Problem 47 - Essential Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

\(\mathbf{r}^{\prime}(t) = \langle 1, 2t, 3t^2 \rangle\), \( \mathbf{T}(1) = \frac{1}{\sqrt{14}}\langle 1, 2, 3 \rangle \), \(\mathbf{r}^{\prime\prime}(t) = \langle 0, 2, 6t \rangle\), \(\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) = \langle 12t - 6t^2, -6t, 2 \rangle\).

Step 1: Find the Derivative \( \mathbf{r}^{\prime}(t) \)

To find \( \mathbf{r}^{\prime}(t) \), take the derivative of each component of \( \mathbf{r}(t) \). \[ \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \] So, \( \mathbf{r}^{\prime}(t) \) becomes: \[ \mathbf{r}^{\prime}(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}(t^2), \frac{d}{dt}(t^3) \right\rangle = \left\langle 1, 2t, 3t^2 \right\rangle \]

Step 2: Calculate \( \mathbf{T}(1) \)

The unit tangent vector \( \mathbf{T}(t) \) is given by: \[ \mathbf{T}(t) = \frac{\mathbf{r}^{\prime}(t)}{\| \mathbf{r}^{\prime}(t) \|} \] Compute \( \mathbf{T}(1) \):1. Find \( \mathbf{r}^{\prime}(1) \): \[ \mathbf{r}^{\prime}(1) = \langle 1, 2(1), 3(1)^2 \rangle = \langle 1, 2, 3 \rangle \]2. Find \( \| \mathbf{r}^{\prime}(1) \| \): \[ \| \mathbf{r}^{\prime}(1) \| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \]3. Therefore, \( \mathbf{T}(1) \) is: \[ \mathbf{T}(1) = \frac{\langle 1, 2, 3 \rangle}{\sqrt{14}} = \left\langle \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right\rangle \]

Step 3: Find the Second Derivative \( \mathbf{r}^{\prime\prime}(t) \)

Compute the second derivative by finding the derivative of \( \mathbf{r}^{\prime}(t) \).\[ \mathbf{r}^{\prime}(t) = \langle 1, 2t, 3t^2 \rangle \]The derivative of this is: \[ \mathbf{r}^{\prime\prime}(t) = \left\langle \frac{d}{dt}(1), \frac{d}{dt}(2t), \frac{d}{dt}(3t^2) \right\rangle = \left\langle 0, 2, 6t \right\rangle \]

Step 4: Calculate the Cross Product \( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \)

The cross product of \( \mathbf{r}^{\prime}(t) \) and \( \mathbf{r}^{\prime\prime}(t) \) is computed using:\[ \mathbf{r}^{\prime}(t) = \langle 1, 2t, 3t^2 \rangle \] and\[ \mathbf{r}^{\prime\prime}(t) = \langle 0, 2, 6t \rangle \]The cross product is:\[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2t & 3t^2 \ 0 & 2 & 6t \end{vmatrix} \]Expanding the determinant:\[ \mathbf{i}(2(6t) - 3t^2(2)) - \mathbf{j}(1(6t) - 0) + \mathbf{k}(1(2) - 0) \]\[ = \mathbf{i}(12t - 6t^2) - \mathbf{j}(6t) + \mathbf{k}(2) \]\[ = \langle 12t - 6t^2, -6t, 2 \rangle \]

Key Concepts

Derivatives of Vector Functions

Derivatives of vector functions are an integral part of vector calculus. They provide information about the rate of change of vector quantities. For a vector function \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \), the derivative \( \mathbf{r}^{\prime}(t) \) is determined by differentiating each scalar component with respect to \( t \). This results in a new vector \( \langle f^{\prime}(t), g^{\prime}(t), h^{\prime}(t) \rangle \).
In the original exercise, the function \( \mathbf{r}(t) = \langle t, t^2, t^3 \rangle \) was given. By applying the derivative operation to each component, you arrive at \( \mathbf{r}^{\prime}(t) = \langle 1, 2t, 3t^2 \rangle \). These derivatives reflect the speed and direction of change for each component of the vector function.

Unit Tangent Vector

The unit tangent vector provides a sense of direction along a path, normalizing the tangent into a vector of unit length. It is computed by dividing the derivative of the vector function by its magnitude. The expression for the unit tangent vector \( \mathbf{T}(t) \) is given by \( \mathbf{T}(t) = \frac{\mathbf{r}^{\prime}(t)}{\| \mathbf{r}^{\prime}(t) \|} \).

For the exercise at hand, we focused on \( \mathbf{T}(1) \). By first evaluating \( \mathbf{r}^{\prime}(1) = \langle 1, 2, 3 \rangle \) and its magnitude \( \| \mathbf{r}^{\prime}(1) \| = \sqrt{14} \), the unit tangent vector is calculated as \( \mathbf{T}(1) = \left\langle \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right\rangle \). This resulting vector points in the direction the curve is heading at \( t = 1 \), with a normalized length of 1.

Cross Product

The cross product, a significant operation in vector calculus, results in a vector perpendicular to two original vectors in three-dimensional space. It's often used to find orthogonal directions or calculate areas spanned by vectors. The formula for the cross product of two vector functions \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:

• \( \mathbf{i}(a_2b_3 - a_3b_2) \)
• \(- \mathbf{j}(a_1b_3 - a_3b_1) \)
• \(+ \mathbf{k}(a_1b_2 - a_2b_1) \)

For the vectors \( \mathbf{r}^{\prime}(t) = \langle 1, 2t, 3t^2 \rangle \) and \( \mathbf{r}^{\prime\prime}(t) = \langle 0, 2, 6t \rangle \), the cross product is \( \langle 12t - 6t^2, -6t, 2 \rangle \). This new vector highlights the perpendicular component to the tangent and the rate of change vectors.

Second Derivative

The second derivative of a vector function gives us information about the concavity and acceleration of the vector path. Applying the derivative operation to the first derivative \( \mathbf{r}^{\prime}(t) \) results in the second derivative \( \mathbf{r}^{\prime\prime}(t) \). For a vector function \( \langle f^{\prime}(t), g^{\prime}(t), h^{\prime}(t) \rangle \), its second derivative is \( \langle f^{\prime\prime}(t), g^{\prime\prime}(t), h^{\prime\prime}(t) \rangle \).

In the given exercise, starting with the first derivative \( \mathbf{r}^{\prime}(t) = \langle 1, 2t, 3t^2 \rangle \), the second derivative becomes \( \mathbf{r}^{\prime\prime}(t) = \langle 0, 2, 6t \rangle \). This tells us how the velocity \( \mathbf{r}^{\prime}(t) \) is changing, indicating whether the vector path is curving upwards, downwards, or maintaining a constant arc at any point \( t \).