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Chapter 13: Problem 20

Describe the domain and range of the function. $$ f(x, y)=\arccos (y / x) $$

Short Answer

Expert verified
The domain of the function \(f(x, y) = \arccos (y / x)\) is all the (x, y) such that \(x 鈮� 0\), and when \(x > 0\) it is true that \(-x 鈮� y 鈮� x\), and when \(x < 0\) it is \(-x 鈮� y 鈮� x\). The range of the function is \([0, \pi]\).

Step by step solution

01

Determining the denominator restriction

Firstly, we need to recall a simple rule: the denominator cannot be zero. Therefore, for this exercise the denominator is \(x\), which means \(x\) cannot be zero. This is because division by zero is undefined in mathematics.
02

Considering the arccos function

Recall that the arccos function is defined for values from -1 to 1. This means \(\frac{y}{x}\) should be in the interval [-1, 1]. Hence, our function is defined when \(-x \leq y \leq x\), for \(x > 0\) and when \(-x \geq y \geq x\) ,for \(x < 0\).
03

Combining the conditions

When we combine the conditions described in step 1 and step 2, we find that the domain of \(f(x, y)\) is all the \(x, y\) such that \(x =谈 0\), and when \(x > 0\) it is true that \(-x \leq y \leq x\), and when \(x < 0\) it is true that \(-x \geq y \geq x\).
04

Finding the range of the function.

The range of the arccos function is \([0, \pi]\). This means the set of all possible output values the function can produce is between 0 and \(\pi\), inclusive.

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Key Concepts

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

Domain of a function
The domain of a function is all about determining the set of possible input values (also called arguments) that the function can accept without running into mathematical problems. For a given function, determining its domain is a crucial step because it helps us understand where the function is defined and behaves nicely.
  • Understanding restrictions: In many functions, particularly those involving division or roots, there are restrictions to be considered. For example, the function \(f(x, y) = \arccos (y / x)\) has a restriction that \(x\) cannot be zero. This is because dividing by zero is undefined in mathematics.
  • Handling specific functions: When analyzing functions like \(\arccos\), we also need to consider the domain specific to the trigonometric functions. The arccos function is only defined when its input is between -1 and 1. So, for \(f(x, y)\), the expression \(\frac{y}{x}\) must lie within this interval. This leads to the condition \(-x \leq y \leq x\) when \(x > 0\) and \(-x \geq y \geq x\) when \(x < 0\).
Range of a function
The range of a function refers to the set of all possible output values the function can produce. Once we have the input values (domain) sorted out, we can start thinking about what values the function can return. This gives us insight into the behavior of the function over its entire domain.
  • Output possibilities: For our example \(f(x, y) = \arccos (y / x)\), the nature of the \(\arccos\) function helps us immediately identify the range. The arccos function produces outputs from 0 to \(\pi\) (inclusive) for inputs within its domain. Thus, these are the possible outputs for our function.
  • Graphical representation: Visualizing the range can often help in understanding. If you graph \(f(x, y)\), you will see the outputs smoothly transitioning between 0 and \(\pi\). This visualization reinforces the understanding of \([0, \pi]\) as the range.
Inverse trigonometric functions
Inverse trigonometric functions are crucial in both pure and applied mathematics because they help us find angles when we know the sine, cosine, or tangent of these angles. They are essentially the inverse operations of standard trigonometric functions, providing angles in a standard range.
  • Understanding arccos: The \(\arccos\) function, or inverse cosine, gives the angle whose cosine value is a given number. Its domain is restricted to inputs from -1 to 1, and it returns an angle in radians between 0 and \(\pi\).
  • Application: In problems like \(f(x, y) = \arccos (y / x)\), the \(\arccos\) function finds the angle whose cosine corresponds to the value \(\frac{y}{x}\), provided this falls within the allowable range.
Inverse trigonometric functions often appear in scenarios involving triangles, periodic phenomena, or even in calculus problems where angle determination is critical. Understanding their properties and limitations is key to applying them effectively.

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Most popular questions from this chapter

Show that the tangent plane to the quadric surface at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) can be written in the given form.Ellipsoid: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1\) Plane: \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}+\frac{z_{0} z}{c^{2}}=1\)

Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point.\(z=\arctan \frac{y}{x}, \quad\left(1,1, \frac{\pi}{4}\right)\)

Use Lagrange multipliers to find any extrema of the function subject to the constraint \(x^{2}+y^{2} \leq 1\). $$ f(x, y)=x^{2}+3 x y+y^{2} $$

Use Lagrange multipliers to find the indicated extrema, assuming that \(x\) and \(y\) are positive. Minimize \(f(x, y)=x^{2}-y^{2}\) Constraint: \(x-2 y+6=0\)

Use Lagrange multipliers to find any extrema of the function subject to the constraint \(x^{2}+y^{2} \leq 1\). $$ f(x, y)=e^{-x y / 4} $$
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