Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

a parametric representation of a curve is given. $$ x=2 \sqrt{t-2}, y=3 \sqrt{4-t} ; 2 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The curve starts at (0, 3√2) and ends at (2√2, 0), tracing a path likely part of a parabolic or elliptic shape.

Step 1: Understand the Parametric Equations

This curve is given by the parametric equations \( x = 2 \sqrt{t-2} \) and \( y = 3 \sqrt{4-t} \). The parameter \( t \) varies from 2 to 4. This specifies a path in the coordinate plane when you change the value of \( t \) within the given range.

Step 2: Determine the Domain of the Parametric Equations

For the expressions \( \sqrt{t-2} \) and \( \sqrt{4-t} \) to be valid, \( t \) must satisfy: \( t-2 \geq 0 \) and \( 4-t \geq 0 \). Solving these inequalities gives \( t \geq 2 \) and \( t \leq 4 \). Hence, the range of \( t \) given, \( 2 \leq t \leq 4 \), is consistent with the domain of the square root expressions.

Step 3: Analyze the Endpoints of the Curve

Evaluate the parametric equations at the endpoints of \( t \). At \( t=2 \), \( x=2 \sqrt{2-2} = 0 \) and \( y=3 \sqrt{4-2} = 3 \sqrt{2} \). At \( t=4 \), \( x = 2 \sqrt{4-2} = 2 \sqrt{2} \) and \( y = 3 \sqrt{4-4} = 0 \). The endpoints of the curve in the \( (x, y) \) plane are \( (0, 3\sqrt{2}) \) and \( (2\sqrt{2}, 0) \).

Step 4: Describe the Shape of the Curve

As \( t \) increases from 2 to 4, \( x \) increases from 0 to \( 2 \sqrt{2} \), while \( y \) decreases from \( 3\sqrt{2} \) to 0. This means that the curve moves from the point \( (0, 3\sqrt{2}) \) to the point \( (2\sqrt{2}, 0) \). Hence, the curve is a segment of a path that is possibly part of a broader parabolic or elliptic shape.

Key Concepts

Curve Analysis

Curve analysis involves examining how a curve behaves between its endpoints by using parametric equations. In our exercise, the curve is represented by two parametric equations:

• \( x = 2 \sqrt{t-2} \)
• \( y = 3 \sqrt{4-t} \)

The parameter \( t \) is a key aspect of this analysis, varying from 2 to 4. Changing \( t \) in this interval traces a path through the coordinate plane. By examining the values of \( x \) and \( y \) for these \( t \) values, we figure out how the curve progresses.

Initially, when \( t = 2 \), the calculated point is \( (0, 3\sqrt{2}) \), and when \( t = 4 \), it is \( (2\sqrt{2}, 0) \). As \( t \) increases between these points, \( x \) moves from 0 to \( 2\sqrt{2} \), while \( y \) decreases to 0. This provides us an insight into how the curve behaves overall, forming a segment that likely belongs to a parabolic or elliptic shape.

Coordinate Plane

The coordinate plane is where we map the curve by plotting points calculated from parametric equations. Think of the coordinate plane as a grid where you can visually see the path traced by changing \( t \). This way, you can understand the curve’s reality better.

In our scenario, as \( t \) changes from 2 to 4, the points \( (x, y) \) are plotted on this two-dimensional grid. The x-axis and y-axis help us to see how the curve moves across different coordinates.

Plotting the start \( (0, 3\sqrt{2}) \) and end \( (2\sqrt{2}, 0) \) points, the gradual pixels or dots filled by intermediate values of \( t \), give a sense of the continuous curve. This makes the coordinate plane an indispensable tool for visualizing and understanding the trajectory of parametric equations.

Domain of a Function

Understanding the domain of a function is crucial for ensuring the validity of its calculations. In this context, the domain concerns the acceptable values for the parameter \( t \). A parametric expression like \( \sqrt{t-2} \) and \( \sqrt{4-t} \) requires that these square roots yield real numbers.

The inequalities \( t-2 \geq 0 \) and \( 4-t \geq 0 \) need to be satisfied:

• The first inequality simplifies to \( t \geq 2 \).
• The second leads to \( t \leq 4 \).

Together, they dictate that \( t \) must lie between 2 and 4, inclusive. These constraints make sure that inputs into the square roots are non-negative, ensuring real outputs.

This domain defines not just the curve's valid interval but also aligns with the endpoints of the curve in the coordinate plane. Knowing the domain thus offers a boundary within which the parametric equations can confidently be used to map out the curve.