Problem 18 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the level curves of the function corresponding to each value of \(z\). \(h(u, v)=\sqrt{4-u^{2}-v^{2}}\)

Step-by-Step Solution

Answer

For the given function \(h(u, v) = \sqrt{4-u^2-v^2}\), we set it equal to a constant value z: \(z = \sqrt{4-u^2-v^2}\). Squaring both sides and isolating the u term gives us \(u^2 = 4 - z^2 - v^2\). Now, we plot the level curves on the xy-plane for different values of z (e.g., -2, -1, 0, 1, 2). The level curves represent circles centered at the origin with different radii and a vertical line through the origin.

Step 1: Set the function equal to a constant value (z)

Let's set the function \(h(u, v) = z\). So we have: \[z = \sqrt{4-u^2-v^2}\]

Step 2: Solve the equation for u or v in terms of the other

To solve for u or v, first square both sides of the equation: \[z^2 = 4 - u^2 -v^2\] Now, let's isolate the u term. We'll have: \[u^2 = 4 - z^2 - v^2\] Since we're asked to sketch the level curves and not find the exact points, we can stop here and proceed to the next step.

Step 3: Plot the level curves on the xy-plane for different values of z

To plot the level curves, we just need to assign values to z and see how the equation behaves. Let's plot the level curves for the following values of z: -2, -1, 0, 1, and 2. 1. For \(z = -2\), \(u^2 = 0\). This means that u = 0. The level curve will be a vertical line at u = 0. 2. For \(z = -1\), \(u^2 = 3 - v^2\), which is a circle with radius \(\sqrt{3}\) centered at the origin. 3. For \(z = 0\), \(u^2 = 4 - v^2\), which is a circle with radius 2 centered at the origin. 4. For \(z = 1\), \(u^2 = 3 - v^2\), which is a circle with radius \(\sqrt{3}\) centered at the origin. 5. For \(z = 2\), \(u^2 = 0\). This means that u = 0. The level curve will be a vertical line at u = 0. Now, you can sketch these level curves on the xy-plane. You'll see that for different values of z, the level curves represent circles centered at the origin with different radii and a vertical line through the origin.

Key Concepts

Multivariable Calculus

Multivariable calculus involves the study of functions with more than one variable. In this exercise, the function \[ h(u, v) = \sqrt{4 - u^2 - v^2} \]is a prime example as it involves two variables, \(u\) and \(v\). Function with multiple variables enables us to analyze surfaces and curves that exist in higher-dimensional spaces, beyond the typical line or parabola in single-variable calculus.

When we explore such multivariable functions, we consider many outcomes of how variables interact and affect the function's value. Important concepts in multivariable calculus include partial derivatives, gradients, and vectors, which help in understanding the behavior of such functions.

• Partial Derivatives: Finds how a function changes as one variable changes, holding others constant.
• Gradients: Provide the direction of fastest increase of a function and can help understand its nature at a point.

These concepts form the groundwork necessary to work with higher-dimensional functions and are crucial when dealing with topics like level curves.

Function Sketching

Sketching functions is a fundamental skill in calculus, aiding in visualizing a function's behavior without computing many values. In this particular problem, we consider level curves, which are 2D representations for a 3D effect.

Level curves are generated by setting the function equal to constant values, which was done in the exercise by letting \(h(u, v) = z\). This results in curves such as circles or vertical lines on the plane, each corresponding to different constant values of \(z\).

• Choosing the right constants: Select values of \(z\) that help reveal the symmetry and pattern in the curves.
• Outcome of plotted curves: These curves give insight into the function's topography, like representing a mountain's height on a map.

Function sketching connects algebraic expressions to visual models, providing an intuitive grasp of mathematical concepts.

Level Set Analysis

Level set analysis allows one to understand and draw functions through graphic reasoning and conceptual visualization. There are several key ideas attached to level sets which manifest in the exercise:

In the given function, \[ z = \sqrt{4 - u^2 - v^2} \]this assumes the role of level curves on the \(uv\)-plane when \(z\) takes on different constant values. Each constant \(z\) value generates a unique set of points satisfying the equation.

• Interpretation: Each level curve is associated with a unique height or value of \(z\).
• Geometry: Identifies points where the variables \(u\) and \(v\) result in the same outcome, describing circles or lines as seen in the exercise.

Level set analysis, in essence, offers a way to 'slice' a function into understandable sections, each revealing a part of the whole mathematical story.