Question

If \(\mathbf{r}(t)=\sin 2 t \mathbf{i}+\cosh t \mathbf{j}\) and \(h(t)=\ln (3 t-2)\), find \(D_{t}[h(t) \mathbf{r}(t)] .\)

Short answer

The derivative is: \[ \left(\frac{3\sin(2t)}{3t-2} + 2\ln(3t-2) \cos(2t)\right) \mathbf{i} + \left(\frac{3\cosh(t)}{3t-2} + \ln(3t-2) \sinh(t)\right) \mathbf{j} \]

Step-by-step solution

Identify the Functions

The problem involves two functions: \( \mathbf{r}(t) = \sin(2t) \mathbf{i} + \cosh(t) \mathbf{j} \) and \( h(t) = \ln(3t-2) \). We need to find the derivative of their product, \( h(t) \mathbf{r}(t) \).

Apply the Product Rule

The derivative of a product of two functions \( f(t) = h(t) \) and \( \mathbf{g}(t) = \mathbf{r}(t) \) is given by \( D_t[f(t) \mathbf{g}(t)] = f'(t) \mathbf{g}(t) + f(t) \mathbf{g}'(t) \). We'll compute \( f'(t) \) and \( \mathbf{g}'(t) \) separately.

Differentiate \( h(t) \)

Find the derivative of \( h(t) = \ln(3t-2) \). Using the chain rule, \( h'(t) = \frac{1}{3t-2} \cdot 3 = \frac{3}{3t-2} \).

Differentiate \( \mathbf{r}(t) \)

Compute the derivatives of each component of \( \mathbf{r}(t) = \sin(2t) \mathbf{i} + \cosh(t) \mathbf{j} \).- The derivative of \( \sin(2t) \) is \( 2\cos(2t) \), so the \( \mathbf{i} \) component of \( \mathbf{r}'(t) \) is \( 2\cos(2t) \mathbf{i} \).- The derivative of \( \cosh(t) \) is \( \sinh(t) \), so the \( \mathbf{j} \) component is \( \sinh(t) \mathbf{j} \).Thus, \( \mathbf{r}'(t) = 2\cos(2t) \mathbf{i} + \sinh(t) \mathbf{j} \).

Substitute into the Product Rule Formula

Using the derivatives calculated, substitute into the product rule formula: \( D_t[h(t) \mathbf{r}(t)] = h'(t) \mathbf{r}(t) + h(t) \mathbf{r}'(t) \).- \( h'(t) \mathbf{r}(t) = \frac{3}{3t-2} [\sin(2t) \mathbf{i} + \cosh(t) \mathbf{j}] \).- \( h(t) \mathbf{r}'(t) = \ln(3t-2) [2\cos(2t) \mathbf{i} + \sinh(t) \mathbf{j}] \).

Simplify the Expression

Combine the results from the previous step:\[D_t[h(t) \mathbf{r}(t)] = \frac{3\sin(2t)}{3t-2} \mathbf{i} + \frac{3\cosh(t)}{3t-2} \mathbf{j} + \ln(3t-2) [2\cos(2t) \mathbf{i} + \sinh(t) \mathbf{j}]\]Distribute \( \ln(3t-2) \) across the terms:\[D_t[h(t) \mathbf{r}(t)] = \left(\frac{3\sin(2t)}{3t-2} + 2\ln(3t-2) \cos(2t)\right) \mathbf{i} + \left(\frac{3\cosh(t)}{3t-2} + \ln(3t-2) \sinh(t)\right) \mathbf{j}\]

Final Answer

The derivative of the product \( h(t) \mathbf{r}(t) \) is:\[D_t[h(t) \mathbf{r}(t)] = \left(\frac{3\sin(2t)}{3t-2} + 2\ln(3t-2) \cos(2t)\right) \mathbf{i} + \left(\frac{3\cosh(t)}{3t-2} + \ln(3t-2) \sinh(t)\right) \mathbf{j}\]

Key concepts

Product Rule in Vector Calculus

In vector calculus, calculating derivatives when functions are multiplied involves the product rule. This rule states that the derivative of a product of two functions is not just the product of their derivatives. Instead, it requires us to take the derivative of one function while keeping the other constant, then vice versa, and sum them up.

The formula for the product rule is as follows:

• For functions \( f(t) \) and \( \mathbf{g}(t) \), the derivative \( D_t[f(t) \mathbf{g}(t)] \) is given by:
• \( f'(t) \mathbf{g}(t) + f(t) \mathbf{g}'(t) \)

In our example, we applied the product rule to the functions \( h(t) \) and \( \mathbf{r}(t) \), where each function's derivative had to be calculated separately, considering it's a vector function.

Chain Rule Simplified

The chain rule is crucial when dealing with composite functions in calculus. It allows us to differentiate a function based on its inner and outer functions. In essence, you differentiate the outer function, keep the inner function unchanged, and then multiply by the derivative of the inner function.

In this exercise, the function \( h(t) = \ln(3t-2) \) requires using the chain rule. The outer function is \( \ln(u) \), and the inner function is \( u = 3t-2 \). The derivative using the chain rule is:

• Differentiate the outer function: \( \frac{1}{u} \); keep \( u = 3t-2 \).
• Multiply by the derivative of the inner function, \( 3 \), giving \( \frac{3}{3t-2} \).

This approach simplifies handling functions nested within others.

Differentiating Vectors

Differentiation in vectors doesn't diverge too far from ordinary functions, but it involves working with each component separately. Vectors are often expressed as a combination of unit vectors \( \mathbf{i} \) and \( \mathbf{j} \). Each component of the vector needs individual differentiation.

For the vector \( \mathbf{r}(t) = \sin(2t) \mathbf{i} + \cosh(t) \mathbf{j} \), we compute:

• For \( \sin(2t) \mathbf{i} \), the derivative is \( 2 \cos(2t) \mathbf{i} \)
• For \( \cosh(t) \mathbf{j} \), the derivative is \( \sinh(t) \mathbf{j} \)

Thus, deriving vector functions involves applying standard differentiation rules separately to each unit vector component.

Understanding Hyperbolic Functions

Hyperbolic functions, such as \( \cosh(t) \) and \( \sinh(t) \), are analogues of trigonometric functions but are linked to hyperbolas instead of circles.

The basic hyperbolic functions are:

• \( \cosh(t) = \frac{e^t + e^{-t}}{2} \)
• \( \sinh(t) = \frac{e^t - e^{-t}}{2} \)

These functions have properties similar to their trigonometric counterparts but with some variations, such as:

• Derivatives: The derivative of \( \cosh(t) \) is \( \sinh(t) \), and vice versa, \( \frac{d}{dt} \sinh(t) = \cosh(t) \).
• Identities: \( \cosh^2(t) - \sinh^2(t) = 1 \) mimics the Pythagorean identity.

Comprehending hyperbolic functions aids in solving problems where these functions arise, making them integral in many calculus applications.