Problem 31 - Advanced Engineering Mathematics

Question

In Problems, 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 to be Found

We are asked to find \[ \frac{d}{d t}\left[\mathbf{r}_{1}(2t)+\mathbf{r}_{2}\left(\frac{1}{t}\right)\right]\] where \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are vector functions of their respective arguments. Our goal is to differentiate this expression with respect to \(t\).

Step 2: Apply the Sum Rule

The derivative of a sum is the sum of the derivatives. We can break it down as follows: \[\frac{d}{d t}\left[\mathbf{r}_{1}(2t) + \mathbf{r}_{2}\left(\frac{1}{t}\right)\right] = \frac{d}{d t}\mathbf{r}_{1}(2t) + \frac{d}{d t}\mathbf{r}_{2}\left(\frac{1}{t}\right)\] Now, we need to compute the derivative of each term separately.

Step 3: Use the Chain Rule for \( \, \mathbf{r}_{1}(2t) \, \)

The first term requires the chain rule. Let \( u = 2t \), so \[\frac{d}{d t} \mathbf{r}_{1}(2t) = \frac{d}{d u} \mathbf{r}_{1}(u) \cdot \frac{d u}{d t} = \mathbf{r}_{1}'(2t) \cdot 2\] because \(\frac{d u}{d t} = 2\).

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

For the second term, apply the chain rule again. Let \(v = \frac{1}{t}\), so \[\frac{d}{d t} \mathbf{r}_{2}\left(\frac{1}{t}\right) = \frac{d}{d v} \mathbf{r}_{2}(v) \cdot \frac{d v}{d t} = \mathbf{r}_{2}'\left(\frac{1}{t}\right) \cdot \left(-\frac{1}{t^2}\right)\] because \(\frac{d v}{d t} = -\frac{1}{t^2}\).

Step 5: Assemble the Final Derivative Expression

Combine the results from Steps 3 and 4 to form the complete derivative: \[\mathbf{r}_{1}'(2t) \cdot 2 - \mathbf{r}_{2}'\left(\frac{1}{t}\right) \cdot \frac{1}{t^2}\] This is the indicated derivative of the given vector function expression.

Key Concepts

Derivative of Vector Functions

Differentiating vector functions is similar to differentiating scalar functions, but it happens component-wise. A vector function \( \mathbf{r}(t) \) can be represented as \( \langle f(t), g(t), h(t) \rangle \), where each component is a scalar function. The derivative of \( \mathbf{r}(t) \) with respect to \( t \) is \( \mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle \). This means you differentiate each component separately, treating the vector as a collection of individual functions rather than a single entity. This concept underpins many rules used for vector differentiation.

Chain Rule

The chain rule is essential when dealing with compositions of functions. For vector functions, it allows us to differentiate when an inner variable is dependent on the outer one. If we have a composite vector function \( \mathbf{r}(u) \), and \( u \) is a function of \( t \), we use the chain rule:

• Differentiate \( \mathbf{r} \) with respect to \( u \)
• Multiply by the derivative of \( u \) with respect to \( t \)

For example, in our exercise, \( \mathbf{r}_1(2t) \) required us to set \( u = 2t \). Then, we calculate \( \frac{d}{dt} \mathbf{r}_1(2t) = \mathbf{r}_1'(2t) \cdot 2 \). Understanding the chain rule is crucial for handling transformations and parameterizations in vector calculus.

Sum Rule

The sum rule in calculus states that the derivative of a sum of functions is the sum of their derivatives. This rule applies directly to vector functions too. Given two differentiable vector functions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), we have:\[\frac{d}{dt}[\mathbf{u}(t) + \mathbf{v}(t)] = \frac{d}{dt}\mathbf{u}(t) + \frac{d}{dt}\mathbf{v}(t) \]This equation shows that you take the derivative of each part separately and then add them together. In the given problem, the composite vector function includes \( \mathbf{r}_1(2t) \) and \( \mathbf{r}_2\left(\frac{1}{t}\right) \), so we apply the sum rule to handle these separately.

Vector Differentiation

Vector differentiation involves applying rules of differentiation to vector-valued functions. Each component of the vector is differentiated concerning its parameter, usually time \( t \).Key points to remember include:

• Each vector component requires separate differentiation.
• Rules applicable to scalar functions hold true, such as the chain rule and sum rule.
• Vector differentiation is helpful in physics and engineering, especially when dealing with position, velocity, and acceleration vectors.

In our problem, we differentiated the vector sum \( \mathbf{r}_1(2t) + \mathbf{r}_2\left(\frac{1}{t}\right) \) by handling each term with chain and differentiation rules, showcasing the intricacies of vector differentiation.