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Chapter 11: Problem 19

\(x=t, \quad y=\sqrt{t^{2}-2 t+1} ; \quad 0 \leq t \leq 4\)

Short Answer

Expert verified
The function y is piecewise: \( y = 1-t \) for 0 \( \leq t < 1 \) and \( y = t-1 \) for \( t \geq 1 \).

Step by step solution

01

Analyze the equation

The given equation for y is \( y = \sqrt{t^2 - 2t + 1} \). Notice that the expression inside the square root is \( t^2 - 2t + 1 \).
02

Recognize the perfect square

The expression \( t^2 - 2t + 1 \) can be rewritten as \( (t-1)^2 \). This is a perfect square, making the equation \( y = \sqrt{(t-1)^2} \).
03

Simplify the square root

Since \( y = \sqrt{(t-1)^2} \), the square root simplifies to an absolute value, resulting in \( y = |t-1| \).
04

Determine the expression for y based on the domain

Considering the domain \( 0 \leq t \leq 4 \), the absolute value function \( |t-1| \) should be evaluated as: for \( 0 \leq t < 1 \), \( y = 1 - t \); for \( t \geq 1 \), \( y = t - 1 \).
05

Conclusion

The piecewise function for \( y \) is: \( y = 1-t \) for \( 0 \leq t < 1 \) and \( y = t-1 \) for \( t \geq 1 \). This reflects the V-shaped graph characteristic to absolute value functions.

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Key Concepts

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

Understanding Absolute Value Functions
An absolute value function takes a number and turns it into its non-negative value. For example, the absolute value of both \( 3 \) and \( -3 \) is 3, as it measures the distance from zero on the number line.
Absolute value functions often have a characteristic V-shape on a graph. This is due to how they convert negative results into positive ones.
In the example, the function \( y = |t-1| \) means:
  • If \( t \) is less than \( 1 \), the value inside the absolute value becomes negative, leading to \( y = 1-t \).
  • For \( t \) greater than or equal to \( 1 \), the expression remains positive, thus \( y = t-1 \).
This results in a graph that changes direction at \( t = 1 \). This is a crucial trait seen with absolute value functions. It splits the domain based on the behavior of the expression within the absolute value.
Perfect Square Expressions
The expression \( t^2 - 2t + 1 \) is known as a perfect square. A perfect square comes from multiplying a binomial by itself. Recognizing these can simplify many mathematical processes.
Think of it like \( (t-1)^2 \), which when expanded, becomes \( t^2 - 2t + 1 \). This recognition is key to simplifying expressions.
When dealing with square roots of perfect squares, it simplifies dramatically. For instance:
  • \( \sqrt{(t-1)^2} = |t-1| \).
This connects the concept of absolute value functions because it morphs the square root into an absolute value. Perfect squares simplify progression in algebraic problems by making equations easier to handle.
Domain and Range in Functions
When analyzing functions, "domain" refers to all the possible input values (typically \( t \), \( x \), etc.), and "range" pertains to all the possible outputs (\( y \)).
For the function \( y = |t-1| \), and given \( 0 \leq t \leq 4 \), the domain is \( 0 \) to \( 4 \). This specifies the allowed values for \( t \).
With absolute value functions, the range is typically every non-negative number the function can output.
  • In this case, since \( |t-1| \) is never negative, the range will be from \( 0 \) to the maximum output when \( t = 4 \), which would be \( 3 \).
The piecewise function explicitly states how the function behaves within the domain. This not only determines how the graph looks but also ensures the function remains true to its defined constraints.

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Most popular questions from this chapter

Prove that the graph of each polar equation is a circle, and find its center and radius. (a) \(r=a \sin \theta, a \neq 0\) (b) \(r=b \cos \theta, b \neq 0\) (c) \(r=a \sin \theta+b \cos \theta, a \neq 0\) and \(b \neq 0\)

Exer. 45-78: Sketch the graph of the polar equation. $$ \theta=\pi / 4 $$

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Describe the part of a hyperbola given by the equation. $$x=\frac{5}{4} \sqrt{y^{2}+16}$$

Exer. 45-78: Sketch the graph of the polar equation. $$ r=-4 \cos ^{2}(\theta / 2) $$
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