Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

x=t \ln t, y=t-1 ; 1 \leq t \leq 3

Step-by-Step Solution

Answer

The parametric curve starts at (0,0) and ends at (3.2958,2).

Step 1: Parametric Equations

The given equations are parametric, where both \( x \) and \( y \) are defined in terms of the parameter \( t \). The equations are \( x = t \ln t \) and \( y = t - 1 \) with \( 1 \leq t \leq 3 \).

Step 2: Determine Range of x and y

Since \( 1 \leq t \leq 3 \), we substitute to find the range of \( y \): \( y = t - 1 \) gives \( 0 \leq y \leq 2 \) as \( t \) varies. For \( x = t \ln t \), evaluate at \( t=1 \), \( x=1 \cdot \ln 1 = 0 \) and at \( t=3 \), \( x = 3 \ln 3 \). Calculate \( \ln 3 \approx 1.0986 \), so \( x = 3 \cdot 1.0986 \approx 3.2958 \). Hence, \( 0 \leq x \leq 3.2958 \).

Step 3: Plotting the Curve

The parametric curve described by \( x = t \ln t \) and \( y = t - 1 \) can be plotted for \( t \) in the interval from 1 to 3. This curve starts at the point (0,0) when \( t=1 \) and ends at the point approximately (3.2958,2) when \( t=3 \). The parameter \( t \) increases along the curve.

Key Concepts

Range of Parametric Curve

Understanding the range of a parametric curve involves determining the set of possible values that variables can take. In this case, both \( x \) and \( y \) are expressed in terms of the parameter \( t \).
To find these ranges for \( x = t \ln t \) and \( y = t - 1 \), substitute the given values of \( t \). The interval for \( t \) is \([1, 3]\).

• For \( y = t - 1 \), substituting the limits of \( t \) gives \( y \) values ranging from 0 (when \( t = 1 \)) to 2 (when \( t = 3 \)).
• Similarly, \( x = t \ln t \) produces \( x \) values ranging from 0 to approximately 3.2958. This results from evaluating the equation at \( t = 1 \) and \( t = 3 \), respectively. \( x \) increases as \( t \) increases and reaches its maximum at the endpoint of the interval.

This understanding of parameter ranges is crucial for correctly interpreting what values the curve will exhibit, providing insights into its spread across an axis.

Curve Plotting in Parametric Form

Plotting a parametric curve means drawing the path created by the equations as the parameter changes. In this exercise, it refers to the path traced by the equations \( x = t \ln t \) and \( y = t - 1 \) as \( t \) moves from 1 to 3.
Begin by calculating key points on the curve. For instance:

• When \( t = 1 \), both \( x \) and \( y \) equal zero, giving the point (0,0).
• When \( t = 3 \), you get approximately the point (3.2958, 2) after calculation.

These benchmarks provide the start and end of the curve. In between, the curve is smoothly sketched based on similar intermediate calculations.
Visualizing in parametric form requires recognizing how \( x \) and \( y \) vary together. This creates context that Cartesian plotting lacks, offering insights into how the curve's shape and direction develop between defined points.

Parameter Variations

Exploring parameter variations involves observing changes in the path of a curve when the parameter \( t \) is adjusted. This exercise reveals how changes in \( t \) influence the values of \( x \) and \( y \).
Imagine \( t \) as a time parameter that alters where a point on the curve is drawn at any moment. As \( t \) progresses:

• The curve starts progressing from the initial point (0,0), following the route plotted by the equations.
• Increases in \( t \) result in shifts of both \( x \) and \( y \), gradually reaching (3.2958,2) as \( t \) approaches 3.

By viewing \( t \) as a slider along the curve, one gains valuable perspectives on how digit values are interconnected. It indicates not just position but orientation, providing a comprehensive grasp of the parametric relation's dynamic behavior.