Problem 1 - Calculus with Concepts in Calculus

Question

Find the domain of the function. \(f(x, y)=\sqrt{x}+\sqrt{y}\)

Step-by-Step Solution

Answer

The domain is \(D = \{(x, y) \mid x \geq 0, y \geq 0\}\).

Step 1: Understand the Function Components

The function given is \(f(x, y) = \sqrt{x} + \sqrt{y}\). It consists of two square root terms: \(\sqrt{x}\) and \(\sqrt{y}\).

Step 2: Consider the Domain of \(\sqrt{x}\)

For \(\sqrt{x}\) to be defined in real numbers, \(x\) must be non-negative. Therefore, the condition for \(x\) is \(x \geq 0\).

Step 3: Consider the Domain of \(\sqrt{y}\)

Similarly, for \(\sqrt{y}\) to be defined in real numbers, \(y\) must be non-negative. Therefore, the condition for \(y\) is \(y \geq 0\).

Step 4: Combine the Domain Constraints

The constraints for \(x\) and \(y\) can be combined. The function is defined for all points \((x, y)\) such that \(x \geq 0\) and \(y \geq 0\).

Step 5: Conclusion on Domain

The domain of the function \(f(x, y) = \sqrt{x} + \sqrt{y}\) is all points in the first quadrant of the plane, including the axes, represented by \(D = \{(x, y) \mid x \geq 0, y \geq 0\}\).

Key Concepts

Square Root Function

The square root function is a fundamental component in many mathematical expressions. Mathematically, a square root of a number is a value that, when multiplied by itself, gives the original number. For instance, the square root of 9 is 3, because 3 times 3 equals 9.
When it comes to functions involving square roots, such as \(f(x, y) = \sqrt{x} + \sqrt{y}\), it's important to consider the concept of domain. The domain represents all the possible input values for which the function is defined.

For a square root to be defined in the set of real numbers, the expression under the square root must be non-negative. This is because the square root of a negative number isn't defined in the real number system. Therefore, for \(\sqrt{x}\) and \(\sqrt{y}\), \(x\) and \(y\) must be greater than or equal to zero. This results in the domain being defined in terms of these non-negativity constraints.

Multivariable Calculus

Multivariable calculus is an extension of calculus that deals with functions of multiple variables. While single-variable calculus focuses on functions with one input, multivariable calculus explores functions like \(f(x, y)\) where there are multiple inputs.
In the function \(f(x, y) = \sqrt{x} + \sqrt{y}\), both \(x\) and \(y\) are variables that can change independently. The concepts of limits, derivatives, and integrals extend to these multivariable functions, providing powerful techniques to analyze and solve complex problems.

Understanding the domain of these functions is crucial in multivariable calculus. The inputs need to lie within a specific region for the function to be well-defined. In our example, this region is the first quadrant of the plane \((x \geq 0, y \geq 0)\). This ensures both square roots are calculated with non-negative values, preserving real number computation and consistency.

Function Constraints

Function constraints are conditions that must be satisfied for a function to exist. In mathematical terms, these constraints often limit the domain and define the set of permissible inputs for the function.
For \(f(x, y) = \sqrt{x} + \sqrt{y}\), there are specific constraints tied to the components \(\sqrt{x}\) and \(\sqrt{y}\). Both require their inputs to be non-negative values. Consequently, this dictates the domain of the function, confining it to the first quadrant.

The combined effect of non-negative constraints on \(x\) and \(y\) results in a clear domain guideline: \((x, y)\) such that \(x \geq 0, y \geq 0\). These constraints ensure the function behaves predictably, providing a basis for further exploration in topics like optimization and differential equations in multivariable contexts. Constraints like these lay the foundation for deeper understanding and analysis, aligning domain specifications with practical computations.