Question

If x = u (1+v) , y = v (1+u), find the Jacobian of x, y with respect to u, v.

Answer 1

Hint: Find the partial derivatives $ \dfrac{\partial x}{\partial u},\dfrac{\partial y}{\partial u},\dfrac{\partial x}{\partial v},\dfrac{\partial y}{\partial v} $ . In the operation $ \dfrac{\partial }{\partial u} $ , differentiate the function with respect to u keeping v constant while in the operation $ \dfrac{\partial }{\partial v} $ , differentiate the function with respect to v keeping u constant. Now, apply the formula for the Jacobian of x, y with respect to u, v given as $ J\dfrac{(x,y)}{(u,v)}=\left| \begin{matrix}
   \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\
   \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \\
\end{matrix} \right| $ and expand the determinant to get the answer.

Complete step by step answer:
Here, we are provided with the functions x = u (1+v) and y = v (1+u) and we have been asked to find the Jacobian of x, y with respect to u, v. First let us see what is meant by Jacobian.
Now, if x and y are functions of independent variables x and v, then Jacobian of x, y with respect to u, v is given as $ J\dfrac{(x,y)}{(u,v)}=\dfrac{\partial (x,y)}{\partial (u,v)}=\left| \begin{matrix}
   \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\
   \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \\
\end{matrix} \right| $ . In this formula the terms $ \dfrac{\partial x}{\partial u},\dfrac{\partial y}{\partial u},\dfrac{\partial x}{\partial v},\dfrac{\partial y}{\partial v} $ denotes the partial derivative of x and y with respect to u and partial derivative of x and y with respect to v respectively. In the partial derivative of the functions x and y with respect to u we have to keep v constant we have to keep u constant. So, let us find these partial derivatives. Therefore, we have,
 $ \begin{align}
  & \Rightarrow \dfrac{\partial x}{\partial u}=\dfrac{\partial \left[ u\left( 1+v \right) \right]}{\partial u}=\left( 1+v \right) \\
 & \Rightarrow \dfrac{\partial x}{\partial v}=\dfrac{\partial \left[ u\left( 1+v \right) \right]}{\partial v}=u \\
 & \Rightarrow \dfrac{\partial y}{\partial u}=\dfrac{\partial \left[ v\left( 1+u \right) \right]}{\partial u}=v \\
 & \Rightarrow \dfrac{\partial y}{\partial v}=\dfrac{\partial \left[ v\left( 1+u \right) \right]}{\partial v}=\left( 1+u \right) \\
\end{align} $
Now, substituting the above obtained values of partial derivative in the formula of Jacobian, we get,
 $ \Rightarrow J\dfrac{(x,y)}{(u,v)}=\left| \begin{matrix}
   \left( 1+v \right) & v \\
   u & \left( 1+u \right) \\
\end{matrix} \right| $
Now, expanding the above determinant, we get,
 $ \begin{align}
  & \Rightarrow J\dfrac{(x,y)}{(u,v)}=\left( 1+v \right)\left( 1+u \right)-uv \\
 & \Rightarrow J\dfrac{(x,y)}{(u,v)}=1+u+v+uv-uv \\
\end{align} $
Cancelling the like terms, we get,
 $ \Rightarrow J\dfrac{(x,y)}{(u,v)}=1+u+v $
Hence, the above expression is our answer.

Note:
 You must remember the formula of Jacobian to solve the above question. You may see that in the above question there were two functions and two variables and that is why we obtained a determinant, with order 2. If there would have been three functions (x, y, z) and three variables (u, v, w) then we would have obtained a determinant of order 3. Remember the process of finding partial derivatives of functions. Lastly, remember that just like the derivative of a constant is 0, the partial derivative of a constant is also 0.