Problem 11 - The Calculus with Analytic Geometry

Question

In Exercises 8 through 17, determine the region of continuity of \(f\) and draw a sketch showing as a shaded region in \(R^{2}\) the region of continuity of \(f\). f(x, y)=\frac{x}{\sqrt{4 x^{2}+9 y^{2}-36}}

Step-by-Step Solution

Answer

The function \( f(x, y) = \frac{x}{\sqrt{4x^2 + 9y^2 - 36}} \) is continuous outside the ellipse \[ \frac{x^2}{9} + \frac{y^2}{4} = 1. \]

Step 1 - Determine the domain of the function

Identify the set of points \( (x, y) \) for which the function \( f(x, y) = \frac{x}{\sqrt{4x^2 + 9y^2 - 36}} \) is defined. The function is defined everywhere the denominator is non-zero and the expression under the square root is positive. Thus, we need \[ 4x^2 + 9y^2 - 36 > 0. \]

Step 2 - Solve the inequality

Solve the inequality \[ 4x^2 + 9y^2 - 36 > 0 \] to determine the region where the function is continuous. Rewrite the inequality as \[ 4x^2 + 9y^2 > 36. \] Divide both sides by 36 to get \[ \frac{4x^2}{36} + \frac{9y^2}{36} > 1, \] which simplifies to \[ \frac{x^2}{9} + \frac{y^2}{4} > 1. \]

Step 3 - Identify the region of continuity

The inequality \[ \frac{x^2}{9} + \frac{y^2}{4} > 1 \] represents the region outside an ellipse centered at the origin with semi-major axis 3 along the x-axis and semi-minor axis 2 along the y-axis. The function \( f(x, y) \) is continuous in this region since the denominator remains non-zero and positive.

Step 4 - Sketch the region of continuity

Draw the ellipse given by the equation \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] in the \( xy \)-plane. Shade the area outside this ellipse to represent the region of continuity of the function \( f(x, y) \).

Key Concepts

Domain of a Function

To understand the region where a function is continuous, we first need to determine its domain. The domain of a function represents all possible input values \(x, y\) for which the function is defined.
For the function \(f(x, y) = \frac{x}{\sqrt{4x^2 + 9y^2 - 36}}\), it is defined only where the denominator is non-zero and the expression under the square root is positive.
This means we require: \[ 4x^2 + 9y^2 - 36 > 0. \]
Let's solve this inequality to find the domain.

Inequality Solving

Solving inequalities involves finding the set of values that satisfy the inequality.
For the inequality \[ 4x^2 + 9y^2 - 36 > 0, \] we first rewrite it as \[ 4x^2 + 9y^2 > 36. \]
Now, simplifying by dividing both sides by 36, we get \[ \frac{4x^2}{36} + \frac{9y^2}{36} > 1, \] which simplifies to \[ \frac{x^2}{9} + \frac{y^2}{4} > 1. \]
This inequality now tells us about an ellipse.

Ellipse Properties

The equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] represents a standard ellipse centered at the origin. An ellipse has two axes:

• Semi-major axis (longest one) along the x-axis
• Semi-minor axis (shortest one) along the y-axis

For our inequality \[ \frac{x^2}{9} + \frac{y^2}{4} > 1, \] the ellipse has:

• Semi-major axis = 3 (\( \sqrt{9} \))
• Semi-minor axis = 2 (\( \sqrt{4} \))

So, the ellipse is centered at the origin, stretching 3 units along the x-axis and 2 units along the y-axis.
The inequality tells us about the region outside this ellipse because it specifies \[ \frac{x^2}{9} + \frac{y^2}{4} > 1. \]

Multivariable Functions

Multivariable functions take multiple inputs (in this case, x and y) to produce an output. For the continuity of a multivariable function, it must not have any undefined points in the specified region.
For \[ f(x, y) = \frac{x}{\sqrt{4x^2 + 9y^2 - 36}}, \] the denominator must be positive and non-zero for the function to remain defined.
This means our solution domain, \[ \frac{x^2}{9} + \frac{y^2}{4} > 1, \] specifies the region where the function is continuous. This region lies entirely outside the ellipse centered at the origin with semi-major and semi-minor axes as 3 and 2.
By knowing this, we can sketch the ellipse and shade the outside area to visualize the region of continuity.