Problem 63 - Calculus 3

Question

Find \(\mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime \prime}(t)\) for \(\mathbf{r}(t)=-3 t^{5} \mathbf{i}+5 t \mathbf{j}+2 t^{2} \mathbf{k}\)

Step-by-Step Solution

Answer

\( \mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime\prime}(t) = 900t^7 + 16t \).

Step 1: Find the First Derivative

Differentiate each component of the vector function \( \mathbf{r}(t) = -3t^5 \mathbf{i} + 5t \mathbf{j} + 2t^2 \mathbf{k} \) with respect to \( t \). - The derivative of \( -3t^5 \) with respect to \( t \) is \( -15t^4 \).- The derivative of \( 5t \) with respect to \( t \) is \( 5 \).- The derivative of \( 2t^2 \) with respect to \( t \) is \( 4t \).Thus, \( \mathbf{r}^{\prime}(t) = -15t^4 \mathbf{i} + 5 \mathbf{j} + 4t \mathbf{k} \).

Step 2: Find the Second Derivative

Differentiate \( \mathbf{r}^{\prime}(t) = -15t^4 \mathbf{i} + 5 \mathbf{j} + 4t \mathbf{k} \) with respect to \( t \).- The derivative of \( -15t^4 \) is \( -60t^3 \).- The derivative of \( 5 \) is \( 0 \).- The derivative of \( 4t \) is \( 4 \).Thus, \( \mathbf{r}^{\prime\prime}(t) = -60t^3 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} \).

Step 3: Compute the Dot Product

Now compute the dot product \( \mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime\prime}(t) \) by multiplying corresponding components and adding the results:- Multiply \( -15t^4 \) and \( -60t^3 \), resulting in \( 900t^7 \).- Multiply \( 5 \) and \( 0 \), resulting in \( 0 \).- Multiply \( 4t \) and \( 4 \), resulting in \( 16t \).The dot product is: \( 900t^7 + 0 + 16t = 900t^7 + 16t \).

Key Concepts

Vector Differentiation

Vector differentiation plays a fundamental role in vector calculus, allowing us to find the rate of change of vector functions. A vector function consists of multiple components, often a mixing of unit vectors like \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), each associated with their own functions of a variable, commonly \( t \). By taking the derivative of each component of the vector, we can understand how that vector changes overall with respect to \( t \).

For instance, consider the vector function \( \mathbf{r}(t) = -3t^5 \mathbf{i} + 5t \mathbf{j} + 2t^2 \mathbf{k} \). To compute its derivative, we can apply the power rule to each term:

• The derivative of \( -3t^5 \) is \( -15t^4 \).
• The derivative of \( 5t \) yields \( 5 \).
• The derivative of \( 2t^2 \) is \( 4t \).

The derivative vector, therefore, becomes \( \mathbf{r}^{\prime}(t) = -15t^4 \mathbf{i} + 5 \mathbf{j} + 4t \mathbf{k} \). Differentiating vectors enables mathematicians and scientists to model the dynamic behaviors of systems like moving particles or flowing fluids.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number. It is widely used in physics and engineering to project vectors, calculate work done, and determine angles between vectors.

In mathematical terms, the dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = A_x \cdot B_x + A_y \cdot B_y + A_z \cdot B_z \] where \( A_x, A_y, A_z \) are the components of vector \( \mathbf{A} \) and \( B_x, B_y, B_z \) are the components of vector \( \mathbf{B} \).

For our derivatives \( \mathbf{r}^{\prime}(t) \) and \( \mathbf{r}^{\prime\prime}(t) \), the dot product is calculated by multiplying corresponding components:

• Multiply \( -15t^4 \) and \( -60t^3 \) to get \( 900t^7 \).
• Multiply \( 5 \) and \( 0 \) to obtain \( 0 \).
• Multiply \( 4t \) and \( 4 \) to get \( 16t \).

Thus, the dot product \( \mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime\prime}(t) \) is \( 900t^7 + 16t \). The result reveals not only the summed magnitude of these projections but also interactions stemming from their directional alignment.

Derivatives of Vector Functions

Calculating derivatives of vector functions involves taking the derivative of each individual component separately, treating them like ordinary functions. This process simplifies handling multidimensional data and is integral in analyzing paths, velocities, or changes in vector quantities across time or other variables.

In our example, once we derived the first derivative \( \mathbf{r}^{\prime}(t) \), we continued by differentiating it again to find the second derivative \( \mathbf{r}^{\prime\prime}(t) \). Let's break it down:

• Differentiating \( -15t^4 \) yields \( -60t^3 \).
• Differentiating \( 5 \) results in \( 0 \), as the derivative of a constant remains 0.
• Differentiating \( 4t \) results in \( 4 \).

Hence, \( \mathbf{r}^{\prime\prime}(t) = -60t^3 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} \). Grasping these derivative processes translates into understanding dynamic systems more intuitively, aiding in the study of acceleration or curvature in fields like kinematics and electromagnetism.