Problem 59 - Calculus 3

Question

Let $$\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}-t^{4} \mathbf{k}$$ and $$s(t)=\sin (t) \mathbf{i}+e^{t} \mathbf{j}+\cos (t) \mathbf{k}$$ Here is the graph of the function: Find the following. $$\frac{d}{d t}\left[r\left(t^{2}\right)\right]$$

Step-by-Step Solution

Answer

The derivative is \( 2t \mathbf{i} + 4t^3 \mathbf{j} - 8t^7 \mathbf{k} \).

Step 1: Understand the Problem

In this problem, we're given a vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \). We need to find the derivative of \( \mathbf{r}(t^2) \) with respect to \( t \). This will involve applying the chain rule for differentiation.

Step 2: Parameter Substitution

Substitute \( t^2 \) for \( t \) in the function \( \mathbf{r}(t) \). This means replacing every instance of \( t \) in \( \mathbf{r}(t) \) with \( t^2 \). So, \( \mathbf{r}(t^2) = (t^2) \mathbf{i} + (t^2)^2 \mathbf{j} - (t^2)^4 \mathbf{k} = t^2 \mathbf{i} + t^4 \mathbf{j} - t^8 \mathbf{k} \).

Step 3: Differentiate the Substituted Function

Now we differentiate each component of \( \mathbf{r}(t^2) = t^2 \mathbf{i} + t^4 \mathbf{j} - t^8 \mathbf{k} \) with respect to \( t \), using the power rule. - The derivative of \( t^2 \mathbf{i} \) is \( 2t \mathbf{i} \).- The derivative of \( t^4 \mathbf{j} \) is \( 4t^3 \mathbf{j} \).- The derivative of \( -t^8 \mathbf{k} \) is \( -8t^7 \mathbf{k} \).

Step 4: Write the Final Result

Combine the derivatives to get the final result: \[ \frac{d}{dt}[\mathbf{r}(t^2)] = 2t \mathbf{i} + 4t^3 \mathbf{j} - 8t^7 \mathbf{k} \].

Key Concepts

Vector Differentiation

Vector differentiation is the process of finding the derivative of a vector function with respect to a parameter, usually time or another variable. In vector calculus, a vector function assigns a vector to each point in its domain. In our case, the vector function \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \) denotes how a point moves in three-dimensional space as parameter \( t \) changes.

When differentiating vector functions, we treat each component separately. It's akin to differentiating scalar functions, but applied to the individual components of the vector function. Here are the general steps:

1. Identify each vector component, such as \( t \mathbf{i} \), \( t^2 \mathbf{j} \), and \( -t^4 \mathbf{k} \).
2. Differentiate each component with respect to \( t \).
3. Combine the derivatives to form a new vector.

This technique is essential in physics and engineering, where you often need to compute velocity and acceleration vectors by differentiating position vectors.

Parametric Functions

Parametric functions are functions that express coordinates as functions of one or more parameters. In our context, the functions \( \mathbf{r}(t) \) and \( s(t) \) express the coordinates \( (x, y, z) \) in terms of the parameter \( t \).

This allows for a more flexible representation of curves and surfaces, compared to the usual Cartesian coordinates. Parametric functions can describe complex shapes and motions by varying the parameter.

In this exercise, substituting \( t^2 \) for \( t \) in \( \mathbf{r}(t) \) changes the parameters of the original function, modifying the output path or shape. Once you have the new parametric function, you can differentiate it using the same calculus rules, but applied to the new parameter output.

Power Rule in Differentiation

The power rule is a basic differentiation rule used extensively when dealing with polynomial functions. It states that if \( f(t) = t^n \), then its derivative is \( f'(t) = nt^{n-1} \).

In our solution, this rule is applied to each component of the modified vector function \( \mathbf{r}(t^2) = t^2 \mathbf{i} + t^4 \mathbf{j} - t^8 \mathbf{k} \). For example, for \( t^2 \mathbf{i} \), using the power rule, the derivative is \( 2t \mathbf{i} \).

Applying the power rule simplifies the process of finding derivatives, especially when dealing with each component of a vector separately. This makes it very useful in calculus, as it turns the work of differentiation into a straightforward task with a simple formula.