Problem 310 - Calculus Volume 3

Question

For the following exercises, find all critical points. $$f(x, y)=1+x^{2}+y^{2}$$

Step-by-Step Solution

Answer

The only critical point is \((0, 0)\).

Step 1: Understanding Critical Points

Critical points of a function occur where its gradient is zero or undefined. For our function, the gradient is the vector of the first partial derivatives.

Step 2: Finding Partial Derivatives

Calculate the partial derivatives of the function. The partial derivative with respect to \(x\) is:\[ \frac{\partial f}{\partial x} = 2x \] The partial derivative with respect to \(y\) is:\[ \frac{\partial f}{\partial y} = 2y \]

Step 3: Setting Gradients to Zero

To find the critical points, set the first partial derivatives equal to zero:1. \( 2x = 0 \)2. \( 2y = 0 \)

Step 4: Solving for Variables

Solve the equations:1. \( 2x = 0 \) implies \( x = 0 \).2. \( 2y = 0 \) implies \( y = 0 \).

Step 5: Concluding Critical Points

The critical points occur where both partial derivatives are zero at the same time. Thus, the only critical point for the function is \((x, y) = (0, 0)\).

Key Concepts

Partial Derivatives

Partial derivatives are like the backbone of multivariable calculus. They help us understand how a function changes as one variable changes, while keeping others constant. If you imagine a function like a landscape, partial derivatives tell us about the slope of that landscape in different directions. For example, if you have a function of two variables like our function here, the partial derivative with respect to \(x\), noted as \( \frac{\partial f}{\partial x} \), gives the rate of change of the function as \(x\) changes, while keeping \(y\) fixed.

In the context of our exercise, we calculated the partial derivatives as:

• \( \frac{\partial f}{\partial x} = 2x \)
• \( \frac{\partial f}{\partial y} = 2y \)

These tell us how \(f(x, y)\) changes as \(x\) or \(y\) alone changes. Partial derivatives are crucial when we're trying to find critical points, as they help form the gradient vector.

Gradient Vector

The gradient vector is a powerful tool in multivariable calculus. It's like a compass that points in the direction of the steepest ascent of a function. In simpler terms, if you were standing on a hill represented by the function, the gradient tells you which way to walk to climb up the steepest slope.

For our function \(f(x, y) = 1 + x^2 + y^2\), the gradient vector is formed by combining the partial derivatives. So, in vector form, it looks like this:

• \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2x, 2y) \)

The gradient vector helps us find critical points by identifying where the gradient is zero or undefined. In our exercise, since the gradient vector is \((0, 0)\) when \(x = 0\) and \(y = 0\), this indicates a critical point at \((0, 0)\).

Multivariable Calculus

Multivariable calculus extends the concepts of ordinary calculus to functions with more than one variable. It helps us understand and solve real-world problems where several factors can change simultaneously. In our function \(f(x, y) = 1 + x^2 + y^2\), there are two variables, \(x\) and \(y\), influencing the function's value.

In this realm, derivatives become partial derivatives, and functions can be visualized in 3D as surfaces or landscapes with hills and valleys. Critical points found through partial derivatives and the gradient vector reveal peaks, troughs, or flat plains of these surfaces.

• These concepts allow us to optimize solutions in fields such as physics, economics, and engineering, where multiple variables need to be considered.

By learning to calculate partial derivatives and work with gradient vectors, students can better understand how multiple variables interact in a system.

Optimization in Calculus

Optimization involves finding the best solution or outcome in a given situation. In calculus, this often means identifying maxima or minima of a function. The critical points discovered with partial derivatives and gradient vectors play a central role in this process.

For instance, in the context of our exercise, we might want to find the highest or lowest point on the surface of our function in the given region. To do that:

• First, we locate critical points by setting the gradient vector to zero.
• Next, we analyze these points to determine whether they are maxima, minima, or saddle points, often using second derivative tests.

Understanding optimization in calculus allows students to solve practical problems, like minimizing cost, maximizing efficiency, or strategically managing resources. This concept is fundamental not only in mathematics but in various applications where optimal outcomes are desired.