Problem 36 - Precalculus: Functions and Graphs

Question

The parametric equations specify the position of a moving point \(P(x, y)\) at time \(t\). Sketch the graph, and indicate the motion of \(P\) as \(t\) increases. (a) \(x=t^{2}, \quad y=1-t^{2} ; \quad 0 \leq t \leq 1\) (b) \(x=1-\ln t, \quad y=\ln t, \quad 1 \leq t \leq e\) (c) \(x=\cos ^{2} t, \quad y=\sin ^{2} t, \quad 0 \leq t \leq 2 \pi\)

Step-by-Step Solution

Answer

(a) Line from (0,1) to (1,0); (b) Curve from (1,0) to (0,1); (c) Circle centered at origin.

Step 1: Analyze Equation (a)

The given parametric equations are \(x=t^2\) and \(y=1-t^2\). We are given that \(0 \leq t \leq 1\). By substituting \(t=0\) into the equations, we get the starting point \((x, y) = (0, 1)\). At \(t=1\), \(x=1\) and \(y=0\), giving us the ending point \((1, 0)\). Since \(x\) and \(y\) are quadratic expressions of \(t\), the path traced by \(P\) is straight from \((0, 1)\) to \((1, 0)\). This is part of the line \(x+y=1\).

Step 2: Sketch and Analyze Motion for (a)

Plot the points \((0, 1)\) and \((1, 0)\). As \(t\) increases from 0 to 1, the point \(P\) will move linearly downwards along the line from \((0, 1)\) to \((1, 0)\), indicating a straight-line motion.

Step 3: Analyze Equation (b)

The parametric equations are \(x=1-\ln t\) and \(y=\ln t\) with \(1 \leq t \leq e\). When \(t=1\), \(x=1-0=1\) and \(y=0\). When \(t=e\), \(x=1-1=0\) and \(y=1\). Therefore, the path travels from \((1, 0)\) to \((0, 1)\). The expressions involve logarithmic functions, which change more rapidly at smaller values and slow down as values increase.

Step 4: Sketch and Analyze Motion for (b)

Plot the points \((1, 0)\) and \((0, 1)\). As \(t\) increases from 1 to \(e\), the point \(P\) moves along a curve from \((1, 0)\) to \((0, 1)\), indicating a logarithmic curve due to the nature of the \(ln\) functions.

Step 5: Analyze Equation (c)

The parametric equations are \(x=\cos^2 t\) and \(y=\sin^2 t\). We can use the identity \(x+y=\cos^2 t + \sin^2 t = 1\), confirming that the path traced is a unit circle, with \(0 \leq t \leq 2 \pi\). At \(t=0\), \(x=1\), \(y=0\), while at \(t=\pi/2\), \(x=0\), \(y=1\), and this continues for \(t=\pi\) and \(t=3\pi/2\) completing the circle.

Step 6: Sketch and Analyze Motion for (c)

Sketch a circle with radius 1 centered at the origin. \(P\) will start at \((1,0)\), move counterclockwise to \((0,1)\), then to \((-1,0)\), \((0,-1)\), and back to \((1,0)\), completing a full revolution as \(t\) goes from 0 to \(2\pi\).

Key Concepts

Motion of a Point

In the context of parametric equations, the motion of a point is a way to describe how a point travels across a coordinate plane over time. Let's consider a point \( P(x, y) \) whose position is determined by two equations. These equations describe how the \( x \)-coordinate and \( y \)-coordinate of the point move as the parameter \( t \) changes.

For instance, in exercise (a), we have \( x = t^2 \) and \( y = 1 - t^2 \). This means as \( t \) increases from 0 to 1, the point \( P \) moves from the position \( (0, 1) \) to \( (1, 0) \). This path represents a linear motion on the line \( x + y = 1 \).

Understanding this motion is critical because it tells us not only where the point is at any given time, but also how it gets there. This is important for plotting graphs and analyzing the nature of these paths.

Graph Sketching

Graph sketching in parametric equations involves plotting the path that a point \( P(x, y) \) takes as the parameter \( t \) changes. This relies heavily on understanding how the separate \( x \) and \( y \) functions react over \( t \).

For example, when sketching the path for exercise (b): \( x = 1 - \ln t \) and \( y = \ln t \), it's important to detect the start and end points. Here, \( P \) starts at \( (1, 0) \) when \( t = 1 \) and ends at \( (0, 1) \) when \( t = e \).

Consider graphing this by marking these specific points and remembering that since these equations contain logarithms, the curve will steeply ascend initially and then slowly as \( t \) increases. Always note the nature of the functions involved for creating accurate sketches.

Analyzing Parametric Paths

Analyzing parametric paths involves understanding how the equations describe movement on a coordinate plane. The paths are not simply lines; they can be curves or loops that provide rich information about motion. Various characteristics such as direction, speed at different points, and the overall shape can be derived.

In the case of exercise (c), where \( x = \cos^2 t \) and \( y = \sin^2 t \), the path is a perfect circle because it follows the identity \( \cos^2 t + \sin^2 t = 1 \), which is the equation for a unit circle.

This analysis tells us that regardless of the value of \( t \), the total distance from the origin stays constant, providing insight into the constant speed of movement along the circle. Recognizing these critical path characteristics aids in not only sketching the correct path but also understanding the movement's nature and intricacies.

Unit Circle

The unit circle is fundamental in trigonometry and parametric equations. It is a circle with a radius of one, centered at the origin of the coordinate system. This makes it a perfect tool for describing trigonometric functions.

In exercise (c), the parametric equations \( x = \cos^2 t \) and \( y = \sin^2 t \) are perfect examples of paths traced on the unit circle. From \( t = 0 \) to \( t = 2 \pi \), the path completes a full circle.

Understanding the unit circle helps in interpreting these equations by recognizing how each \( t \) value corresponds to a specific point on the circle, predicting the path as the point \( P(t) \) moves. It’s not just a part of these equations, but an essential part of trigonometric applications, making it a staple in mathematics.