Problem 21 - Calculus Early Transcendentals

Question

If $$ \mathbf{r}(t)=\left\langle t, t^{2}, t^{3}\right\rangle, \text { find } \mathbf{r}^{\prime}(t), \mathbf{T}(1), \mathbf{r}^{\prime \prime}(t), \text { and } \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) $$

Step-by-Step Solution

Answer

\( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \); \( \mathbf{T}(1) = \left\langle \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right\rangle \); \( \mathbf{r}''(t) = \langle 0, 2, 6t \rangle \); \( \mathbf{r}'(t) \times \mathbf{r}''(t) = \langle 6t^2, -6t, 2 \rangle \).

Step 1: Find the first derivative of \( \mathbf{r}(t) \)

To find the first derivative \( \mathbf{r}'(t) \), differentiate each component of the vector \( \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \). The derivative of \( t \) is 1, the derivative of \( t^2 \) is \( 2t \), and the derivative of \( t^3 \) is \( 3t^2 \). Therefore, \( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \).

Step 2: Evaluate the unit tangent vector \( \mathbf{T}(1) \)

First, substitute \( t = 1 \) into \( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \) to get \( \mathbf{r}'(1) = \langle 1, 2, 3 \rangle \). Next, calculate the magnitude of \( \mathbf{r}'(1) \): \( \| \mathbf{r}'(1) \| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \). Then, \( \mathbf{T}(1) = \frac{\mathbf{r}'(1)}{\| \mathbf{r}'(1) \|} = \left\langle \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right\rangle \).

Step 3: Find the second derivative \( \mathbf{r}''(t) \)

To find \( \mathbf{r}''(t) \), differentiate \( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \). The derivative of 1 is 0, the derivative of \( 2t \) is 2, and the derivative of \( 3t^2 \) is \( 6t \). Therefore, \( \mathbf{r}''(t) = \langle 0, 2, 6t \rangle \).

Step 4: Calculate the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \)

Use the formula for the cross product of two vectors: if \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), then \( \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \). Applying this to \( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \) and \( \mathbf{r}''(t) = \langle 0, 2, 6t \rangle \), we have:1. First component: \( 2t(6t) - 3t^2(2) = 12t^2 - 6t^2 = 6t^2 \).2. Second component: \( 3t^2(0) - 1(6t) = -6t \).3. Third component: \( 1(2) - 2t(0) = 2 \).Hence, \( \mathbf{r}'(t) \times \mathbf{r}''(t) = \langle 6t^2, -6t, 2 \rangle \).

Key Concepts

Derivatives

In vector calculus, derivatives are essential for understanding how a vector function changes with respect to a parameter. For a parametric vector function such as \( \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \), finding the derivative involves differentiating each component of the vector separately. The result, \( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \), provides the rate of change of the vector function as the parameter \( t \) changes.
Understanding derivatives is crucial for analyzing the motion and changes occurring in vector fields and is foundational to many applications in physics and engineering. Here are some basic things to keep in mind:

• Each component of the vector is differentiated separately.
• The result is a new vector that shows the direction and rate of change of the original vector.

Calculating derivatives helps us determine tangent vectors and other important quantities when working with curves defined parametrically.

Cross Product

The cross product is a vector operation that takes two vectors and returns another vector that is perpendicular to both. This is especially useful in physics and engineering for finding a direction normal to a plane defined by two vectors. For the vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the cross product is given by:
\[ \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \]
In the exercise above, finding \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) involves computing this for \( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \) and \( \mathbf{r}''(t) = \langle 0, 2, 6t \rangle \), resulting in \( \langle 6t^2, -6t, 2 \rangle \).
Important points include:

• The result is always orthogonal, meaning both perpendicular, to the original vectors.
• It's mainly used to find torques, rotational force, and directional normals.
• Magnitude of the cross product relates to the area of the parallelogram formed by the original vectors.

Understanding the cross product mechanism enhances our ability to tackle three-dimensional vector problems.

Tangent Vector

Tangent vectors reveal the direction of a curve at any given point. For a parametric curve defined by \( \mathbf{r}(t) \), the tangent vector at a point is just \( \mathbf{r}'(t) \). For example, at \( t = 1 \), the tangent vector \( \mathbf{r}'(1) = \langle 1, 2, 3 \rangle \). Calculating tangent vectors help differentiate the curve's geometry at specific points, describing the path's immediate direction.
The unit tangent vector normalizes the tangent vector to unit length, providing only directional information. The unit tangent vector is:

Calculated as \( \mathbf{T}(1) = \frac{\mathbf{r}'(1)}{\| \mathbf{r}'(1) \|} \).

Represented as \( \langle \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \rangle \) in this case.

These vectors are crucial in physics for defining motion direction, and path curvature, and in animations and computer graphics for rendering objects smoothly.

Parametric Curves

Parametric curves are described by equations that express the coordinates of points as functions of a parameter, usually denoted by \( t \). This approach provides flexibility in modeling curves in space and is powerful for applications like animation, physics, and computer graphics.
An example is \( \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \), where each component is a function of \( t \). This representation allows for detailed analysis and manipulation of the curve by varying \( t \). Key characteristics include:

• Each point on the curve is determined by a single parameter.
• Useful in defining curves that cannot be easily expressed as \( y = f(x) \).
• Enables easy computation of derivatives, which help in finding curve properties like tangent vectors.

Overall, parametric representations give a comprehensive way to understand and work with complex curves in multiple dimensions.