Problem 34 - 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 \(t\) . $$ \mathbf{r}(t)=\left\langle t^{2}, t^{3}\right\rangle, \quad t=1 $$

Step-by-Step Solution

Answer

The curve is a parabola; the tangent at \( t = 1 \) is \( \langle 2, 3 \rangle \).

Step 1: Understanding the Problem

We are given the vector equation \( \mathbf{r}(t) = \langle t^2, t^3 \rangle \) and need to complete several tasks: sketch the curve, find the derivative \( \mathbf{r}^{\prime}(t) \), and sketch both \( \mathbf{r}(t) \) and \( \mathbf{r}^{\prime}(t) \) at a specific point \( t = 1 \).

Step 2: Sketching the Plane Curve

The curve is defined by \( x(t) = t^2 \) and \( y(t) = t^3 \). As \( t \) varies, we plot points \( (t^2, t^3) \) and connect them smoothly to sketch the curve. This shape generally follows a parabolic trajectory as \( t \) increases from negative to positive values.

Step 3: Finding the Derivative

To find the derivative \( \mathbf{r}^{\prime}(t) \), compute the derivatives of each component: \( \frac{d}{dt}(t^2) = 2t \) and \( \frac{d}{dt}(t^3) = 3t^2 \). Thus, \( \mathbf{r}^{\prime}(t) = \langle 2t, 3t^2 \rangle \).

Step 4: Evaluating at Specific Point

Evaluate both \( \mathbf{r}(t) \) and \( \mathbf{r}^{\prime}(t) \) at \( t = 1 \). For \( \mathbf{r}(1) = \langle 1^2, 1^3 \rangle = \langle 1, 1 \rangle \) and \( \mathbf{r}^{\prime}(1) = \langle 2 \times 1, 3 \times 1^2 \rangle = \langle 2, 3 \rangle \).

Step 5: Sketching Vectors at \( t = 1 \)

Plot the position vector \( \mathbf{r}(1) = \langle 1, 1 \rangle \) from the origin to point \( (1, 1) \) on the graph. Then, sketch the tangent vector \( \mathbf{r}^{\prime}(1) = \langle 2, 3 \rangle \), starting from the end of \( \mathbf{r}(1) \) and extending in the direction \( (2, 3) \).

Key Concepts

Vector Equation

A vector equation describes a curve or a path in space using vector expressions. It is often represented in the form \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) for three-dimensional space, or simply \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \) in two dimensions. Here, \( x(t) \) and \( y(t) \) (and \( z(t) \) if in 3D) are functions of the parameter \( t \), which can represent time or any other dimension of change.
For example, the vector equation \( \mathbf{r}(t) = \langle t^2, t^3 \rangle \) specifies a curve in the plane. The parameter \( t \) affects how the x-coordinate (as \( t^2 \)) and y-coordinate (as \( t^3 \)) change. By varying \( t \), we trace out the curve.

• \( x(t) = t^2 \)
• \( y(t) = t^3 \)

This shows that as \( t \) varies, the point \( (x(t), y(t)) \) moves along a specific trajectory, which for this equation, is smooth and not closed.

Plane Curve Sketching

Sketching a plane curve involves tracing out the path of a curve described by a vector equation. To sketch our curve \( \mathbf{r}(t) = \langle t^2, t^3 \rangle \), we need to plot points by choosing values for \( t \) and calculating \( x(t) \) and \( y(t) \).
Start by selecting several values of \( t \) (e.g., -2, -1, 0, 1, 2) and determine the corresponding coordinates.

• For \( t = -2 \): \( (x, y) = (4, -8) \)
• For \( t = -1 \): \( (x, y) = (1, -1) \)
• For \( t = 0 \): \( (x, y) = (0, 0) \)
• For \( t = 1 \): \( (x, y) = (1, 1) \)
• For \( t = 2 \): \( (x, y) = (4, 8) \)

Connect these points smoothly. The shape of this particular curve resembles a parabola, since it follows a more polynomial pattern defined by the powers in \( t^2 \) and \( t^3 \). This can be sketched on the xy-plane by connecting the dots smoothly.

Derivative of a Vector

The derivative of a vector function represents how the curve changes with respect to the parameter \( t \). It is found by taking the derivative of each component of the vector equation separately.
For the vector \( \mathbf{r}(t) = \langle t^2, t^3 \rangle \), we find its derivative \( \mathbf{r}^{\prime}(t) \) by calculating:

• Derivative of \( t^2 \) is \( 2t \)
• Derivative of \( t^3 \) is \( 3t^2 \)

Therefore, \( \mathbf{r}^{\prime}(t) = \langle 2t, 3t^2 \rangle \). This derivative gives the slope or direction of the tangent to the curve at any point \( t \). Calculating the derivatives separately and combining helps understand the rate of change in both x and y directions separately and together.

Tangent Vector

The tangent vector to a curve at a given point is a vector that "touches" the curve at that point and points in the direction that the curve is heading. It is critical because it provides insight into the behavior of the curve at a specific location.
For the given vector function \( \mathbf{r}(t) = \langle t^2, t^3 \rangle \), we have its derivative \( \mathbf{r}^{\prime}(t) = \langle 2t, 3t^2 \rangle \). Evaluating this at \( t = 1 \) gives the tangent vector:

• \( \mathbf{r}^{\prime}(1) = \langle 2 \times 1, 3 \times 1^2 \rangle = \langle 2, 3 \rangle \)

The point \( \mathbf{r}(1) = \langle 1, 1 \rangle \) is where the tangent vector originates.
To sketch it, draw the position vector from the origin to the point (1, 1), and from that point, draw the tangent in the direction of \( (2, 3) \). This visually demonstrates how the curve moves through that point.