Problem 31 - Advanced Engineering Mathematics

Question

Find the indicated derivative. Assume that all vector functions are differentiable. $$ \frac{d}{d t}\left[\mathbf{r}_{1}(2 t)+\mathbf{r}_{2}\left(\frac{1}{t}\right)\right] $$

Step-by-Step Solution

Answer

The derivative is \( 2\mathbf{r}_{1}'(2t) - \frac{1}{t^2}\mathbf{r}_{2}'\left(\frac{1}{t}\right) \).

Step 1: Identify the Derivative Components

We need to find the derivative of the expression \( \mathbf{r}_{1}(2t)+\mathbf{r}_{2}\left(\frac{1}{t}\right) \). Identify that this is a sum of two vector functions: \( \mathbf{r}_{1}(2t) \) and \( \mathbf{r}_{2}\left(\frac{1}{t}\right) \).

Step 2: Apply the Sum Rule of Differentiation

According to the sum rule, the derivative of a sum is the sum of the derivatives. Therefore, take the derivative of each vector function separately: \[ \frac{d}{d t}\left[\mathbf{r}_{1}(2t)\right] + \frac{d}{d t}\left[\mathbf{r}_{2}\left(\frac{1}{t}\right)\right] \]

Step 3: Derivative of \( \mathbf{r}_{1}(2t) \) Using the Chain Rule

Use the chain rule to differentiate \( \mathbf{r}_{1}(2t) \). Let \( u = 2t \), then the derivative becomes \[ \frac{d}{d t} \mathbf{r}_{1}(u) = \mathbf{r}_{1}'(u) \cdot \frac{d}{d t}(2t) = \mathbf{r}_{1}'(2t) \cdot 2 \]

Step 4: Derivative of \( \mathbf{r}_{2}\left(\frac{1}{t}\right) \) Using the Chain Rule

Also apply the chain rule for \( \mathbf{r}_{2}\left(\frac{1}{t}\right) \). Let \( v = \frac{1}{t} \), the derivative becomes \[ \frac{d}{d t} \mathbf{r}_{2}(v) = \mathbf{r}_{2}'(v) \cdot \frac{d}{d t}\left(\frac{1}{t}\right) = \mathbf{r}_{2}'\left(\frac{1}{t}\right) \cdot \left(-\frac{1}{t^2}\right) \]

Step 5: Combine the Derivatives

Combine the derivatives obtained from Steps 3 and 4 to get the final derivative: \[ 2\mathbf{r}_{1}'(2t) - \frac{1}{t^2}\mathbf{r}_{2}'\left(\frac{1}{t}\right) \]

Key Concepts

Differentiation of Vector Functions

In calculus, vector functions are functions that take one or more variables and return a vector. Differentiating these functions requires extending the ideas from scalar-valued function differentiation to vector-valued functions. This involves differentiating each component of the vector function separately.
When you encounter an expression like \( \mathbf{r}_1(2t) + \mathbf{r}_2\left(\frac{1}{t}\right) \), you are dealing with a vector-valued function. Here, \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) are vector functions dependent on different transformations of the variable \( t \). To differentiate such functions, you focus on:

• Identifying each vector function component.
• Differentiating each component individually.
• Using rules like the sum rule and the chain rule.

This is a critical step as these derivations are foundational in physics and engineering, where they describe changes in direction and magnitude of vectors, such as velocity and acceleration.

Chain Rule in Calculus

The chain rule is an essential concept in calculus for finding the derivative of composite functions. When a function is nested within another function, like \( \mathbf{r}_1(2t) \), the chain rule becomes necessary. It helps us by calculating the derivative of the outer function with respect to the inner function, then multiplying by the derivative of the inner function itself.
For example, see how it applies to \( \mathbf{r}_1(2t) \):

• Identify inner function: \( u = 2t \).
• Differentiate the outer function as if \( u \) were the variable: \( \frac{d}{du} \mathbf{r}_1(u) \).
• Multiply by the derivative of the inner function: \( \frac{d}{dt}(2t) = 2 \).

This results in \( \mathbf{r}_1'(2t) \cdot 2 \). Similarly, for \( \mathbf{r}_2\left(\frac{1}{t}\right) \):

• Let \( v = \frac{1}{t} \).
• Find \( \frac{d}{dt} \left(\frac{1}{t}\right) = -\frac{1}{t^2} \).
• Give \( \mathbf{r}_2'(v) \cdot \frac{d}{dt}(v) = \mathbf{r}_2'\left(\frac{1}{t}\right) \cdot \left(-\frac{1}{t^2}\right) \).

This powerful rule simplifies complex derivatives, allowing us to handle nested functions efficiently.

Sum Rule in Differentiation

The sum rule in differentiation is a straightforward but powerful rule. It states that the derivative of a sum of functions is simply the sum of the derivatives of those functions. This rule is particularly useful when dealing with vector functions composed of multiple components.
Consider our expression \( \mathbf{r}_1(2t) + \mathbf{r}_2\left(\frac{1}{t}\right) \). Using the sum rule means we can calculate the derivative of each part separately and then simply add the results together:

• Find the derivative of \( \mathbf{r}_1(2t) \) using the chain rule as explained earlier: \( 2 \mathbf{r}_1'(2t) \).
• Find the derivative of \( \mathbf{r}_2\left(\frac{1}{t}\right) \) also using the chain rule: \( -\frac{1}{t^2} \mathbf{r}_2'\left(\frac{1}{t}\right) \).
• Sum the derivatives: \( 2 \mathbf{r}_1'(2t) - \frac{1}{t^2} \mathbf{r}_2'\left(\frac{1}{t}\right) \).

The sum rule is a handy tool that enables the simplification and calculation of the derivative of multiple-function expressions.