Problem 23 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the level curves of the function corresponding to each value of \(z\). \(f(x, y)=\sqrt{16-x^{2}-y^{2}} ; z=0,1,2,3,4\)

Step-by-Step Solution

Answer

The level curves of the function \(f(x, y)=\sqrt{16-x^{2}-y^{2}}\) for the given values of \(z\) are circles centered at the origin with the following radii: 4, \(\sqrt{15}\), 2\(\sqrt{3}\), \(\sqrt{7}\), and 0 (single point at the origin).

Step 1: Analyze the given function and values of z

We are given the function \(f(x, y)=\sqrt{16-x^{2}-y^{2}}\) and the constant values for \(z\), namely \(0,1,2,3\), and \(4\). We need to create equations based on these constant values.

Step 2: Obtain equations for each value of z

For each of the given values of \(z\), we will set the function equal to \(z\) and solve: 1. For \(z = 0\), we have \(0 = \sqrt{16-x^{2}-y^{2}}\) 2. For \(z = 1\), we have \(1 = \sqrt{16-x^{2}-y^{2}}\) 3. For \(z = 2\), we have \(2 = \sqrt{16-x^{2}-y^{2}}\) 4. For \(z = 3\), we have \(3 = \sqrt{16-x^{2}-y^{2}}\) 5. For \(z = 4\), we have \(4 = \sqrt{16-x^{2}-y^{2}}\) Square both sides of these equations to remove the square root.

Step 3: Square both sides

Sqaure both sides in the obtained equations: 1. \(0^2 = (16-x^{2}-y^{2}) \Rightarrow x^2 + y^2 = 16\) 2. \(1^2 = (16-x^{2}-y^{2}) \Rightarrow x^2 + y^2 = 15\) 3. \(2^2 = (16-x^{2}-y^{2}) \Rightarrow x^2 + y^2 = 12\) 4. \(3^2 = (16-x^{2}-y^{2}) \Rightarrow x^2 + y^2 = 7\) 5. \(4^2 = (16-x^{2}-y^{2}) \Rightarrow x^2 + y^2 = 0\) Now we have obtained five equations in terms of \(x\) and \(y\).

Step 4: Plot the level curves

Now we will plot these equations on the Cartesian plane to obtain the level curves for each value of \(z\). 1. \(x^2 + y^2 = 16\) is a circle with a radius of 4 and centered at the origin. 2. \(x^2 + y^2 = 15\) is a circle with a radius of \(\sqrt{15}\) and centered at the origin. 3. \(x^2 + y^2 = 12\) is a circle with a radius of 2\(\sqrt{3}\) and centered at the origin. 4. \(x^2 + y^2 = 7\) is a circle with a radius of \(\sqrt{7}\) and centered at the origin. 5. \(x^2 + y^2 = 0\) represents a single point, the origin (0,0). The level curves of the function \(f(x, y)=\sqrt{16-x^{2}-y^{2}}\) for the given constant values of \(z\) are circles centered at the origin with radii corresponding to each equation obtained in step 3.

Key Concepts

Contour Lines

Contour lines, also known as level curves, represent a powerful tool in visualizing functions of two variables in multivariable calculus. Imagine you are on a hike and you're using a map that shows the terrain's elevation through various closed loops. These loops are contour lines, with each loop indicating a constant elevation.

Let's apply this concept to the mathematical function from our exercise, where we have the function ( f(x, y)=√(16-x^{2}-y^{2}))). When asked to sketch the level curves for each value of ( z), we equate ( z) to the function to obtain specific 'heights'. For example, when ( z = 2), the level curve we trace on the ( x)-( y) plane will contain all points where the function's value is equal to 2. This is akin to tracing a loop on a map that defines a constant elevation above sea level.

During our step-by-step solution, we encountered circular level curves centered at the origin, illustrating one way in which complex multivariable functions can have very familiar geometric representations.

Multivariable Calculus

Multivariable calculus extends the reach of traditional calculus to functions of more than one variable. Instead of working with functions that depend on ( x) alone, such as ( f(x)), we consider functions like ( f(x, y)), where both ( x) and ( y) influence the output. This requires a different set of tools compared to single-variable calculus.

The concept of a derivative must be generalized to partial derivatives, where the rate of change is analysed in relation to each independent variable while keeping the others constant. Similarly, integration becomes a more complex task, often requiring techniques such as double or triple integrals depending on the number of variables.

Our exercise dives into one such aspect of multivariable calculus: finding and understanding level curves. By setting the function equal to a series of constant values, we are able to glimpse into the structure and behavior of the function across two dimensions, setting the stage for deeper analysis and visualization techniques.

Function Visualization

Function visualization is aimed at providing a graphical representation of mathematical functions, which in turn helps to convey complex information in a tangible manner. In multivariable calculus, where we deal with functions of multiple variables, visual tools like plots and graphs are indispensable for understanding the relationship between variables and the overall behavior of the function.

In our exercise, function visualization takes the form of sketching level curves. These curves transform an abstract function into a series of understandable shapes. For the function ( f(x, y)=√(16-x^{2}-y^{2})), we visualized the function’s output for specific values of ( z) as circles with varying radii. This immediately tells us that as we move away from the center point, the value of our function decreases until it reaches zero - the maximum value being at the origin where the radius is zero.

Through function visualization, we can quickly identify critical points, comprehend the rate of change, and analyze the geometric properties of the function, providing an enhanced understanding that algebraic expressions alone may not readily convey.