Chapter 3: Problem 78
Describe and sketch the curve represented by the vector-valued function \(\mathbf{r}(t)=\left\langle 6 t, 6 t-t^{2}\right\rangle\).
Short Answer
Expert verified
The curve is a downward-opening parabola \( y = x - \frac{x^2}{36} \).
Step by step solution
01
Understand the Vector-Valued Function
The function is given as \( \mathbf{r}(t) = \left\langle 6t, 6t - t^2 \right\rangle \). This means that the vector \( \mathbf{r}(t) \) represents a point \( (x(t), y(t)) \) on the plane where \( x(t) = 6t \) and \( y(t) = 6t - t^2 \).
02
Express the Components as Parametric Equations
From the vector-valued function, the parametric equations can be written as:1. \( x(t) = 6t \)2. \( y(t) = 6t - t^2 \).These represent the horizontal and vertical positions of the curve at any parameter \( t \).
03
Eliminate the Parameter
To describe the curve, eliminate the parameter \( t \) by expressing both \( x \) and \( y \) in terms of \( x \) alone. As \( x = 6t \), solve for \( t \):\[ t = \frac{x}{6} \].Substitute \( t = \frac{x}{6} \) into the equation for \( y \):\[ y = 6 \left( \frac{x}{6} \right) - \left( \frac{x}{6} \right)^2 = x - \frac{x^2}{36} \].
04
Analyze the Shape of the Curve
The equation \( y = x - \frac{x^2}{36} \) is a quadratic equation representing a downward-opening parabola with vertex at \( (0,0) \). The parabola opens downwards because the coefficient of \( x^2 \) is negative.
05
Sketch the Curve
To sketch the curve, plot the vertex \( (0, 0) \) and choose a few values for \( x \) to calculate corresponding \( y \) values. For example:- When \( x = 0 \), \( y = 0 \).- When \( x = 6 \), \( y = 6 - 1 = 5 \).- When \( x = 12 \), \( y = 12 - 4 = 8 \).- When \( x = -6 \), \( y = -6 - 1 = -7 \).Plot these points and draw a smooth downward-opening parabola through them.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a way to express the coordinates of the points that make up a curve. Instead of describing the curve with a single equation in terms of x and y, parametric equations express both the x and y coordinates as functions of a third variable, usually denoted as t, known as the parameter.
In the vector-valued function \( \mathbf{r}(t) = \left\langle 6t, 6t - t^2 \right\rangle \), the parametric equations are:
The advantage of parametric equations is that they can easily represent complex curves and motions which might be difficult to express as a single equation relating x and y. They also show the trajectory and position over time, making them especially useful in physics and engineering applications.
In the vector-valued function \( \mathbf{r}(t) = \left\langle 6t, 6t - t^2 \right\rangle \), the parametric equations are:
- \( x(t) = 6t \)
- \( y(t) = 6t - t^2 \)
The advantage of parametric equations is that they can easily represent complex curves and motions which might be difficult to express as a single equation relating x and y. They also show the trajectory and position over time, making them especially useful in physics and engineering applications.
Curve Sketching
Curve sketching involves plotting a curve by determining the key characteristics and points along the curve. It helps in visualizing how a curve behaves and changes. For our example, the simplified equation obtained from eliminating the parameter \(t\) was \(y = x - \frac{x^2}{36}\).
Key Points to Analyze:
- Identify the vertex: The vertex of the parabola is at \((0, 0)\). This means the point where the curve changes its direction.
- Determine direction: The negative coefficient of \(x^2\) indicates that the parabola opens downward.
- Intercepts: When \(x = 0\), \(y = 0\). This is the intersection point at the y-axis.
- Additional Points: By substituting values like \(x = 6\), \(x = 12\) and \(x = -6\), we derive more points to plot.
Quadratic Equations
A quadratic equation is any equation that can be written in the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. These equations create parabolas on a graph. Depending on the sign of \(a\), the parabola will open upwards (\(a > 0\)) or downwards (\(a < 0\)). In our case, the quadratic equation derived was \(y = x - \frac{x^2}{36}\).
Because the term \(- \frac{x^2}{36}\) carries a negative coefficient, the parabola opens downwards.
Because the term \(- \frac{x^2}{36}\) carries a negative coefficient, the parabola opens downwards.
Analyzing a Quadratic Equation:
- Vertex Form: The equation already suggested that the vertex is at \((0, 0)\), as the standard vertex form \(y = a(x-h)^2 + k\) points towards \(h = 0\) and \(k = 0\).
- Axis of Symmetry: This is the vertical line \(x = h\), here \(x = 0\), meaning the curve is symmetric about the y-axis.
- End Behavior: As \(x\) moves further from 0 in either direction, \(y\) becomes more negative, clearly indicating the downward opening.