^actually I read your post again and I'm not even sure what you are asking.

9/6/2007 9:50:47 AM

*Find two unit vectors orthogonal to both a and b* a and b being <x,y,z> (just not including the numbers) - so I defined both a and b, took the cross product of them (say the cross product was <d,e,f> - then I took <d,e,f>/sqrt( <d^2+e^2+f^2> ) - found the evaluated answer of that and got m<-q,r,-s> - put that for ans1, which was counted correctly, then I took the negative of that -m<q,-r,s> which was wrong for ans2 - but I thought the second unit vector was the negative of the first - probably wrong tho - sorry for the way I explained it but I'm not trying to get someone to do my homework for me and everyone I asked doesn't know what to do

[Edited on September 6, 2007 at 12:44 PM. Reason : ]

9/6/2007 12:44:12 PM

umm...

You can't find two (non-coincident) unit vectors orthogonal to two other vectors in three space. Unless of course the first two vectors you were given were coincident.

The problem is that any two non-coincident vectors span a 2-D plane in three space, so all you can do is find a vector that is orthogonal to that plane (which you did by taking the cross product).


I think the question was probably more like "given two vectors a and b, find two orthonormal vectors that span the same space". In which case my first response of doing a gram-schmidt process would be right.

If you want you can post the question verbatim and I will try to help.

[Edited on September 6, 2007 at 2:26 PM. Reason : .]

9/6/2007 2:21:47 PM

Find two unit vectors orthogonal to both a= [11, 14, 1] and b = [6, 2, -8]. Assign your answers to ans1 and ans2.
This is my exact work. ans1 is right, but ans2 is wrong.

> with(linalg):

> a:=[11, 14 ,1];

> b:=[6, 2 ,-8];

> p:= crossprod(a,b);

> z:=(crossprod(a,b))/sqrt((-114^2)+(94^2)+(-62^2));

> evalf(z);



> ans1:= -.01117754585*[-114,94,-62];



> ans2:= .01117754585*[114,-94,62];

9/6/2007 2:29:19 PM

I agree that ans1 is right. But the way you wrote ans2, it is actually the exact same vector. Just like -1*(-5) = 1*(5). All you did was distribute the minus sign.


Your answer would have been right if you wrote

ans2= -ans1

ans2:=.01117754585*[-114,94,-62];


[Edited on September 6, 2007 at 2:46 PM. Reason : OK I retract that(the bit about your teacher being an idiot). Although I still think it's a dumb que]

9/6/2007 2:40:59 PM

it still didnt work that way, but its alrite it was only worth like 4 pts and I'm out of submissions, thanks anyway

9/6/2007 4:30:00 PM

I really don't see how you can get two vectors orthognal to two other vectors in three space. I think the question is wrong. You should bring this up with your teacher.

Who is your teacher by the way?

9/6/2007 5:39:41 PM

I'm taking it through DE with Norris, soooo it kinda doesnt help considering he is 'Mr. Maple'

9/6/2007 7:23:09 PM

eh, yeah. if you remember, post the answer when you get it.

9/6/2007 9:10:45 PM

given A and B (let assume they're nonzero and nonparallel) there are two vectors perpendicular to both A and B

A x B
B x A

These are simply the normal vectors to the plane spanned by A and B. In fact there are
infinitely many such vectors, for any constant c != 0 the following is perpendicular to both A and B

c(A x B)

But you only want the two vectors with unit length, the unit normal "up" and the unit
normal "down". These are,

(A x B) / (|A x B|)
(B x A) / (|A x B|)

where |A x B| is the length of A x B.

No maple needed.

9/6/2007 10:33:45 PM

well thats completely understandable - but it was a maple assignment, kinda sux tho

9/6/2007 10:44:52 PM

^^ If that's what the professor wanted then I (And the intintion of tdwhitlo) would have been right about putting the negative in front of the other vector. But he said that was wrong....

Recall: (A x B)= -(B x A)

hence, why I am still confused about what the professor wanted.

9/6/2007 11:06:02 PM

bwahahahaha...i remember when it was so easy to cheat on maple.

9/6/2007 11:13:46 PM

Maybe he meant the Maple was wrong... Otherwise we need more dimensions...
If Maple distracts from the math then Maple fails.

9/7/2007 1:56:30 AM