Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

x=3 t^{2}+2, y=2 t^{3}-1 / 2 ; 1 \leq t \leq 4

Step-by-Step Solution

Answer

The range of \(x\) is from 5 to 50, and the range of \(y\) is from 1.5 to 63.5.

Step 1: Understanding the Parameterization

The variables \(x\) and \(y\) are expressed in terms of a parameter \(t\). Specifically, \(x = 3t^2 + 2\) and \(y = 2t^3 - \frac{1}{2}\). The range for \(t\) is given as \(1 \leq t \leq 4\). This means that as \(t\) varies in this range, \(x\) and \(y\) will trace a curve in the \((x, y)\)-plane.

Step 2: Determine x-values

To find the range of \(x\), substitute the boundary values of \(t\) into the equation for \(x\):- For \(t=1\), \(x = 3(1)^2 + 2 = 5\).- For \(t=4\), \(x = 3(4)^2 + 2 = 50\).Therefore, the range of \(x\) is from 5 to 50.

Step 3: Determine y-values

To find the range of \(y\), substitute the boundary values of \(t\) into the equation for \(y\):- For \(t=1\), \(y = 2(1)^3 - \frac{1}{2} = \frac{3}{2}\).- For \(t=4\), \(y = 2(4)^3 - \frac{1}{2} = 63.5\).Thus, the range of \(y\) is from \(1.5\) to \(63.5\).

Step 4: Using the Calculated Ranges

The curve described by these parameterizations will occupy the section of the plane where \(5 \leq x \leq 50\) and \(1.5 \leq y \leq 63.5\). Since the problem does not ask for a specific function of \(y\) in terms of \(x\), identifying the ranges of \(x\) and \(y\) is sufficient to understand the extent of the curve.

Key Concepts

Range of Parameters

In parametric equations, like the ones in the exercise, we often have a parameter, commonly denoted by \(t\), that controls the values of \(x\) and \(y\). This parameter can take on values within a specified range, called the range of parameters.

• Here, the parameter \(t\) ranges from 1 to 4.
• This range is crucial because it limits the parts of the curve that we are interested in.

For example, when \(t\) moves from 1 to 4, the resulting \(x\) and \(y\) values change, thereby tracing different parts of the curve.
In practice, determining this range helps us to focus on how the curve behaves over this specific interval.

Parameterization

The concept of parameterization involves defining a curve using a parameter, \(t\), that expresses \(x\) and \(y\) in terms of it.

• In the given problem, \(x = 3t^2 + 2\) and \(y = 2t^3 - \frac{1}{2}\).
• These equations show how \(x\) and \(y\) change as \(t\) changes.

Parameterization is especially useful when dealing with curves that can't be easily described using \(y\) in terms of \(x\) or vice versa.
It allows us to methodically study each part of the curve, one value of \(t\) at a time.

Curve Tracing

Curve tracing is the process of understanding how a curve behaves as \(t\) changes from the start to the end of its range.

• As \(t\) increases from 1 to 4, the curve's path in the \(x-y\) plane becomes apparent.
• By calculating \(x\) and \(y\) for specific \(t\) values, we can see how the curve evolves.

In the exercise, we calculate endpoints: - For \(t = 1\), \(x = 5\) and \(y = \frac{3}{2}\).- For \(t = 4\), \(x = 50\) and \(y = 63.5\).
This guides us to understand which portion of the plane the curve occupies.
Understanding the curve's trace helps in graphing the curve and predicting its behavior.

X-Y Plane

The \(x-y\) plane, also known as the Cartesian plane, is a two-dimensional plane where each point is defined by a pair of coordinates \((x, y)\).

• It's helpful for visualizing the path of parametric curves.
• The calculated range of \(x\) and \(y\) in the exercise helps us understand where this path lies.

In this specific case, given the equations, the curve would lie between:- \(5 \leq x \leq 50\), for \(x\)- \(1.5 \leq y \leq 63.5\), for \(y\)
Mapping this on the \(x-y\) plane lets us see the shape and direction of the curve, aiding in further geometric and graphical analysis.