Problem 317 - Calculus 3

Question

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test. $$ f(x, y)=\sqrt{x^{2}+y^{2}}+1 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Function

The function given is \[ f(x, y) = \sqrt{x^2 + y^2} + 1 \].This function describes a cone in three-dimensional space with a vertex shifted along the z-axis. Critical points are points where the partial derivatives of the function either equal zero or do not exist.

Step 2: Calculate Partial Derivatives

To find critical points, we first need to compute the partial derivatives of the function with respect to \(x\) and \(y\).

Step 3: Calculate \(\frac{\partial f}{\partial x}\)

Compute the partial derivative with respect to \(x\):\[ \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \]This partial derivative exists except at the origin \((x, y) = (0, 0)\).

Step 4: Calculate \(\frac{\partial f}{\partial y}\)

Compute the partial derivative with respect to \(y\):\[ \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \]This partial derivative also exists except at the origin \((x, y) = (0, 0)\).

Step 5: Determine the Critical Points

Set the partial derivatives to zero to find potential critical points:- From \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = 0\): \(x = 0\)- From \(\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} = 0\): \(y = 0\)Thus, the only point where both derivatives are zero is \((x, y) = (0, 0)\).

Step 6: Verify Result Using Partial Derivatives

Examine the behaviour of the partial derivatives at the origin:At \((0,0)\), the partial derivatives\[ \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \text{ and } \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \] are undefined because the denominator is zero, confirming that there is no directional slope.

Step 7: Conclusion

The critical point of the function \(f(x, y) = \sqrt{x^2 + y^2} + 1\) is at \((0, 0)\), confirmed by the undefined nature of the partial derivatives at this point.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental tool for analyzing multivariable functions. They measure how a function changes as one variable changes, keeping the other variables constant. In this context, partial derivatives help identify critical points. Critical points occur where the gradient of the function is zero or undefined, indicating potential maximum, minimum, or saddle points. When dealing with a function like \( f(x, y) = \sqrt{x^2 + y^2} + 1 \), the partial derivatives with respect to \( x \) and \( y \) are:

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

These derivatives are simple enough, yet they offer insight into the function's behavior. Importantly, they are undefined at the origin \((0,0)\), guiding us toward a potential critical point where further investigation is needed. Calculating partial derivatives is crucial in exposing where and how a function's surface changes.

Function Analysis

Function analysis involves understanding the behavior of a function based on its algebraic form and derivatives. For \( f(x, y) = \sqrt{x^2 + y^2} + 1 \), this function represents a cone with a rounded base shifted up by 1 unit in the z-direction. The shape gives an initial impression of how the surface behaves globally.By examining its first derivatives, we can pinpoint critical points or possible changes in slope and curvature. The term \( \sqrt{x^2 + y^2} \) suggests rotational symmetry around the origin. Therefore, both \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) approach zero at the origin. At this point, the behavior of the surface is undefined, which strengthens the impression of the origin as a critical point. In function analysis, both visual and mathematical interpretations play a crucial role.

Algebraic Techniques

Algebraic techniques can simplify the discovery of critical points, especially by manipulating the function's form. Completing the square, for instance, is a technique often used to make quadratic equations more manageable. While this approach isn't directly applicable to \( f(x, y) = \sqrt{x^2 + y^2} + 1 \), understanding the generalized equation of a circle as \( x^2 + y^2 \) shows the base of a cone centered at the origin. Instead of transforming this function algebraically, we rely on examining its derivatives.For simple functions like ours, setting the derivatives equal to zero or examining where they are undefined gives us the location of potential critical points. At \((x, y) = (0, 0)\), the undefined nature of the partial derivatives confirms our analysis. Algebraic techniques are essential for breaking down complex functions, but sometimes the straightforward analysis of derivatives is more effective.