Finding and Classifying Stationary Points for a Function of Three Variables Quiz

Finding and Classifying Stationary Points for a Function of Three Variables

Finding and classifying stationary points for a function of three variables is a fundamental task in multivariable calculus. This involves locating points where the gradient of the function is equal to the zero vector and determining their nature using second derivative test or other methods. Here's how you can approach this problem:

Part 1: Finding the Partial Derivatives

The first step is to find the partial derivatives of your given function with respect to each variable. For example, if your function is denoted by f(x, y, z) = x^2 + 2y^2 - 3z^2 + 5xy + z^3, then we need to compute all six possible partial derivatives:

∂f/∂x = 2x + 5y ∂f/∂y = 4y + 5x ∂f/∂z = -6z + z^2

These represent the instantaneous rate of change of the function along each coordinate axis.

Part 2: Setting up the Equations

Next, set these partial derivatives equal to zero to find the critical points, which will be our candidates for stationary points. In our case, we get three equations:

1. 2x + 5y = 0
2. 4y + 5x = 0
3. -6z + z^2 = 0

Solving these sets of simultaneous linear equations gives us the following solutions:

x = 0, y = 0, z = 0 x = -5y / 2, y = -x/5, z = 0 z = 0, y = -\frac{1}{6}z, z \neq 0

Since the third solution has a nonzero value for z, only the first two correspond to potential stationary points.

Part 3: Determining Nature

To determine whether these potential stationary points are local minima, maxima, or saddle points, we use the second derivative test. First, compute the Hessian matrix of the function:

H = |∂²f/∂x² ∂²f/∂x∂y ∂²f/∂x∂z| |∂²f/∂y∂x ∂²f/∂y² ∂²f/∂y∂z| |∂²f/∂z∂x ∂²f/∂z∂y ∂²f/∂z³|

For the first point (x = 0, y = 0), the Hessian is:

H = |2 5 0| |5 4 0| |0 0 -6|

At the second point, the Hessian is:

H = |-1/(2y) -10/(2y^2) 0| |-20/(2y) (20/(2y) + 1) -6y| |0 -6y -6(1+y)|

Now, evaluate the determinant of the Hessian at each point:

Det_H(0, 0) = 3 ≠ 0 Det_H(-5y/2, y = -x/5) = -147100y^3 ≠ 0

Since both determinants are nonzero, neither of these points corresponds to a local extremum. Therefore, there are no stationary points for the given function.

Test your understanding of finding and classifying stationary points for a function of three variables in multivariable calculus. Learn how to find partial derivatives, set up equations to locate critical points, and determine their nature using the second derivative test. This quiz will help you master the process of analyzing stationary points.