Problem 3 - Thomas Calculus

Question

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the plane curves. \(\mathbf{r}(t)=(2 t+3) \mathbf{i}+\left(5-t^{2}\right) \mathbf{j}\)

Step-by-Step Solution

Answer

\(\mathbf{T}(t) = \frac{1}{\sqrt{1+t^2}}(\mathbf{i} - t\mathbf{j})\), \(\mathbf{N}(t) = -t\mathbf{i} - \mathbf{j}\), and \(\kappa(t) = \frac{1}{2(1+t^2)}\).

Step 1: Compute the Derivative of the Position Vector

To find the tangent vector \(\mathbf{T}\), we first need to compute the derivative of the position vector \(\mathbf{r}(t) = (2t + 3)\mathbf{i} + (5 - t^2)\mathbf{j}\). The derivative is \(\mathbf{r}'(t) = \frac{d}{dt}[(2t+3)\mathbf{i} + (5-t^2)\mathbf{j}] = 2\mathbf{i} - 2t\mathbf{j}\).

Step 2: Determine the Unit Tangent Vector

The unit tangent vector \(\mathbf{T}(t)\) is found by normalizing \(\mathbf{r}'(t)\). First, calculate the magnitude: \(||\mathbf{r}'(t)|| = \sqrt{(2)^2 + (-2t)^2} = \sqrt{4 + 4t^2} = 2\sqrt{1 + t^2}\). Thus, the unit tangent vector \(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} = \frac{2\mathbf{i} - 2t\mathbf{j}}{2\sqrt{1+t^2}} = \frac{1}{\sqrt{1+t^2}}(\mathbf{i} - t\mathbf{j})\).

Step 3: Calculate the Derivative of the Unit Tangent Vector

Now we find the derivative of the unit tangent vector \(\mathbf{T}(t)\): \(\mathbf{T}'(t) = \frac{d}{dt}\left[\frac{1}{\sqrt{1+t^2}}(\mathbf{i} - t\mathbf{j})\right]\). Using the product and chain rules, \(\mathbf{T}'(t) = \left(-\frac{t}{(1+t^2)^{3/2}}\right)\mathbf{i} - \left(\frac{1}{(1+t^2)^{3/2}}\right)\mathbf{j}\).

Step 4: Compute the Principal Unit Normal Vector

The principal unit normal vector \(\mathbf{N}(t)\) is \(\frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||}\). First, find \(||\mathbf{T}'(t)|| = \sqrt{\left(-\frac{t}{(1+t^2)^{3/2}}\right)^2 + \left(-\frac{1}{(1+t^2)^{3/2}}\right)^2} = \frac{\sqrt{1+t^2}}{1+t^2} = \frac{1}{\sqrt{1+t^2}}\). Thus, \(\mathbf{N}(t) = \frac{(\mathbf{T}'(t))}{\frac{1}{\sqrt{1+t^2}}} = -t\mathbf{i} - \mathbf{j}\).

Step 5: Calculate the Curvature

Curvature \(\kappa(t)\) is given by \(\kappa(t) = \frac{||\mathbf{T}'(t)||}{||\mathbf{r}'(t)||}\). Thus, \(\kappa(t) = \frac{\frac{1}{\sqrt{1+t^2}}}{2\sqrt{1+t^2}} = \frac{1}{2(1+t^2)}\).

Key Concepts

Understanding the Tangent Vector

In the study of plane curves, a tangent vector plays a crucial role. It's essentially a vector that touches a curve at just one point and indicates the direction in which the curve is heading. To find the tangent vector for a curve described by the position vector \( \mathbf{r}(t) = (2t + 3)\mathbf{i} + (5 - t^2)\mathbf{j} \), we start by taking the derivative with respect to \( t \). This gives us the velocity of a particle moving along the curve. Applying the derivative yields a new vector, \( \mathbf{r}'(t) = 2\mathbf{i} - 2t\mathbf{j} \), which is the tangent vector. This shows how fast and in which direction the position changes as \( t \) changes. Remember, the tangent vector is not necessarily of unit length, so further steps are needed if we're interested in its unit version.

Exploring the Unit Normal Vector

While the tangent vector \( \mathbf{T}(t) \) shows the direction along the curve, the unit normal vector \( \mathbf{N}(t) \) gives insight into the direction of the curve's bending away from the tangent. To find the unit normal vector, we take the derivative of the unit tangent vector. In this case, the derivative \( \mathbf{T}'(t) \) results in \( -\frac{t}{(1+t^2)^{3/2}}\mathbf{i} - \frac{1}{(1+t^2)^{3/2}}\mathbf{j} \). We then normalize this derivative by dividing by its magnitude: \( \frac{1}{\sqrt{1+t^2}} \). Thus, the unit normal vector becomes \( \mathbf{N}(t) = -t\mathbf{i} - \mathbf{j} \). This unit normal vector is perpendicular to the tangent vector, always pointing outward relative to the curve.

Curvature and its Significance

Curvature, denoted as \( \kappa(t) \), quantifies how sharply a curve bends at any point. It offers a measure of the rate at which the tangent vector changes direction as one moves along the curve. The calculation of curvature involves both the derivative of the tangent vector and the original derivative of the position vector \( \mathbf{r}'(t) \). Here, it is given as \( \kappa(t) = \frac{||\mathbf{T}'(t)||}{||\mathbf{r}'(t)||} \). Using our specific functions, we find \( \kappa(t) = \frac{1}{2(1+t^2)} \). Curvature is crucial in disciplines such as physics and engineering, where it helps to understand phenomena like motion and dynamics on paths.

The Essence of Derivatives

Derivatives are foundational in calculus and are central to understanding rates of change. When applied to a position vector, like \( \mathbf{r}(t) \), the derivative \( \mathbf{r}'(t) \) uncovers the velocity vector, indicating how position changes over time. This provides a pathway to finding tangent and normal vectors. In our context, \( \mathbf{r}'(t) = 2\mathbf{i} - 2t\mathbf{j} \) describes how the curve moves through the plane, guiding us to the \( \mathbf{T}(t) \) once normalized. It's through derivatives that we traverse from a mere description of a path to an analytical comprehension of its behavior and properties.