Problem 37 - Calculus Early Transcendentals

Question

Evaluate the Jacobians \(J(u, v, w)\) for the following transformations. $$x=v+w, y=u+w, z=u+v$$

Step-by-Step Solution

Answer

Question: Determine the Jacobian for the given transformation: x = v + w, y = u + w, z = u + v. Answer: The Jacobian for the given transformation is J(u,v,w) = 2.

Step 1: Determine the partial derivatives

First, we'll write down the transformation equations and find the partial derivatives of x, y, and z with respect to u, v, and w. Given transformation equations: $$x = v + w$$ $$y = u + w$$ $$z = u + v$$ Now find the partial derivatives: $$\frac{\partial x}{\partial u} = 0, \frac{\partial x}{\partial v} = 1, \frac{\partial x}{\partial w} = 1$$ $$\frac{\partial y}{\partial u} = 1, \frac{\partial y}{\partial v} = 0, \frac{\partial y}{\partial w} = 1$$ $$\frac{\partial z}{\partial u} = 1, \frac{\partial z}{\partial v} = 1, \frac{\partial z}{\partial w} = 0$$

Step 2: Form the matrix of partial derivatives

Next, we form the matrix of partial derivatives using the values we found in step 1: $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} $$

Step 3: Evaluate the determinant of the matrix

Now, we'll evaluate the determinant of the matrix, which represents the Jacobian J(u,v,w): $$ J(u, v, w) = \begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} $$ Expanding along the first row, we get: $$J(u, v, w) = 0(0 - 1) - 1(1(0) - 1(1)) + 1(1 - 1(1))$$ $$J(u,v,w) = - (-1) - 1(-1) + 1(0)$$ $$J(u, v, w) = 1 + 1 = 2$$ Thus, the Jacobian for the given transformation is J(u,v,w) = 2.