Problem 33 - Essential Calculus Early Transcendentals

Question

\(33-38=\) (a) Sketch the plane curve with the given vector equation. (b) Find \(\mathbf{r}^{\prime}(t) .\) (c) Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}^{\prime}(t)\) for the given value of \(t\) . $$\mathbf{r}(t)=\left\langle t-2, t^{2}+1\right\rangle, \quad t=-1$$

Step-by-Step Solution

Answer

1. Sketch the parabola. 2. Derivative: \( \langle 1, 2t \rangle \). 3. Vectors at \( t = -1 \): \( \mathbf{r}(-1) = \langle -3, 2 \rangle \), \( \mathbf{r}^{\prime}(-1) = \langle 1, -2 \rangle \).

Step 1: Sketch the Plane Curve

The given vector equation is \( \mathbf{r}(t) = \langle t-2, t^2+1 \rangle \). To sketch the curve, note that the first component \( x = t - 2 \) and the second component \( y = t^2 + 1 \). This describes a parabolic curve where \( x \) depends linearly on \( t \) and \( y \) quadratically. However, converting this directly to a parametric form, \( y = (x+2)^2 + 1 \), gives the correspondence between \( x \) and \( y \). Plot this on the Cartesian plane.

Step 2: Find the Derivative

To find the derivative \( \mathbf{r}^{\prime}(t) \), differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \):- The derivative of the first component \( t - 2 \) is \( 1 \).- The derivative of the second component \( t^2 + 1 \) is \( 2t \).Thus, \( \mathbf{r}^{\prime}(t) = \langle 1, 2t \rangle \).

Step 3: Evaluate Vectors at t = -1

Evaluate both \( \mathbf{r}(t) \) and \( \mathbf{r}^{\prime}(t) \) at \( t = -1 \).- For \( \mathbf{r}(t) \), substitute \( t = -1 \) into \( \langle t - 2, t^2 + 1 \rangle \): - First component: \( -1 - 2 = -3 \). - Second component: \( (-1)^2 + 1 = 2 \). Therefore, \( \mathbf{r}(-1) = \langle -3, 2 \rangle \).- For \( \mathbf{r}^{\prime}(t) \), substitute \( t = -1 \) into \( \langle 1, 2t \rangle \): - First component: \( 1 \). - Second component: \( 2(-1) = -2 \). Therefore, \( \mathbf{r}^{\prime}(-1) = \langle 1, -2 \rangle \).

Step 4: Sketch Vectors at t = -1

Sketch the position vector \( \mathbf{r}(-1) = \langle -3, 2 \rangle \) and tangent vector \( \mathbf{r}^{\prime}(-1) = \langle 1, -2 \rangle \) on the same set of axes.1. Position Vector: Start at the origin and draw an arrow to the point \( (-3, 2) \).2. Tangent Vector: Start at \( (-3, 2) \) and draw an arrow indicating the direction and relative magnitude \( \langle 1, -2 \rangle \) from this point, ending at \( (-2, 0) \).This visually represents the location and direction of change along the curve at \( t = -1 \).

Key Concepts

Parametric Equations

Parametric equations are a way of representing a curve in the plane by using a parameter, typically denoted as \( t \). Instead of describing the relationship between \( x \) and \( y \) directly, parametric equations express both \( x \) and \( y \) in terms of \( t \). For instance, in the exercise shared, \( x \) is given as \( t - 2 \) and \( y \) is given as \( t^2 + 1 \). These parametric equations mean that as \( t \) varies, \( x \) and \( y \) will follow the path of the curve described.

• Parametric equations can describe various types of curves including lines, parabolas, circles, and more complex shapes.
• They are useful as they offer a way to represent curves that cannot be expressed as a single function \( y = f(x) \).
• They also allow us to easily compute derivatives and examine the behavior of the curve at specific parameter values.

Plane Curves

A plane curve is a curve that lies on a flat, two-dimensional plane. When we use parametric equations to plot a plane curve, we can visualize how both components \( x \) and \( y \) change with \( t \).
In our case, the parametric equation \( \mathbf{r}(t) = \langle t - 2, t^2 + 1 \rangle \) represents a parabolic plane curve. As \( t \) changes, it traces out a path that resembles a parabola when plotted on Cartesian coordinates. The component \( t-2 \) causes horizontal movement while \( t^2+1 \) accounts for vertical movement.

• To sketch such a curve, it's important to understand the relationship between \( x \) and \( y \), which is derived from eliminating the parameter \( t \).
• By substituting \( x = t - 2 \), you can derive that \( y = (x+2)^2 + 1 \), confirming the parabolic nature in the Cartesian plane.
• This transformation helps in visualizing the original parametric curve as a standard function graph.

Tangent Vectors

Tangent vectors are essential for understanding how a curve changes at a specific point. For any given point on the curve, the tangent vector indicates the direction and rate of the curve's immediate movement.
By differentiating the parametric equations, we find the tangent vector \( \mathbf{r}'(t) = \langle 1, 2t \rangle \). This vector tells us the slope and direction of the curve at any parameter \( t \).

• The first component of the tangent vector, 1, is constant, meaning at any point, the curve moves uniformly in the horizontal direction.
• The second component, \( 2t \), shows vertical movement dependent on \( t \), dictating the steepness of the curve at each point.
• Understanding tangent vectors is critical for applications like physics and engineering, where knowing the rate of change at a point is needed.

Derivatives

Derivatives in the context of parametric equations help us determine how each component of the function changes as \( t \) changes. In this exercise, the derivative \( \mathbf{r}'(t) = \langle 1, 2t \rangle \) is found by differentiating each component of \( \mathbf{r}(t) = \langle t-2, t^2+1 \rangle \).
The derivative tells us the gradient or slope of the tangent to the curve at a particular point:

• The derivative of the first component \( t-2 \) is 1, indicating constant horizontal speed along the curve.
• The derivative of the second component \( t^2+1 \) is \( 2t \), showing how the vertical speed of the curve changes with \( t \).
• This process of taking derivatives helps us in locating points of interest on the curve, such as inflection points, maxima, and minima.

Cartesian Coordinates

Cartesian coordinates are the most common coordinate system used to define points on a plane using two numerical values. These are typically labeled as \( x \) and \( y \) axes. In this exercise, we used the parametric equations to describe paths on the Cartesian plane by establishing horizontal and vertical components.

• The Cartesian form lets us visualize how the parametric components interact to create familiar shapes, like parabolas or circles.
• By converting the given parametric equations to a Cartesian form \( y = (x+2)^2 + 1 \), the curve is more interpretable as it aligns with our usual geometric representations.
• Understanding of Cartesian coordinates, combined with parametric forms, enhances our ability to plot and analyze complex trajectories with ease.