Question

\begin{array}{l} \text { 17. If } \mathbf{r}(t)=\sqrt{t-1} \mathbf{i}+\ln \left(2 t^{2}\right) \mathbf{j} \text { and } h(t)=e^{-3 t}, \text { find }\\\ D_{t}[h(t) \mathbf{r}(t)] . \end{array}

Short answer

Apply product rule and simplify: \( D_t[h(t)\mathbf{r}(t)] = [-3e^{-3t}\sqrt{t-1} + \frac{e^{-3t}}{2\sqrt{t-1}}] \mathbf{i} + [-3e^{-3t}\ln(2t^2) + \frac{2e^{-3t}}{t}] \mathbf{j} \).

Step-by-step solution

Understand the Objective

We are tasked with finding the derivative \( D_t[h(t) \mathbf{r}(t)] \) with respect to \( t \). The function \( \mathbf{r}(t) \) is a vector function given by \( \mathbf{r}(t) = \sqrt{t-1} \mathbf{i} + \ln(2t^2) \mathbf{j} \), and \( h(t) = e^{-3t} \) is a scalar function.

Apply the Product Rule for Derivatives

Since we need the derivative of the product of a scalar \( h(t) \) and a vector \( \mathbf{r}(t) \), we will apply the product rule for derivatives of vector-valued functions: \( D_t[h(t) \mathbf{r}(t)] = h'(t) \mathbf{r}(t) + h(t) \mathbf{r}'(t) \).

Differentiate the Scalar Function

First, we differentiate the scalar function \( h(t) = e^{-3t} \). The derivative is given by: \( h'(t) = -3e^{-3t} \).

Differentiate the Vector Function

Next, we differentiate the vector function \( \mathbf{r}(t) = \sqrt{t-1} \mathbf{i} + \ln(2t^2) \mathbf{j} \):- Derivative of \( \sqrt{t-1} \) is \( \frac{1}{2\sqrt{t-1}} \) with respect to \( t \). - Derivative of \( \ln(2t^2) \) is \( \frac{4}{2t^2} = \frac{2}{t} \) with respect to \( t \). Thus, \( \mathbf{r}'(t) = \frac{1}{2\sqrt{t-1}} \mathbf{i} + \frac{2}{t} \mathbf{j} \).

Substitute into the Product Rule Formula

Now substitute \( h'(t) \), \( h(t) \), \( \mathbf{r}(t) \), and \( \mathbf{r}'(t) \) back into the formula: \[ D_t[h(t) \mathbf{r}(t)] = [-3e^{-3t}] (\sqrt{t-1} \mathbf{i} + \ln(2t^2) \mathbf{j}) + e^{-3t} \left( \frac{1}{2\sqrt{t-1}} \mathbf{i} + \frac{2}{t} \mathbf{j} \right) \]

Simplify the Expression

Multiply out the terms:\[-3e^{-3t}\sqrt{t-1} \mathbf{i} - 3e^{-3t}\ln(2t^2) \mathbf{j} + \frac{e^{-3t}}{2\sqrt{t-1}} \mathbf{i} + \frac{2e^{-3t}}{t} \mathbf{j}\].Combine like terms if possible to simplify further.

Key concepts

Product rule

The product rule is a key concept in vector calculus essential when dealing with the derivative of a product involving functions. When considering a product of two functions, such as a scalar and a vector function, the rule guides us on how to differentiate effectively. This rule is not only applicable to scalar functions but extends seamlessly to vector-valued functions.

Importantly, the product rule for differentiating the product of a scalar function, say \( h(t) \), and a vector-valued function, \( \mathbf{r}(t) \), is expressed as follows:

• \[ D_t[h(t) \mathbf{r}(t)] = h'(t) \mathbf{r}(t) + h(t) \mathbf{r}'(t) \]

This formula is helpful because it allows you to break down the problem into parts: the derivative of the scalar term, multiplied by the original vector, and the original scalar term, multiplied by the derivative of the vector.

By understanding this rule, you can solve problems involving complex expressions of scalar and vector combinations, enhancing your ability to dissect and comprehend interconnected mathematical functions.

Derivative of vector-valued functions

In calculus, the derivative of a vector-valued function is analogous to differentiating scalar functions but tailored for vectors. A vector-valued function, such as \( \mathbf{r}(t) = \sqrt{t-1} \mathbf{i} + \ln(2t^2) \mathbf{j} \), consists of multiple components, each potentially requiring separate differentiation with respect to \( t \).To differentiate a vector-valued function:

• Find the derivative of each component individually.
• For the component \( \sqrt{t-1} \), the derivative is \( \frac{1}{2\sqrt{t-1}} \).
• For the component \( \ln(2t^2) \), apply the chain rule to differentiate to \( \frac{2}{t} \).

The result is a new vector whose components are the derivatives of the original components:

• \( \mathbf{r}'(t) = \frac{1}{2\sqrt{t-1}} \mathbf{i} + \frac{2}{t} \mathbf{j} \)

This process allows us to manage how the vector function changes within its domain straightforwardly.

Scalar multiplication in vectors

Scalar multiplication in vectors involves scaling a vector by a scalar quantity, which, in calculus, can vary with respect to time or another variable. In our problem, \( h(t) = e^{-3t} \) provides this scalar effect on the vector \( \mathbf{r}(t) \).

To understand scalar multiplication:

• This operation stretches or shrinks the vector depending on the scalar's magnitude.
• When multiplying \( e^{-3t} \) with \( \mathbf{r}(t) \), each component of \( \mathbf{r}(t) \) is multiplied by \( e^{-3t} \).
• This combination modifies the direction and magnitude of the vector \( \mathbf{r}(t) \), resulting in a new scaled vector function.

Scalar multiplication is a foundational concept leading to more intricate operations, such as applying the product rule and differentiating as demonstrated.