Calc III - Graphing a Function of Multiple Variables by hand

Cloudless
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I come across questions where I have to match the equation with its contour map and graph.

Examples:

z = sin(xy)

z = sin(x-y)


Right now I'm using Wolfram Alpha for all of these, but supposing these appear on an exam... how do I graph it by hand? :confused: For example, for z = sin(xy) I'm just plugging in random values for x and y... but doesn't that mean the entire xy graph should be filled then?
 
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A contour of (for example) sin(xy) is a curve on the x, y plane where this function is constant. This in turn means that the product xy must be constant. So you could write

c = xy
y = c/x

And for each c you get a different contour. Plotting the curves once you have them in that form is probably a little more familiar to you.
 
Cloudless said:
I come across questions where I have to match the equation with its contour map and graph.

Examples:

z = sin(xy)

z = sin(x-y)


Right now I'm using Wolfram Alpha for all of these, but supposing these appear on an exam... how do I graph it by hand? :confused: For example, for z = sin(xy) I'm just plugging in random values for x and y... but doesn't that mean the entire xy graph should be filled then?
Set z equal to some value (-1 ≤ z ≤ 1, why?), then solve for y in terms of x.
 
If you're doing it by hand I think it's easier to do what I wrote, for a few simple c values and afterwards evaluate what z is on each curve. Of course your way works too.
 
A contour of (for example) sin(xy) is a curve on the x, y plane where this function is constant. This in turn means that the product xy must be constant. So you could write

c = xy
y = c/x

And for each c you get a different contour. Plotting the curves once you have them in that form is probably a little more familiar to you.

Wow, this is brilliant @_@ Thank you

Set z equal to some value (-1 ≤ z ≤ 1, why?), then solve for y in terms of x.
You would need a calculator for this method though, right? Inverse sin of (-1 to 1) = xy.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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