Problem 28 - Advanced Engineering Mathematics

Question

In Problems, find the indicated derivative. Assume that all vector functions are differentiable. $$ \frac{d}{d t}[\mathbf{r}(t) \cdot(t \mathbf{r}(t))] $$

Step-by-Step Solution

Answer

The derivative is \([\mathbf{r}(t) \cdot \mathbf{r}(t)] + 2t [\mathbf{r}'(t) \cdot \mathbf{r}(t)]\)."

Step 1: Identify the Components

First, identify the components of the expression, \( \mathbf{r}(t) \cdot (t \mathbf{r}(t)) \). Here, \( \mathbf{r}(t) \) is a vector function and \( t \cdot \mathbf{r}(t) \) is the scalar multiplication of \( t \) with the vector function \( \mathbf{r}(t) \). This is a dot product of a vector with itself multiplied by a scalar.

Step 2: Apply the Product Rule to the Dot Product

Since the expression involves the dot product of two vector functions, we use the product rule for differentiation of dot products. The product rule states that \( \frac{d}{dt} [\mathbf{a}(t) \cdot \mathbf{b}(t)] = \mathbf{a}'(t) \cdot \mathbf{b}(t) + \mathbf{a}(t) \cdot \mathbf{b}'(t) \). Here, let \( \mathbf{a}(t) = \mathbf{r}(t) \) and \( \mathbf{b}(t) = t \mathbf{r}(t) \).

Step 3: Differentiate Each Component

Differentiate \( \mathbf{a}(t) = \mathbf{r}(t) \), which gives \( \mathbf{a}'(t) = \mathbf{r}'(t) \). For \( \mathbf{b}(t) = t \mathbf{r}(t) \), use the product rule, giving \( \mathbf{b}'(t) = \mathbf{r}(t) + t \mathbf{r}'(t) \) since \( (t)' = 1 \).

Step 4: Substitute into the Product Rule

Substitute the derivatives into the product rule: \( \frac{d}{dt} [\mathbf{r}(t) \cdot (t \mathbf{r}(t))] = \mathbf{r}'(t) \cdot (t \mathbf{r}(t)) + \mathbf{r}(t) \cdot (\mathbf{r}(t) + t \mathbf{r}'(t)) \).

Step 5: Simplify the Expression

Simplify the expression from Step 4: \( \mathbf{r}'(t) \cdot (t \mathbf{r}(t)) = t [\mathbf{r}'(t) \cdot \mathbf{r}(t)] \) and \( \mathbf{r}(t) \cdot \mathbf{r}(t) + \mathbf{r}(t) \cdot (t \mathbf{r}'(t)) = [\mathbf{r}(t) \cdot \mathbf{r}(t)] + t [\mathbf{r}(t) \cdot \mathbf{r}'(t)] \).

Step 6: Combine and Complete the Derivative

Combining results, the derivative is: \( t [\mathbf{r}'(t) \cdot \mathbf{r}(t)] + [\mathbf{r}(t) \cdot \mathbf{r}(t)] + t [\mathbf{r}(t) \cdot \mathbf{r}'(t)] = [\mathbf{r}(t) \cdot \mathbf{r}(t)] + 2t [\mathbf{r}'(t) \cdot \mathbf{r}(t)] \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors resulting in a scalar (a single number). When dealing with vectors, this operation captures how much one vector goes in the direction of another. It is given by the formula: \[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]where \(\mathbf{a}\) and \(\mathbf{b}\) are vectors with respective components. In this context, we are dealing with vector functions.
The dot product helps to simplify expressions involving vectors in calculus. Particularly, it appears in physics when calculating work done by a force or projecting one vector onto another.
Remember that the result of a dot product is a number, not a vector, which highlights the "scalar" nature.

Derivative

Derivatives are a fundamental concept in calculus and measure how a function changes as its input changes. For a real-valued function of one variable, the derivative is simply a rate of change or slope of the function at any point.
When functions involve vectors, derivatives become a bit more involved but still retain their essence of showing how vector functions change with respect to time or another variable. \[\mathbf{r}'(t)\] represents the derivative of a vector function \(\mathbf{r}\) with respect to \(t\). This captures how each component of the vector changes over time.
Understanding derivatives in vector functions helps in applications like physics, where they can represent velocity (as the change in position over time) or acceleration (the change in velocity over time).

Product Rule

The product rule is an essential tool in calculus for differentiating products of two functions. For scalar functions \(f\) and \(g\), the product rule states:\[\frac{d}{dt} [f(t) \cdot g(t)] = f'(t) \cdot g(t) + f(t) \cdot g'(t)\]This rule can be extended to vector functions, which is crucial when dealing with expressions involving dot products of vectors.
For vector functions \(\mathbf{a}(t)\) and \(\mathbf{b}(t)\), the product rule becomes: \[\frac{d}{dt} [\mathbf{a}(t) \cdot \mathbf{b}(t)] = \mathbf{a}'(t) \cdot \mathbf{b}(t) + \mathbf{a}(t) \cdot \mathbf{b}'(t)\]This form of the product rule allows us to differentiate complex vector expressions efficiently and is widely used in physics and engineering problems that involve motion and dynamics.

Vector Function

A vector function assigns a vector to each value in its domain. Typically, you might see a vector function written as \(\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle\), where \(f(t), g(t),\) and \(h(t)\) are individual scalar functions.
These functions are often used to describe paths or trajectories in space, meaning they are pivotal in physics, engineering, and computer graphics.
Each component of the vector function changes with respect to time or another variable, allowing for a rich description of movement or change. For example, in physics, \(\mathbf{r}(t)\) could represent the position of a particle at time \(t\), with each component describing a position along different axes.
Grasping the behavior of vector functions and their derivatives helps in understanding the dynamics of systems, making them a crucial part of vector calculus.