Question

Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathrm{T}},\) and \(a_{\mathrm{N}}\) at the given time \(t\) for the plane curve \(\mathbf{r}(t)\) $$ \mathbf{r}(t)=t^{2} \mathbf{i}+2 t \mathbf{j}, \quad t=1 $$

Short answer

\(\mathbf{T}(1) = \frac{\sqrt{2}}{2}\mathbf{i} + \frac{\sqrt{2}}{2}\mathbf{j}\), \(a_{\mathrm{T}} = \sqrt{2}\), \(a_{\mathrm{N}} = 1\), \(\mathbf{N}(1) = \mathbf{j}\).

Step-by-step solution

Find the derivative of \(\mathbf{r}(t)\)

Differentiate the given \(\mathbf{r}(t)\) \(t^{2} \mathbf{i}+2 t \mathbf{j}\) with respect to \(t\) where \(t = 1\). This results in \(\mathbf{r'}(t) = 2t\mathbf{i} + 2\mathbf{j}\). Substituting \(t = 1\) gives \(\mathbf{r'}(1) = 2\mathbf{i} + 2\mathbf{j}\). The unit tangent vector \(\mathbf{T}(t)\) is given by the formula \(\mathbf{T}(t) = \frac{\mathbf{r'}(t)}{||\mathbf{r'}(t)||}\). Calculate the magnitude of \(\mathbf{r'}(1)\). This is \(\sqrt{(2)^{2} + (2)^{2}} = 2\sqrt{2}\). Hence, \(\mathbf{T}(1) = \frac{2\mathbf{i} + 2\mathbf{j}}{2\sqrt{2}} = \frac{\sqrt{2}}{2}\mathbf{i} + \frac{\sqrt{2}}{2}\mathbf{j}\).

Find the second derivative of \(\mathbf{r}(t)\) and calculate \(a_{\mathrm{N}}\)

Next, find the acceleration vector by deriving \(\mathbf{r'}(t) = 2t\mathbf{i} + 2\mathbf{j}\) with respect to \(t\) to get \(\mathbf{r''}(t) = 2\mathbf{i}\). Substituting \(t = 1\) gives \(\mathbf{r''}(1) = 2\mathbf{i}\). \(a_{\mathrm{N}}\) is given by the magnitude of the component of the acceleration perpendicular to \(\mathbf{T}(t)\), which is \(||a\mathbf{r''}(t) - a_{\mathrm{T}}\mathbf{T}(t)||\). Here \(a_{\mathrm{T}}\) is the component of the acceleration in the direction of \(\mathbf{T}(t)\), given by \(\mathbf{r''}(t) \cdot \mathbf{T}(t)\). By substituting the known values, \(a_{\mathrm{T}} = 2\mathbf{i} \cdot (\frac{\sqrt{2}}{2}\mathbf{i} + \frac{\sqrt{2}}{2}\mathbf{j}) = \sqrt{2}\), then \(a_{\mathrm{N}} = ||2\mathbf{i} - \sqrt{2}(\frac{\sqrt{2}}{2}\mathbf{i} + \frac{\sqrt{2}}{2}\mathbf{j})|| = 1\).

Find \(\mathbf{N}(t)\)

\(\mathbf{N}(t)\) is given by \(\frac{\mathbf{r''}(t) - a_{\mathrm{T}}\mathbf{T}(t)}{a_{\mathrm{N}}}\). Substituting the calculated values gives \(\mathbf{N}(1) = \frac{2\mathbf{i} - \sqrt{2}(\frac{\sqrt{2}}{2}\mathbf{i} + \frac{\sqrt{2}}{2}\mathbf{j})}{1} = \mathbf{j}\).

Key concepts

Unit Tangent Vector

When analyzing the motion of an object along a plane curve, it's essential to understand its direction at any given time. The unit tangent vector, denoted as \( \mathbf{T}(t) \), provides us with this information by indicating the direction of the curve at a specific point and being of unit length.

To obtain \( \mathbf{T}(t) \), we first determine the velocity vector \( \mathbf{r'}(t) \) by differentiating the position vector \( \mathbf{r}(t) \) with respect to time. We then normalize it to have a magnitude of one. This process involves dividing the velocity vector by its magnitude, denoted by \( ||\mathbf{r'}(t)|| \) for the given time. The normalization step ensures that \( \mathbf{T}(t) \) points in the same direction as the velocity but with a fixed length of 1.

Why is \( \mathbf{T}(t) \) important?

• It helps us understand the direction in which an object is moving at any instant.
• It is crucial for studying the properties of curves, like curvature and torsion.
• It provides a foundation for deriving other important vectors, such as the normal and binormal vectors, which describe the geometry of the space curve in three dimensions.

Acceleration Vector Calculus

To fully capture the dynamics of an object in motion along a path described by parametric equations, the acceleration vector calculus comes into play. Acceleration is the rate of change of velocity with respect to time and is denoted as \( \mathbf{r''}(t) \), obtained by differentiating the velocity vector \( \mathbf{r'}(t) \) with respect to time.

Once we have the acceleration vector, we can decompose it into components parallel and perpendicular to the direction of the velocity vector. These components are represented as \( a_{\mathrm{T}} \) and \( a_{\mathrm{N}} \) for tangential and normal acceleration, respectively.

Determining Tangential and Normal Acceleration

To find \( a_{\mathrm{T}} \) and \( a_{\mathrm{N}} \) from the acceleration vector, we proceed as follows:

• \(a_{\mathrm{T}}\) is computed as the dot product of the acceleration vector with the unit tangent vector. It gives the rate at which the speed of the particle changes.
• \(a_{\mathrm{N}}\) is found by subtracting the component of acceleration in the direction of \(\mathbf{T}(t)\) from the total acceleration and taking the magnitude of this difference. It's associated with the change in direction of the particle's velocity, contributing to its curvature.

Understanding both \( a_{\mathrm{T}} \) and \( a_{\mathrm{N}} \) is crucial because they collectively convey the complete acceleration experienced by the object, offering insights into forces responsible for changing its speed and direction of motion.

Parametric Equations Differentiation

Parametric equations differentiation is a fundamental technique used to find the rate at which one variable changes with respect to another within the framework of parametric curves. In the context of plane curves, these parametric equations often represent the x and y coordinates of a point in terms of a third variable, typically time (t).

By differentiating the parametric equations with respect to time, we acquire the velocity, acceleration, and depending on the complexity of the motion, even higher-order derivatives. This step-by-step differentiation process allows us to not only predict the position of the object at any given time but also its speed, direction, and how these quantities change over time.

For students trying to grasp this concept:

• Begin by understanding how differentiation applies to single-variable functions and then extend this understanding to parametric equations.
• Express the velocity and acceleration vectors in terms of their components and differentiate these components with respect to time.
• Apply the chain rule, when necessary, to take into account the composite nature of some parametric functions.

Mastering parametric equations differentiation is essential for problem-solving in advanced mathematics, physics, engineering, and computer graphics. It's the gateway to understanding motion and change for a wide array of applications where motion is described parametrically.