find the image of the set S under the given transformation S={(u.v)|0<=u<=3, 0<=v<=2} x=2u+3v, y=u-v step by step, please
Isn't this still linear algebra? ^.^
no, it's cal3
part find jacobian
you are very good at cal, right?
Well, \[\huge (u,v) \rightarrow (x,y)\] \[\huge \left[\begin{matrix}x \\ y\end{matrix}\right]=\left[\begin{matrix}2 & 3 \\ 1 & -1\end{matrix}\right]\left[\begin{matrix}u \\ v\end{matrix}\right]\]
so?
Partial derivatives... should be simple, no? ;)
\[\huge \left[\begin{matrix}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{matrix}\right]\]
i know how to find jacobian number, just don't understand how to use it in this part
typo...
ok, ignore that typo, i understand
But the best way to do this is visualise the corners :)
Or, you could just substitute...
Solve for u and v in terms of x and y.
It's interesting to note that the jacobian of a linear transformation is itself :D
so, to find the image of (u,v), first find the jacobian matrix , second, ?
I don't know this method you are using. :/
neither I, that's why I am here to make question:(
I just substitute, and express u in terms of x and y.
and v, too.
ok, show me steps, please
So... \[\huge u = \frac{x+3y}{5}\] \[\huge v = \frac{x-2y}{5}\]
You could arrive at these^ if you solve x = 2u + 3v y = u - v as a system of equations of u and v, considering x and y as constants.
So just replace the u and v with these expressions of x and y \[\large \bar{S}=\left\{(x,y) \ | \ 0\le x+3y \le 15 \ \ , \ \ 0 \le x-2y \le10\right \}\]
that's it? ha!! thanks a lot, talent
It's Terence :P
hey, I saw you use :P twice, what does it mean?
A playful tease :) Like a tongue-out gesture.
ok, young boy, be helpful, especially to older like me. thanks a lot. good night
It's not even noontime yet o.O
night here!!!11:13pm
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