Derivative Of Dot Product Of Vector Functions Explained

In vector calculus, understanding how to differentiate vector functions is crucial for various applications in physics and engineering. This article delves into the process of finding the derivative of a dot product of two vector functions. We will explore the necessary steps, apply the relevant differentiation rules, and compute the final result. Specifically, we will focus on the vector functions $r = t^2 \underline{i} - t \underline{j} + (2t + 1) \underline{k}$ and $s = (2t - 3) \underline{i} + \underline{j} - t \underline{k}$, and evaluate $\frac{d}{dt}(r \cdot s)$ at $t = 1$. This exercise will provide a comprehensive understanding of how to handle such problems.

Problem Statement

Given the vector functions:

$r = t^2 \underline{i} - t \underline{j} + (2t + 1) \underline{k}$ $s = (2t - 3) \underline{i} + \underline{j} - t \underline{k}$

We need to evaluate $\frac{d}{dt}(r \cdot s)$ at $t = 1$.

Step-by-Step Solution

1. Compute the Dot Product $r \cdot s$

The dot product of two vectors is calculated by multiplying corresponding components and summing the results. For vectors $r = r_1 \underline{i} + r_2 \underline{j} + r_3 \underline{k}$ and $s = s_1 \underline{i} + s_2 \underline{j} + s_3 \underline{k}$, the dot product $r \cdot s$ is given by:

$r \cdot s = r_1s_1 + r_2s_2 + r_3s_3$

In our case, we have:

$r = t^2 \underline{i} - t \underline{j} + (2t + 1) \underline{k}$ $s = (2t - 3) \underline{i} + \underline{j} - t \underline{k}$

So, the dot product $r \cdot s$ is:

$r \cdot s = (t^2)(2t - 3) + (-t)(1) + (2t + 1)(-t)$ $r \cdot s = 2t^3 - 3t^2 - t - 2t^2 - t$ $r \cdot s = 2t^3 - 5t^2 - 2t$

2. Differentiate $r \cdot s$ with Respect to $t$

Now, we need to find the derivative of $r \cdot s$ with respect to $t$. We will apply the power rule for differentiation, which states that $\frac{d}{dt}(t^n) = nt^{n-1}$.

$\frac{d}{dt}(r \cdot s) = \frac{d}{dt}(2t^3 - 5t^2 - 2t)$ $\frac{d}{dt}(r \cdot s) = 2 \cdot 3t^2 - 5 \cdot 2t - 2$ $\frac{d}{dt}(r \cdot s) = 6t^2 - 10t - 2$

3. Evaluate the Derivative at $t = 1$

Finally, we will evaluate the derivative at $t = 1$:

$\frac{d}{dt}(r \cdot s)|_{t=1} = 6(1)^2 - 10(1) - 2$ $\frac{d}{dt}(r \cdot s)|_{t=1} = 6 - 10 - 2$ $\frac{d}{dt}(r \cdot s)|_{t=1} = -6$

Thus, the value of $\frac{d}{dt}(r \cdot s)$ at $t = 1$ is $-6$.

Alternative Method: Using the Product Rule for Dot Products

Another approach to solving this problem is by using the product rule for the dot product. The product rule states that if $r(t)$ and $s(t)$ are vector functions, then:

$\frac{d}{dt}(r(t) \cdot s(t)) = r'(t) \cdot s(t) + r(t) \cdot s'(t)$

This method involves finding the derivatives of $r(t)$ and $s(t)$ individually and then applying the formula.

1. Find the Derivatives of $r(t)$ and $s(t)$

First, we find the derivatives of $r(t)$ and $s(t)$ with respect to $t$:

$r(t) = t^2 \underline{i} - t \underline{j} + (2t + 1) \underline{k}$ $r'(t) = \frac{d}{dt}(t^2) \underline{i} - \frac{d}{dt}(t) \underline{j} + \frac{d}{dt}(2t + 1) \underline{k}$ $r'(t) = 2t \underline{i} - \underline{j} + 2 \underline{k}$

$s(t) = (2t - 3) \underline{i} + \underline{j} - t \underline{k}$ $s'(t) = \frac{d}{dt}(2t - 3) \underline{i} + \frac{d}{dt}(1) \underline{j} - \frac{d}{dt}(t) \underline{k}$ $s'(t) = 2 \underline{i} + 0 \underline{j} - \underline{k}$ $s'(t) = 2 \underline{i} - \underline{k}$

