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Female silver67


England
Joined: Jan 10, 2010
Posts: 1
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need help Reply to this Post
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Hi all,

I'm new to the site and congratulations to whoever did it. It's a v useful tool for Maths students and teachers.
But now coming to the point, I've got a cruel doubt with a function sketch in my Maths assignment. Really appreciate if someone could give me a clue on how to get the graphic done.

It's something like: f(x,y) = max {0,x^2+y^2-1} over the domain ||x||, ||y|| < or = 2. I need to roughly sketch the function and its domain.

thanks,
Ulisses
praying
[Jan 10, 2010 12:12:37 PM] Show Post Printable Version        Hidden to Guest [Link] Report threaten post: please login first  Go to top 
Female pskinner

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Joined: Apr 2, 2005
Posts: 797
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applause   Re: need help Reply to this Post
Reply with Quote
 
Hi all,

I'm new to the site and congratulations to whoever did it. It's a v useful tool for Maths students and teachers.
But now coming to the point, I've got a cruel doubt with a function sketch in my Maths assignment. Really appreciate if someone could give me a clue on how to get the graphic done.

It's something like: f(x,y) = max {0,x^2+y^2-1} over the domain ||x||, ||y|| < or = 2. I need to roughly sketch the function and its domain.

thanks,
Ulisses
praying


If you are an A+ member, you can plot x^2+y^2-1 using the interactive function graphing calculator:

www.webgraphing.com/graph3d.jsp

In the x-y plane, you will have the graph of the circle of radius 1 centered at the origin: x^2+y^2-1=0, corresponding to z=0. Every cross section parallel to the x-y plane, z=k^2, for k>0, will be a circle of radius k. Cross sections to the other coordinate planes are parabolas. The base of the 3d solid is the disk in the x-y coordinate plane x^2+y^2<=1.

Also, the domain is the square: -2<=x<=2, -2<=y<=2.
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Principal Skinner
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[Edit 1 times, last edit by pskinner at Jan 11, 2010 9:16:33 PM]
[Jan 11, 2010 8:31:53 AM] Show Post Printable Version     [Link] Report threaten post: please login first  Go to top 
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