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Chapter 14: Problem 10

Let \(F(x, y)=1+\sqrt{4-y^{2}}\) (a) Evaluate \(F(3,1) .\) (b) Find and sketch the domain of \(F\) (c) Find the range of \(F .\)

Short Answer

Expert verified
(a) 1+\sqrt{3}; (b) Domain: -2 \leq y \leq 2; (c) Range: [1, 3]

Step by step solution

01

Evaluate F(3, 1)

Substitute the values \(x = 3\) and \(y = 1\) into the function \(F(x, y) = 1 + \sqrt{4 - y^2}\). This gives \(F(3,1) = 1 + \sqrt{4 - 1^2}\). Simplify the expression to find \(F(3,1) = 1 + \sqrt{3}\).
02

Determine the Domain of F

The domain is determined based on the expression for \(F(x, y)\). The square root \(\sqrt{4 - y^2}\) is only defined for values where \(4 - y^2 \geq 0\), which simplifies to \(-2 \leq y \leq 2\). There are no restrictions on \(x\), so \(x\) can be any real number. Therefore, the domain is all \(x \in \mathbb{R}\) and \(-2 \leq y \leq 2\).
03

Sketch the Domain of F

In the xy-plane, the domain is represented by the strip defined by \(-2 \leq y \leq 2\) without any restrictions on \(x\). This corresponds to horizontal lines at \(y = -2\) and \(y = 2\), with all points \(x\) in between these two lines.
04

Find the Range of F

To determine the range, evaluate the possible values of \(F(x, y) = 1 + \sqrt{4 - y^2}\). Since \(-2 \leq y \leq 2\), \(\sqrt{4-y^2}\) ranges from \(0\) (when \(|y|=2\)) to \(2\) (when \(y=0\)). Thus, the function values \(F(x, y)\) range from \(1 + 0 = 1\) to \(1 + 2 = 3\). Therefore, the range is \([1, 3]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain and Range of Functions
Understanding the domain and range of a function is crucial in multivariable calculus. These concepts help us identify where a function is defined and the possible output values.
To find the domain of a function like \(F(x, y) = 1 + \sqrt{4 - y^2}\), consider the conditions under which the function makes sense. In this case, we look at the expression inside the square root: \(4 - y^2\). The square root is only defined for non-negative inputs, which means \(4 - y^2\geq 0\). Solving this inequality gives us \(-2 \leq y \leq 2\). There are no restrictions on \(x\), so it can be any real number.
  • Domain: All \(x \in \mathbb{R}\) and \(-2 \leq y \leq 2\).
  • Range Determination: With the determined domain, you can find the range by evaluating the function's possible output values. For our function, as \(y\) changes from \(-2\) to \(2\), the expression \(\sqrt{4 - y^2}\) ranges from \(0\) to \(2\).
  • Range: The range of \(F(x, y)\) is \([1, 3]\), calculated from \(1 + 0\) to \(1 + 2\).
Evaluating Functions
When given a function and a set of variables to evaluate, it's a matter of plugging in these variables into the function's formula. Take the function \(F(x, y) = 1 + \sqrt{4 - y^2}\) and say we are given \(x = 3\) and \(y = 1\).
To evaluate this function at \(F(3, 1)\), substitute \(x = 3\) and \(y = 1\) into the function. So it becomes:
  • \(F(3, 1) = 1 + \sqrt{4 - 1^2}\)
  • Calculate \(\sqrt{4 - 1^2} = \sqrt{4 - 1} = \sqrt{3}\)
  • Therefore, \(F(3, 1) = 1 + \sqrt{3}\)
This process highlights the importance of clear substitution and simplification to find a function's value at given points.
Square Root Function
The square root function is a fundamental mathematical function that often appears in different contexts, from simple algebra to complex multivariable calculus problems.
For the function \(F(x, y) = 1 + \sqrt{4 - y^2}\), the square root component \(\sqrt{4 - y^2}\) dictates essential aspects of the function鈥檚 behavior. The square root function has certain key properties:
  • Non-negativity: The expression inside the square root (the radicand) must be zero or positive because you cannot take the square root of a negative number in the real number system.
  • Continuity and Smoothness: The function is continuous and smooth over its domain, except possibly at boundary points, such as where \(y = \pm 2\).
  • Effect on Range: As seen in our function, the range is partly determined by how the square root behaves, going from its minimum value (here, 0) to its maximum value (here, 2).
Understanding how the square root function influences the behavior of complex functions is crucial for solving many calculus problems.

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