Problem 22 - Calculus for Biology and Medicine

Question

Use the definition of continuity to show that $$ f(x, y)=\sqrt{9+x^{2}+y^{2}} $$ is continuous at \((0,0)\).

Step-by-Step Solution

Answer

The function is continuous at \((0,0)\) by the definition of continuity.

Step 1: Recall the Definition of Continuity at a Point

A function \( f(x, y) \) is continuous at a point \( (a, b) \) if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( \sqrt{(x-a)^2 + (y-b)^2} < \delta \), it follows that \( |f(x, y) - f(a, b)| < \epsilon \).

Step 2: Evaluate the Function at the Given Point

Calculate \( f(0, 0) \):\[f(0, 0) = \sqrt{9 + 0^2 + 0^2} = 3\]Thus, \( f(0, 0) = 3 \).

Step 3: Set Up the Absolute Value Inequality

We need \( |f(x, y) - f(0, 0)| = |\sqrt{9 + x^2 + y^2} - 3| < \epsilon \).

Step 4: Simplify the Expression

Rewrite the expression:\[|\sqrt{9 + x^2 + y^2} - 3|\]Using the inequality \( |u - v| = \frac{|u^2 - v^2|}{|u + v|} \):\[= \frac{|(9 + x^2 + y^2) - 9|}{|\sqrt{9 + x^2 + y^2} + 3|}\]= \frac{|x^2 + y^2|}{|\sqrt{9 + x^2 + y^2} + 3|}.

Step 5: Bound the Denominator

Observe that \( \sqrt{9 + x^2 + y^2} \geq 3 \), hence:\[ |\sqrt{9 + x^2 + y^2} + 3| \geq 6\]

Step 6: Choose Delta Sufficiently Small

We want \( \frac{x^2 + y^2}{6} < \epsilon \). Find \( \delta \) such that whenever \( \sqrt{x^2 + y^2} < \delta \), we ensure this inequality holds.Set \( \delta^2 = 6\epsilon \), thus choosing \( \delta = \sqrt{6\epsilon} \) works. Whenever \( \sqrt{x^2 + y^2} < \delta \), then \( x^2 + y^2 < 6\epsilon \), ensuring \( \frac{x^2 + y^2}{6} < \epsilon \).

Step 7: Conclude Continuity

For any \( \epsilon > 0 \), we found a \( \delta = \sqrt{6\epsilon} \) such that \( |\sqrt{9 + x^2 + y^2} - 3| < \epsilon \), which satisfies the definition of continuity. Hence, \( f(x, y) \) is continuous at \( (0,0) \).

Key Concepts

Definition of Continuity

Continuity at a point is a fundamental concept in calculus that ensures a function behaves predictably around that point. For a function of two variables, like \( f(x, y) \), to be continuous at a specific point \( (a, b) \):

• For every positive number \( \epsilon \), no matter how small, there should be a positive number \( \delta \) that defines a radius around \( (a, b) \).
• Within that radius, the distance from the original function value at \( (a, b) \) to any function value within the radius should be less than \( \epsilon \).

This means that small changes in \( x \) and \( y \) should produce small changes in the function's value, assuring that the function doesn't "jump" or behave unpredictably.
The essence of continuity is maintaining a function's value close to its value at \( (a, b) \) as you move closer to that point.

Multivariable Functions

Multivariable functions are functions that depend on more than one input variable. In our example, the function \( f(x,y) = \sqrt{9+x^2+y^2} \) uses both \( x \) and \( y \) as variables. These types of functions are inherently more complex because they describe surfaces or curves in higher dimensions than single-variable functions, which describe lines or simple curves.

• They map a pair of numbers, \( (x, y) \), to a single output value.
• Visualizing such functions often involves graphing in three dimensions.

When evaluating continuity for a multivariable function, like \( f(x, y) \), it's important to remember that changes in any direction in the \( xy \)-plane affect the output. This makes proving continuity slightly more complicated, as you have to ensure the continuity condition is satisfied in all directions around a point.

Epsilon-Delta Proof

The epsilon-delta proof is a formal way to demonstrate the continuity of a function. It might sound complex, but it essentially boils down to showing that:

• For any desired level of closeness, \( \epsilon \), between the function's value at a point and points nearby, there exists a radius \( \delta \) such that all points within this radius yield values close to the original value.
• In our example, we showed that for the given function \( f(x, y) = \sqrt{9+x^2+y^2} \), you can find a \( \delta \) such that if \( \sqrt{x^2 + y^2} < \delta \), then the function value stays within \( \epsilon \) of the value at \( (0, 0) \).

This careful choice of \( \delta \) ensures that as you move towards \( (0, 0) \), the output of \( f(x, y) \) changes gently and predictably, fulfilling the continuity criteria. It's a mathematical certificate verifying a function's smooth behavior.