MHB Jacobian of the transformation T:x=u, y=uv

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To graph the transformation T: x=u, y=uv in Desmos, replace u with x and v with y, resulting in the equations xy=1 and xy=2. Enter the functions x=1, x=2, y=1/x, and y=2/x in Desmos to visualize the graph. It's recommended to limit the axes to values between 0 and 3 for clarity. A shared Desmos plot link provides a visual representation of these equations. This approach effectively demonstrates the transformation and its graphical implications.
karush
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View attachment 8171

how do you graph this in Desmos ?

Assume the rest of the calculation is correct

much thank you ahead...:cool:
 
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Do you mean how to use Desmos to graph u= 1, u= 2, uv= 1, and uv= 2?

First, of course, Desmos graphs either y= f(x) or x= g(y) so you have to replace u with x and v with y and solve xy= 1 and xy= 2 for y (or x). that is, enter the functions x= 1, x= 2, y= 1/x, and y= 2/x. I limited the axes to that x and y were between 0 and 3 to get a good graph.
 
Country Boy said:
Do you mean how to use Desmos to graph u= 1, u= 2, uv= 1, and uv= 2?

First, of course, Desmos graphs either y= f(x) or x= g(y) so you have to replace u with x and v with y and solve xy= 1 and xy= 2 for y (or x). that is, enter the functions x= 1, x= 2, y= 1/x, and y= 2/x. I limited the axes to that x and y were between 0 and 3 to get a good graph.

View attachment 8176

ok here is my eventual typeset

desmos plot is here

https://www.desmos.com/calculator/cmqpsbp85y
 

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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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