Problem 105 - Calculus 3

Question

\([\mathrm{T}]\) a. Use a CAS to draw a contour map of \(z=\sqrt[4]{9-x^{2}-y^{2}}\) b. What is the name of the geometric shape of the level curves? c. Give the general equation of the level curves. d. What is the maximum value of \(z ?\) e. What is the domain of the function? f. What is the domain of the function?

Step-by-Step Solution

Answer

The level curves are circles; max \(z = 3\); domain is \(x^2+y^2 \leq 9\).

Step 1: Understanding the Function

The function given is \(z=\sqrt[4]{9-x^{2}-y^{2}}\). It represents a multivariable expression where \(x\) and \(y\) are independent variables and \(z\) is dependent on them.

Step 2: Contour Map Using CAS

To draw a contour map of the function \(z = \sqrt[4]{9-x^{2}-y^{2}}\), use a Computer Algebra System (CAS) like Desmos or GeoGebra. This will allow you to visualize level curves, which are the cross-sections of the function at different fixed values of \(z\).In a CAS, input the given function and use the contour plot option to visualize the map. Level curves will appear as concentric shapes around a central point, corresponding to different values of \(z\).

Step 3: Identify the Shape of Level Curves

The level curves of the function are circles, since the expression \(9-x^{2}-y^{2}=c^{4}\) (where \(c = z\)) fits the standard form of a circle equation in the \(xy\)-plane.

Step 4: General Equation of Level Curves

Set \(z = c\), where \(c\) is a constant level, then: \(\sqrt[4]{9-x^{2}-y^{2}} = c\). Solving for \(x^2+y^2\), we get:\[ x^2 + y^2 = 9 - c^4 \].These are the equations of circles with radius \(\sqrt{9 - c^4}\).

Step 5: Determine Maximum Value of \(z\)

The maximum value of \(z\) occurs when \(9-x^{2}-y^{2}\) is at its maximum, that is when \(x=0\) and \(y=0\). Thus, the maximum value of \(z = \sqrt[4]{9}\), which evaluates to 3.

Step 6: Find the Domain of the Function

The domain of the function is determined by the expression under the fourth root must be non-negative: \[ 9 - x^2 - y^2 \geq 0 \].This simplifies to:\[ x^2 + y^2 \leq 9 \].So, the domain is the set of all \((x, y)\) such that they lie within or on a circle of radius 3, centered at the origin.

Key Concepts

Level Curves

Level curves are a fundamental concept in multivariable calculus. They are essentially the cross-sections of a multivariable function, sliced parallel to the plane, where the function maintains a constant value. More simply, if you imagine a 3D surface and slice it horizontally at a particular height, the shape you see on the plane is a level curve.

In the context of the exercise, the function is given by \(z = \sqrt[4]{9-x^{2}-y^{2}}\). By setting \(z = c\), where \(c\) is a constant, the expression becomes \(\sqrt[4]{9-x^{2}-y^{2}} = c\). Solving this, we obtain a form of the equation for the level curves \(x^2 + y^2 = 9 - c^4\). These level curves reveal that at any constant \(z\), the resultant curve in the \(xy\)-plane is a circle.

• The radius of these circles is \(\sqrt{9 - c^4}\).
• The center of each circle is at the origin \((0,0)\).

Thus, level curves simplify the visualization of the function's behavior as it helps to see how values change in the \(x-y\) plane.

Contour Map

A contour map is an indispensable tool in visualizing level curves of a scalar field. It's a 2D representation where lines connect points of equal value. For the function \(z = \sqrt[4]{9-x^{2}-y^{2}}\), the contour map is made up of concentric circles formed by its level curves. This graphical representation helps us understand how the function varies with different inputs.

To create this contour map, one can use a Computer Algebra System (CAS) like GeoGebra or Desmos. These tools automatically generate the contour lines based on the function provided, making it easier to identify patterns.

• Each contour line corresponds to a particular \(z\) value, showing us constant-valued paths on the \(x-y\) plane.
• Closer lines indicate a steeper incline or decline in the \(z\)-values.
• With circles in this case, the contour map illustrates symmetry about the center (the origin).

By examining the contour map, one can readily discern not only the shape of the level curves but also understand the spatial orientation of the function as a whole.

Domain of a Function

The domain of a function includes all possible input values that lead to valid outputs. For the given function \(z = \sqrt[4]{9-x^{2}-y^{2}}\), determining the domain involves ensuring the values under the fourth root remain non-negative:
\[ 9 - x^2 - y^2 \geq 0 \]

Solving this inequality provides:
\[ x^2 + y^2 \leq 9 \]
This implies that \((x, y)\) coordinates must lie within or on the boundary of a circle with a radius of 3 and centered at the origin.

• Every point inside this circle satisfies the inequality and thus belongs to the function's domain.
• The circle equation \(x^2 + y^2 = 9\) marks the boundary of the domain.

Recognizing this domain ensures we respect where the function is mathematically viable, avoiding undefined or non-real numbers.

Maximum Value

In multivariable functions, identifying the maximum value is often a key interest. For the function \(z = \sqrt[4]{9-x^2-y^2}\), the maximum value of \(z\) can be found at the point where \(x^2 + y^2\) is minimized, meaning both \(x\) and \(y\) are equal to zero. This corresponds to the very center of the domain circle where the value under the fourth root is maximal.

Calculating for this condition, the maximum value is:
\[ z = \sqrt[4]{9} = 3 \]

• Go to \((x,y) = (0,0)\) for the maximum \(z\).
• This maximum occurs because any positive \(x^2 + y^2\) would reduce \(9 - x^2 - y^2\) directly influencing the value of \(z\).

Knowing the maximum is especially useful in applications involving optimization problems, where pushing the limits of output is often required.