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Mathematics 27 Online
OpenStudy (anonymous):

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

terenzreignz (terenzreignz):

Isn't this still linear algebra? ^.^

OpenStudy (anonymous):

no, it's cal3

OpenStudy (anonymous):

part find jacobian

OpenStudy (anonymous):

you are very good at cal, right?

terenzreignz (terenzreignz):

Well, \[\huge (u,v) \rightarrow (x,y)\] \[\huge \left[\begin{matrix}x \\ y\end{matrix}\right]=\left[\begin{matrix}2 & 3 \\ 1 & -1\end{matrix}\right]\left[\begin{matrix}u \\ v\end{matrix}\right]\]

OpenStudy (anonymous):

so?

terenzreignz (terenzreignz):

Partial derivatives... should be simple, no? ;)

terenzreignz (terenzreignz):

\[\huge \left[\begin{matrix}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{matrix}\right]\]

OpenStudy (anonymous):

i know how to find jacobian number, just don't understand how to use it in this part

terenzreignz (terenzreignz):

typo...

OpenStudy (anonymous):

ok, ignore that typo, i understand

terenzreignz (terenzreignz):

But the best way to do this is visualise the corners :)

terenzreignz (terenzreignz):

Or, you could just substitute...

terenzreignz (terenzreignz):

Solve for u and v in terms of x and y.

terenzreignz (terenzreignz):

It's interesting to note that the jacobian of a linear transformation is itself :D

OpenStudy (anonymous):

so, to find the image of (u,v), first find the jacobian matrix , second, ?

terenzreignz (terenzreignz):

I don't know this method you are using. :/

OpenStudy (anonymous):

neither I, that's why I am here to make question:(

terenzreignz (terenzreignz):

I just substitute, and express u in terms of x and y.

terenzreignz (terenzreignz):

and v, too.

OpenStudy (anonymous):

ok, show me steps, please

terenzreignz (terenzreignz):

So... \[\huge u = \frac{x+3y}{5}\] \[\huge v = \frac{x-2y}{5}\]

terenzreignz (terenzreignz):

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.

terenzreignz (terenzreignz):

So just replace the u and v with these expressions of x and y \[\large \bar{S}=\left\{(x,y) \ | \ 0\le x+3y \le 15 \ \ , \ \ 0 \le x-2y \le10\right \}\]

OpenStudy (anonymous):

that's it? ha!! thanks a lot, talent

terenzreignz (terenzreignz):

It's Terence :P

OpenStudy (anonymous):

hey, I saw you use :P twice, what does it mean?

terenzreignz (terenzreignz):

A playful tease :) Like a tongue-out gesture.

OpenStudy (anonymous):

ok, young boy, be helpful, especially to older like me. thanks a lot. good night

terenzreignz (terenzreignz):

It's not even noontime yet o.O

OpenStudy (anonymous):

night here!!!11:13pm

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