Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

x=t^{2}, y=\sqrt{t} ; 1 \leq t \leq 4

Step-by-Step Solution

Answer

The relation is \( y = x^{1/4} \) for \( 1 \leq x \leq 16 \).

Step 1: Identify the Parametric Equations

The given parametric equations are: 1. \( x = t^2 \)2. \( y = \sqrt{t} \) with the parameter \( t \) ranging from 1 to 4 inclusive.

Step 2: Determine the Range of x

To find the range of \( x \), we substitute the values of \( t \) at the endpoints of its interval into the equation for \( x \).For \( t = 1 \), \( x = 1^2 = 1 \).For \( t = 4 \), \( x = 4^2 = 16 \).Thus, the range for \( x \) is from 1 to 16.

Step 3: Determine the Range of y

To find the range of \( y \), we substitute the values of \( t \) at the endpoints of its interval into the equation for \( y \).For \( t = 1 \), \( y = \sqrt{1} = 1 \).For \( t = 4 \), \( y = \sqrt{4} = 2 \).Thus, the range for \( y \) is from 1 to 2.

Step 4: Eliminate the Parameter t

To express \( y \) in terms of \( x \), eliminate the parameter \( t \).From \( x = t^2 \), solve for \( t \):\[ t = \sqrt{x} \].Substitute \( t = \sqrt{x} \) into \( y = \sqrt{t} \):\[ y = \sqrt{\sqrt{x}} = x^{1/4}. \]

Step 5: State the Relationship Between x and y

The relationship between \( x \) and \( y \) is given by the equation \( y = x^{1/4} \), valid for \( 1 \leq x \leq 16 \).

Key Concepts

Range of Functions

The range of a function describes the set of possible output values (or y-values) that a function can achieve. When dealing with parametric equations, such as when both x and y are expressed in terms of a third variable (the parameter), finding the range can initially seem daunting. However, it's quite manageable.
To determine these ranges separately, we use the interval provided for the parameter. For instance, with the parametric equations given, we evaluate the output of x and y when the parameter t varies over its specified range (from 1 to 4 in this example):

• For x, substituting the minimum value of t gives us the minimum x, and the maximum value of t gives us the maximum x. Here, substituting t = 1 and t = 4 into x = t^2 gives the range 1 to 16.
• Similarly, substituting t = 1 and t = 4 into y = \( \sqrt{t} \) gives us the range 1 to 2 for y.

By finding these endpoints, we determine the range of each function without having to manually compute every single possible pair of (x, y). This process highlights the power of parametric equations in simplifying such tasks.

Eliminating Parameters

Eliminating a parameter in an equation means expressing the function in a non-parametric form, usually by writing y in terms of x directly. This is often the next step after identifying the ranges of x and y. In the context of the example, the parameter t is eliminated by:

• Solving one of the parametric equations for t, such as x = t² to obtain \( t = \sqrt{x} \)
• Substituting this expression for t into the other equation, resulting in the expression for y. Here, y = \( \sqrt{t} \) becomes y = \( \sqrt{\sqrt{x}} = x^{1/4} \).

This conversion is incredibly useful because, by eliminating the parameter, you convert a set of parametric equations into a single, non-parametric equation. This can illustrate how the inputs and outputs directly relate to each other without the intermediary of the parameter. Such simplification often helps with visualization and further analysis.

Relationship Between Variables

Understanding the relationship between the variables x and y transforms our perspective of the system of equations. Upon eliminating the parameter, one central equation manifests that marries the variables, making the analysis straightforward.The derived relationship from example - \( y = x^{1/4} \) - represents how y behaves relative to x over the specified domain (1 ≤ x ≤ 16). It supplies crucial insights into how a change in x influences y, which can be visualized as a curve on the xy-plane.
With this relationship:

• This function, y = \( x^{1/4} \), tells us that y increases slower than x does because of the fourth root operation. Changes in x result in gradual variations in y.
• Knowing specific relationships assists in predicting outcomes without recalculating for each value of x, thereby saving time and enhancing clarity in complex problems.

Establishing this equation gives a concrete characterization of the entire behavior of the system captured by the parametric equations fully shedding light on all possible pairings of x and y.