2. Compute $r'(t) \cdot s(t)$ and $r(t) \cdot s'(t)$

Now, we calculate the dot products $r'(t) \cdot s(t)$ and $r(t) \cdot s'(t)$:

$r'(t) \cdot s(t) = (2t \underline{i} - \underline{j} + 2 \underline{k}) \cdot ((2t - 3) \underline{i} + \underline{j} - t \underline{k})$ $r'(t) \cdot s(t) = 2t(2t - 3) + (-1)(1) + 2(-t)$ $r'(t) \cdot s(t) = 4t^2 - 6t - 1 - 2t$ $r'(t) \cdot s(t) = 4t^2 - 8t - 1$

$r(t) \cdot s'(t) = (t^2 \underline{i} - t \underline{j} + (2t + 1) \underline{k}) \cdot (2 \underline{i} - \underline{k})$ $r(t) \cdot s'(t) = t^2(2) + (-t)(0) + (2t + 1)(-1)$ $r(t) \cdot s'(t) = 2t^2 - 2t - 1$

3. Apply the Product Rule

Using the product rule, we have:

$\frac{d}{dt}(r(t) \cdot s(t)) = r'(t) \cdot s(t) + r(t) \cdot s'(t)$ $\frac{d}{dt}(r(t) \cdot s(t)) = (4t^2 - 8t - 1) + (2t^2 - 2t - 1)$ $\frac{d}{dt}(r(t) \cdot s(t)) = 6t^2 - 10t - 2$

4. Evaluate at $t = 1$

Finally, we evaluate the derivative at $t = 1$:

$\frac{d}{dt}(r(t) \cdot s(t))|_{t=1} = 6(1)^2 - 10(1) - 2$ $\frac{d}{dt}(r(t) \cdot s(t))|_{t=1} = 6 - 10 - 2$ $\frac{d}{dt}(r(t) \cdot s(t))|_{t=1} = -6$

Conclusion

In this detailed exploration, we have successfully evaluated $\frac{d}{dt}(r \cdot s)$ at $t = 1$ using two distinct methods: directly differentiating the dot product and applying the product rule for dot products. Both methods yielded the same result, $-6$. Understanding these techniques is fundamental in vector calculus and is widely applicable in various scientific and engineering domains. The step-by-step solutions provided here offer a comprehensive guide for tackling similar problems, ensuring a strong grasp of the underlying concepts and methodologies. Whether you are a student learning vector calculus or a professional applying these principles, this article serves as a valuable resource for mastering the differentiation of vector functions and their dot products.

By using both the direct differentiation method and the product rule, we have reinforced the importance of having multiple approaches to problem-solving. The direct method is straightforward and efficient when the dot product is easy to compute, while the product rule offers a more structured approach that can be beneficial when dealing with complex vector functions. Understanding when and how to apply each method enhances one's ability to tackle a wider range of problems in vector calculus. Moreover, the consistency in results obtained from both methods validates our calculations and reinforces our understanding of the underlying principles. This thorough analysis not only provides a solution to the specific problem but also equips the reader with the knowledge and skills to approach similar challenges with confidence and precision. The application of these techniques extends beyond theoretical exercises and finds practical use in fields such as physics, engineering, and computer graphics, where vector calculus plays a pivotal role in modeling and analyzing real-world phenomena. Therefore, a solid foundation in these concepts is essential for anyone working in these domains.

Furthermore, the ability to differentiate vector functions is crucial in understanding the dynamics of systems described by vectors. In physics, for instance, vector functions are used to represent the position, velocity, and acceleration of objects moving in space. The derivative of a position vector with respect to time gives the velocity vector, and the derivative of the velocity vector gives the acceleration vector. Similarly, in engineering, understanding how vector fields change is essential for designing systems that interact with these fields, such as electromagnetic systems or fluid dynamics systems. The dot product, in particular, is significant as it provides a scalar quantity that measures the alignment or projection of one vector onto another. The rate of change of this alignment, given by the derivative of the dot product, can provide valuable insights into the interactions and dynamics of the system. Therefore, mastering the techniques to differentiate dot products of vector functions is not just an academic exercise but a fundamental skill with far-reaching applications in various scientific and engineering disciplines. This comprehensive understanding fosters a deeper appreciation for the elegance and power of vector calculus in describing and analyzing the world around us.

Final Answer

The correct answer is (a) -6.