Question Number 7093 by FilupSmith last updated on 10/Aug/16
![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 ı^ unit vector?](https://www.tinkutara.com/question/Q7093.png)
$$\mathrm{If}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{vector}\:\boldsymbol{{v}}=\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix} \\ $$$$\mathrm{and}\:\mathrm{apply}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{tranformation}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\mathrm{new}\:\mathrm{vector}\:\boldsymbol{{v}}^{'} =\begin{bmatrix}{{x}'}\\{{y}'}\end{bmatrix}\:\mathrm{is}\:\mathrm{made}, \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{change}\:\mathrm{of}\:\mathrm{area}\:\mathrm{with} \\ $$$$\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\hat {\imath}\:\mathrm{unit}\:\mathrm{vector}? \\ $$
Commented by FilupSmith last updated on 10/Aug/16
![ı^ =i=unit vector of x axis ȷ^ =j=unit vector of y axis v= [(x),(y) ] v′=kv, k= [(i_x ,j_x ),(i_y ,j_y ) ] (k = transformation matrix) i_x , i_y are new coordinates for i j_x , j_y are new coordinates for j ∴v′= [(i_x ,j_x ),(i_y ,j_y ) ] [(x),(y) ]= [((xi_x +yj_x )),((xi_y +yj_y )) ] therefore: x′=xi_x +yj_x y′=xi_y +yj_y let area be A=(1/2)height×base = (1/2)x′y′ How can we integrate with respect to both i_x and i_y ??? ∣i∣=(√(i_x ^2 +i_y ^2 )) (=1 for original i ∵ i_x =1∧i_y =0)](https://www.tinkutara.com/question/Q7094.png)
$$\hat {\imath}=\boldsymbol{{i}}=\mathrm{unit}\:\mathrm{vector}\:\mathrm{of}\:{x}\:\mathrm{axis} \\ $$$$\hat {\jmath}=\boldsymbol{{j}}=\mathrm{unit}\:\mathrm{vector}\:\mathrm{of}\:{y}\:\mathrm{axis} \\ $$$$ \\ $$$$\boldsymbol{{v}}=\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix} \\ $$$$\boldsymbol{{v}}'={k}\boldsymbol{{v}},\:\:\:{k}=\begin{bmatrix}{{i}_{{x}} }&{{j}_{{x}} }\\{{i}_{{y}} }&{{j}_{{y}} }\end{bmatrix}\:\left({k}\:=\:\mathrm{transformation}\:\mathrm{matrix}\right) \\ $$$${i}_{{x}} ,\:{i}_{{y}} \:\:\mathrm{are}\:\mathrm{new}\:\mathrm{coordinates}\:\mathrm{for}\:\boldsymbol{{i}} \\ $$$${j}_{{x}} ,\:{j}_{{y}} \:\:\mathrm{are}\:\mathrm{new}\:\mathrm{coordinates}\:\mathrm{for}\:\boldsymbol{{j}} \\ $$$$ \\ $$$$\therefore\boldsymbol{{v}}'=\begin{bmatrix}{{i}_{{x}} }&{{j}_{{x}} }\\{{i}_{{y}} }&{{j}_{{y}} }\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\end{bmatrix}=\begin{bmatrix}{{xi}_{{x}} +{yj}_{{x}} }\\{{xi}_{{y}} +{yj}_{{y}} }\end{bmatrix} \\ $$$$\mathrm{therefore}: \\ $$$${x}'={xi}_{{x}} +{yj}_{{x}} \\ $$$${y}'={xi}_{{y}} +{yj}_{{y}} \\ $$$$\mathrm{let}\:{area}\:\mathrm{be}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{height}×\mathrm{base}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}'{y}' \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{we}\:\mathrm{integrate}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to} \\ $$$$\mathrm{both}\:{i}_{{x}} \:\mathrm{and}\:{i}_{{y}} ??? \\ $$$$\mid\boldsymbol{{i}}\mid=\sqrt{{i}_{{x}} ^{\mathrm{2}} +{i}_{{y}} ^{\mathrm{2}} }\:\:\:\:\:\left(=\mathrm{1}\:\mathrm{for}\:\mathrm{original}\:\boldsymbol{{i}}\:\because\:{i}_{{x}} =\mathrm{1}\wedge{i}_{{y}} =\mathrm{0}\right) \\ $$