If-we-have-a-vector-v-x-y-and-apply-a-linear-tranformation-such-that-a-new-vector-v-x-y-is-made-can-we-calculate-the-change-of-area-with-respect-to-the-unit-vector-

Question Number 7093 by FilupSmith last updated on 10/Aug/16

If we have a vector v = [\begin{matrix} x \\ y \end{matrix}]

and apply a linear tranformation such that

a new vector v^{^{'}} = [\begin{matrix} x^{'} \\ y^{'} \end{matrix}] is made,

can we calculate the change of area with

respect to the \hat{ı} unit vector ?

Commented by FilupSmith last updated on 10/Aug/16

\hat{ı} = i = unit vector of x axis

\hat{ȷ} = j = unit vector of y axis

v = [\begin{matrix} x \\ y \end{matrix}]

v^{'} = k v, k = [\begin{matrix} i_{x} & j_{x} \\ i_{y} & j_{y} \end{matrix}] (k = transformation matrix)

i_{x}, i_{y} are new coordinates for i

j_{x}, j_{y} are new coordinates for j

∴ v^{'} = [\begin{matrix} i_{x} & j_{x} \\ i_{y} & j_{y} \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} {x i}_{x} + {y j}_{x} \\ {x i}_{y} + {y j}_{y} \end{matrix}]

therefore :

x^{'} = {x i}_{x} + {y j}_{x}

y^{'} = {x i}_{y} + {y j}_{y}

let a r e a be A = \frac{1}{2} height \times base = \frac{1}{2} x^{'} y^{'}

How can we integrate with respect to

both i_{x} and i_{y} ? ? ?

∣ i ∣= \sqrt{i_{x}^{2} + i_{y}^{2}} (= 1 for original i ∵ i_{x} = 1 \land i_{y} = 0)