Problem 19 - Calculus III

Question

Find the unit tangent vector for the following parameterized curves.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:

Step-by-Step Solution

Answer

The unit tangent vector for \( \mathbf{r}(t) \) is \( \mathbf{T}_r(t) = \frac{1}{\sqrt{1 + 4t^2 + 16t^6}} (\mathbf{i} + 2t \mathbf{j} - 4t^3 \mathbf{k}) \), and for \( s(t) \) is \( \mathbf{T}_s(t) = \frac{1}{\sqrt{1 + e^{2t}}} (\cos(t) \mathbf{i} + e^t \mathbf{j} - \sin(t) \mathbf{k}) \).

Step 1: Differentiate the Curves

First, we need to differentiate both vector-valued functions, \( \mathbf{r}(t) \) and \( s(t) \), with respect to \( t \) to find their derivatives. For \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \), the derivative \( \frac{d\mathbf{r}}{dt} \) is:\[ \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t \mathbf{j} - 4t^3 \mathbf{k} \]For \( s(t) = \sin(t) \mathbf{i} + e^t \mathbf{j} + \cos(t) \mathbf{k} \), the derivative \( \frac{ds}{dt} \) is:\[ \frac{ds}{dt} = \cos(t) \mathbf{i} + e^t \mathbf{j} - \sin(t) \mathbf{k} \]

Step 2: Find the Magnitude of the Derivative

To obtain the unit tangent vector, we need to find the magnitude of each derivative. For \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t \mathbf{j} - 4t^3 \mathbf{k} \), the magnitude \( \Vert \frac{d\mathbf{r}}{dt} \Vert \) is:\[ \Vert \frac{d\mathbf{r}}{dt} \Vert = \sqrt{1^2 + (2t)^2 + (-4t^3)^2} = \sqrt{1 + 4t^2 + 16t^6} \]For \( \frac{ds}{dt} = \cos(t) \mathbf{i} + e^t \mathbf{j} - \sin(t) \mathbf{k} \), the magnitude \( \Vert \frac{ds}{dt} \Vert \) is:\[ \Vert \frac{ds}{dt} \Vert = \sqrt{\cos^2(t) + (e^t)^2 + (-\sin(t))^2} = \sqrt{\cos^2(t) + e^{2t} + \sin^2(t)} \]Since \( \cos^2(t) + \sin^2(t) = 1 \), this becomes \( \sqrt{1 + e^{2t}} \).

Step 3: Divide by the Magnitude to Get Unit Tangent Vectors

Now, divide each component of the derivatives by their respective magnitudes to obtain the unit tangent vectors. For \( \mathbf{r}(t) \):\[ \mathbf{T}_r(t) = \frac{1}{\sqrt{1 + 4t^2 + 16t^6}} (\mathbf{i} + 2t \mathbf{j} - 4t^3 \mathbf{k}) \]For \( s(t) \):\[ \mathbf{T}_s(t) = \frac{1}{\sqrt{1 + e^{2t}}} (\cos(t) \mathbf{i} + e^t \mathbf{j} - \sin(t) \mathbf{k}) \]

Key Concepts

Vector-Valued Functions

Vector-valued functions are interesting because they are functions that have vectors as outputs. Instead of giving a single numeric value for every input, they provide a vector with multiple components. These vectors are usually described in terms of \ \( \mathbf{i} \) (the x-component), \ \( \mathbf{j} \) (the y-component), and \ \( \mathbf{k} \) (the z-component). This means they can represent points in space that change over time or another variable.

• For example, the vector-valued function \ \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \) reveals how a point moves in space as \( t \) changes.
• By using vector-valued functions, we can model many physical situations like the trajectory of a moving object.

Understanding vector-valued functions is a foundation for discussing more complex topics like differentiation and integration in vector calculus.

Differentiation

Differentiation in the context of vector-valued functions involves finding how each component of the vector changes as the input changes. Imagine you're tracking a moving object: differentiation gives us the speed and direction of its movement. This is crucial for understanding the behavior of curves in space.

• For the function \ \( \mathbf{r}(t) \), the derivative \ \( \frac{d \mathbf{r}}{dt} = \mathbf{i} + 2t \mathbf{j} - 4t^3 \mathbf{k} \) shows how each unit of \( t \) affects the components \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \).
• This lets us not only track the position but also the velocity at which each component is changing over time.

Differentiation provides the tools to predict and understand curves and motions represented by vector-valued functions.

Magnitude of Derivative

The magnitude of a derivative describes the speed of motion along a curve. While the derivative itself shows direction, the magnitude singles out the rate of change irrespective of direction. The formula for magnitude of a vector \( \mathbf{v} = (x, y, z) \) is given by \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \). This is very useful in determining the length or speed of a vector.

• In the example, for \ \( \frac{d \mathbf{r}}{dt} \), the magnitude is \( \sqrt{1 + 4t^2 + 16t^6} \).
• Calculating magnitude is essential in stepping towards finding a unit tangent vector.

Ultimately, the concept of magnitude helps us connect geometric notions of length with analytical processes in calculus.

Parametric Curves

Parametric curves are curves that are expressed in terms of a parameter, often denoted as \( t \). Instead of expressing a curve as \( y \) being a function of \( x \) directly, both \( x \) and \( y \) (and potentially \( z \) when in 3D) are expressed as functions of \( t \). This method of representation is powerful as it allows for more complex and intricate curves.

• For instance, \ \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \) traces a specific path through space as \( t \) changes.
• Unlike standard Cartesian equations, parametric curves make it possible to describe curves that loop back on themselves or have multiple orientations.

Parametric curves give us flexibility and power in graphical representation, making them essential in many scientific and engineering fields.