Problem 31 - Calculus an Applied Approach

Question

Find the critical points of the function and, from the form of the function, determine whether a relative maximum or a relative minimum occurs at each point. $$ f(x, y, z)=(x-1)^{2}+(y+3)^{2}+z^{2} $$

Step-by-Step Solution

Answer

The function f(x, y, z) = (x-1)^2 + (y+3)^2 + z^2 has a critical point at (1, -3, 0), which is a relative minimum.

Step 1: Compute the Gradient

First, compute the gradient of the function by taking the partial derivatives of the function with respect to x, y, and z: \( \begin{align*} \frac{\partial f}{\partial x} &= 2(x -1) \\frac{\partial f}{\partial y} &= 2(y+3) \\frac{\partial f}{\partial z} &= 2z \end{align*} \)

Step 2: Find the Critical Points

Set each of the partial derivatives equal to zero and solve for x, y, and z to find the critical points:\[\begin{align*} 2(x-1) &= 0 \ 2(y+3) &= 0 \ 2z &= 0 \end{align*} \]This gives the solutions x = 1, y = -3, and z = 0. Therefore, the function has a critical point at (1, -3, 0)

Step 3: Second Derivative Test

Compute the second order partial derivatives:\[\begin{align*} \frac{\partial^2 f}{\partial x^2} &= 2 \ \frac{\partial^2 f}{\partial y^2} &= 2 \ \frac{\partial^2 f}{\partial z^2} &= 2 \end{align*} \]All these values are positive, which implies that the function has a relative minimum at (1, -3, 0)

Key Concepts

Gradient of a Function

Understanding the gradient of a function is crucial when analyzing its behavior in multivariable calculus. The gradient is a vector representation of the rate of change of a function, pointing in the direction of the steepest ascent. In the context of the function
\( f(x, y, z) = (x-1)^{2}+(y+3)^{2}+z^{2} \),
the gradient would involve taking the derivative with respect to each variable, often termed as \emph{partial derivatives}. For a three-dimensional function like this one, the gradient is composed of three components, each representing the partial derivative in relation to one of the variables. These components answer the question: how does the function change as we move along the x-axis, y-axis, or z-axis independently of the others?
By setting the gradient equal to the zero vector, which is done by equating each component (\(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\)) to zero, we can obtain the critical points. These points are where the function does not increase or decrease in any direction, indicating potential maxima, minima, or saddle points.

Partial Derivatives

The calculation of partial derivatives is a fundamental aspect when studying functions of several variables. A partial derivative of a multivariable function, such as
\( f(x, y, z) = (x-1)^{2}+(y+3)^{2}+z^{2} \),
measures how the function changes as one variable varies, while the others are held constant. To put it in other words, it's like focusing on the slope or rate of change along an axis, temporarily ignoring the rest of the terrain.

Importance in Critical Points

For identifying critical points, each partial derivative is set to zero separately. This operation essentially finds where the 'slope' in each direction levels out—a sign that we might be at the top of a hill (maximum), the bottom of a valley (minimum), or a flat stretch on a mountain side (saddle point). As shown in the solution for the function above, by calculating the partial derivatives, we were able to pinpoint the location of a critical point at \((1, -3, 0)\).

Second Derivative Test

Once critical points are found for a multivariable function, the natural question to ask is: 'Are we at a peak, a valley, or a flat plain?' This is where the second derivative test comes into play, providing information about the concavity at those critical points. For a function like
\( f(x, y, z) = (x-1)^{2}+(y+3)^{2}+z^{2} \),
we apply the test by taking the second derivatives with respect to each variable. These second derivatives tell us whether the function curves up or down along each axis at the critical point.

Applying the Test

If the second derivatives are positive, as they are in this case (\(\frac{\partial^2 f}{\partial x^2} = 2\), \(\frac{\partial^2 f}{\partial y^2} = 2\), and \(\frac{\partial^2 f}{\partial z^2} = 2\)), it suggests that we have a local minimum at the critical point. The function is curving upwards from the critical point in every direction examined separately. This insight, derived from the second derivative test, confirms that the point (1, -3, 0) is indeed a relative minimum for the function.