Answer to Question #297115 in Calculus for Talkiedo

Question

Determine which of the following

pairs of functions are independent. Using the Jacobian matrix

a) u= x cos y and v=x sin y;

b)u= x + y and v= y/x+y;

c)u=x-2y and v=x^2 +4y^2-4xy+3x-6y;

d)u= x + 2y and v= x^2-y^2+2xy-x

Answer 1

ANSWER

Functions continuously differentiable are independent in the domain $D$ if the Jacobian  $J=\frac{\partial (u,v)}{\partial (x,y) }=\begin{vmatrix} \frac{\partial u }{\partial x} &\frac{\partial u }{\partial y} &\\ \frac{\partial v }{\partial x} & \frac{\partial v }{\partial y} & \end{vmatrix}$ is nonzero.

a)

$J=\frac{\partial (u,v)}{\partial (x,y) }=\begin{vmatrix} \ { \cos y } & {\sin y} \\ \ { - x\sin y } & \ {x\cos y} & \end{vmatrix}=x$

Functions are independent in some neighborhood of any point of the region $D=\left \{ \left ( x,y \right ):x\neq{0} \right \}$ .

b)

$J=\frac{\partial (u,v)}{\partial (x,y) }=\begin{vmatrix} \ { 1 } & { 1} \\ \ { \frac{- y }{x^{2}} } & \ {\frac{1}{x}+1} & \end{vmatrix}=\frac{1}{x}+1+ { \frac{ y }{x^{2}} }\\$

Functions are independent in some neighborhood of any point of the region $D=\left \{ \left ( x,y \right ):x\neq0, \frac{1}{x}+1+{ \frac{ y }{x^{2}}\ \neq{0} } \right \}$ .

c)

$J=\frac{\partial (u,v)}{\partial (x,y) }=\begin{vmatrix} \ {1 } & {-2} \\ \ {2x-4y+3 } & \ {8y-4x-6} & \end{vmatrix}= 8y-4x-6+2(2x-4y+3)=0$

  Functions are dependent.

d)

$J=\frac{\partial (u,v)}{\partial (x,y) }=\begin{vmatrix} \ {1 } & { 2} \\ \ {2x+2y-1 } & \ {-2y+2x} & \end{vmatrix}=\\= -2y+2x-4x-4y+2=-2x-6y+2$

Functions are independent in some neighborhood of any point of the region

$D=\left \{ \left ( x,y \right ):-2x-6y+2\neq0 \right \}.$