Problem 40 - Calculus III

Question

Describe and sketch the curve represented by the vector-valued function \(\mathbf{r}(t)=\left\langle 6 t, 6 t-t^{2}\right\rangle\).

Step-by-Step Solution

Answer

The curve is a downward-facing parabola, given by the equation \( y = x - \frac{x^2}{36} \) with vertex at \( (18,18) \).

Step 1: Understanding the Components of the Vector

The given vector-valued function is \( \mathbf{r}(t) = \langle 6t, 6t - t^2 \rangle \). This means the position on the curve at any time \( t \) is given by \( x = 6t \) and \( y = 6t - t^2 \). These are the parametric equations of the curve.

Step 2: Eliminate the Parameter

To eliminate the parameter \( t \), we start with the equation for \( x: x = 6t \). Solving for \( t \), we have \( t = \frac{x}{6} \). Substitute this into the equation for \( y: y = 6t - t^2 \) to get \( y = 6\left(\frac{x}{6}\right) - \left(\frac{x}{6}\right)^2 \). Simplifying, we obtain \( y = x - \frac{x^2}{36} \). This is the equation of the curve in Cartesian coordinates.

Step 3: Determine the Shape of the Curve

The equation \( y = x - \frac{x^2}{36} \) is a quadratic equation, specifically a downward-opening parabola since the coefficient of \( x^2 \) is negative. The vertex form of a parabola, \( y = -\frac{1}{36}x^2 + x \), confirms this. The vertex can be found by comparing to the standard form \( y = ax^2 + bx + c \) and using the vertex formula \( x = -\frac{b}{2a} \); here it simplifies to \( x = 18 \), thus the full vertex is at \( (18, 18) \).

Step 4: Sketch the Curve

Plot the key points and shape for the curve given by the vertex \( (18, 18) \). Since this parabola opens downwards, draw a symmetrical curve about the line \( x = 18 \), passing through the origin as calculated from \( y = 0 \) when \( x = 0 \). For large values, the curve extends infinitely, primarily showing the branches opening downward. Highlight that the curve represents the path traced by \( \mathbf{r}(t) \).

Key Concepts

Parametric Equations

Parametric equations are equations where the coordinates of the points on a curve are expressed as functions of a parameter. In the vector-valued function \( \mathbf{r}(t) = \langle 6t, 6t - t^2 \rangle \), the parameter is \( t \).

Here, we have:

• \( x = 6t \)
• \( y = 6t - t^2 \)

These equations describe how the \( x \) and \( y \) coordinates change over time. As \( t \) varies, the point \( (x, y) \) traces out a path on the plane.

To understand or visualize this path, it is common to eliminate the parameter. By substituting for \( t \) using the given \( x = 6t \), we map the path into a more familiar form in Cartesian coordinates.

Parabolas

Parabolas are U-shaped curves that can open upwards or downwards. They are defined by the quadratic relation \( y = ax^2 + bx + c \). In the exercise, after eliminating \( t \), the equation \( y = x - \frac{x^2}{36} \) emerges, indicating a parabolic path.

Characteristics of This Parabola

- It is a downward-opening parabola since the coefficient of \( x^2 \) is negative \(-\frac{1}{36} \).
- The vertex of the parabola is a critical point that represents the maximum, located at \( (18, 18) \). This is found using the vertex formula \( x = -\frac{b}{2a} \).

This downward-opening shape is symmetric around the vertical line passing through its vertex. Besides providing symmetry, the vertex represents the highest point of the curve.

Cartesian Coordinates

Cartesian coordinates, a common system in mathematics, use \(x\) and \(y\) to define a point's position on a plane. By converting the parametric equations to the Cartesian form, we gain a clear view of the curve's shape and properties.

In this exercise, the parametric equations \( x = 6t \) and \( y = 6t - t^2 \) are transformed into Cartesian coordinates as \( y = x - \frac{x^2}{36} \). This is done by expressing \( t \) in terms of \( x \) and substituting it back into the equation for \( y \).

This transformation helps us sketch the curve by referring to a more typical quadratic equation. Thus, understanding these transformations is crucial for identifying and sketching curves effectively in many mathematical problems.