Problem 58 - Calculus 3

Question

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=t \mathbf{i}+3 t \mathbf{j}+t^{2} \mathbf{k}$$

Step-by-Step Solution

Answer

The unit tangent vector is \(\frac{\mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}}{\sqrt{10 + 4t^2}}\).

Step 1: Differentiate the Position Vector

To find the tangent vector, first differentiate the given position vector function with respect to the parameter \(t\). The given position vector is \(\mathbf{r}(t) = t \mathbf{i} + 3t \mathbf{j} + t^2 \mathbf{k}\). Differentiating each component, we get:\[ \frac{d\mathbf{r}}{dt} = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}. \] This vector \(\frac{d\mathbf{r}}{dt}\) is the tangent vector at any point \(t\).

Step 2: Calculate the Magnitude of the Tangent Vector

Now, calculate the magnitude of the tangent vector \(\frac{d\mathbf{r}}{dt} = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}\). The magnitude is given by:\[ \| \frac{d\mathbf{r}}{dt} \| = \sqrt{1^2 + 3^2 + (2t)^2} = \sqrt{1 + 9 + 4t^2} = \sqrt{10 + 4t^2}. \]

Step 3: Normalize the Tangent Vector

Normalize the tangent vector by dividing it by its magnitude. The unit tangent vector \(\mathbf{T}(t)\) is given by:\[ \mathbf{T}(t) = \frac{\frac{d\mathbf{r}}{dt}}{\| \frac{d\mathbf{r}}{dt} \|} = \frac{\mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}}{\sqrt{10 + 4t^2}}. \] This is the unit tangent vector for the parameterized curve \(\mathbf{r}(t)\).

Key Concepts

Parameterized Curves

When we talk about parameterized curves, we are essentially looking at a way to express a curve using a parameter, often denoted as \( t \). Each value of \( t \) corresponds to a specific point on the curve. Think of \( t \) as a slider that moves along the curve, potentially changing both direction and shape of the curve as it changes.

The position vector \( \mathbf{r}(t) = t \mathbf{i} + 3t \mathbf{j} + t^2 \mathbf{k} \) describes such a parameterized curve in 3-dimensional space. Here, the vectors \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) denote the unit vectors in the x, y, and z directions, respectively.

The components \( t \), \( 3t \), and \( t^2 \) are functions of the parameter \( t \), defining a unique position for any given \( t \). Hence, parameterized curves can provide a dynamic way to describe the motion or path of an object over time.

Differentiation

Differentiation in calculus is a powerful tool that lets us determine the rate at which a function is changing. When applied to parameterized curves, differentiation gives us the tangent vector.

For our curve, the position vector \( \mathbf{r}(t) = t \mathbf{i} + 3t \mathbf{j} + t^2 \mathbf{k} \) was differentiated to find the tangent vector \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k} \).

By differentiating each component of the position vector, we reveal how each part of the curve's position changes with \( t \).

• \( \mathbf{i} \) changes at a constant rate, contributing 1 to the tangent vector.
• \( \mathbf{j} \) also changes at a constant rate, contributing 3.
• The \( \mathbf{k} \) component increases in complexity, contributing \( 2t \), influenced directly by the increase in \( t \).

The result, \( \frac{d\mathbf{r}}{dt} \), provides a glimpse into the direction and speed of the curve at any point.

Magnitude of a Vector

The magnitude of a vector measures its length or size in the space it inhabits. For vectors, this involves using the Pythagorean theorem.

The magnitude of the tangent vector \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k} \) is calculated as follows:\[\| \frac{d\mathbf{r}}{dt} \| = \sqrt{1^2 + 3^2 + (2t)^2} = \sqrt{10 + 4t^2}.\]

• Each component is squared.
• All the squares are summed up.
• The square root of this sum gives the magnitude.

The magnitude represents how long the tangent vector is at any point \( t \). This step is crucial before normalizing the vector to ensure it has a length of 1.

Vector Normalization

Vector normalization is the process of scaling the vector so that its magnitude becomes 1. This is important to convert any vector into a unit vector, which only describes direction.

To normalize a vector, divide the vector by its magnitude. For our tangent vector \( \frac{d\mathbf{r}}{dt} = \mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k} \), normalization is:\[\mathbf{T}(t) = \frac{\frac{d\mathbf{r}}{dt}}{\| \frac{d\mathbf{r}}{dt} \|} = \frac{\mathbf{i} + 3 \mathbf{j} + 2t \mathbf{k}}{\sqrt{10 + 4t^2}}.\]

• The direction remains unchanged.
• The vector's length scales down to 1.

Normalizing allows us to focus solely on the direction of the tangent, providing valuable insights for understanding the nature of the curve at different points.