Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

find the unit tangent vector \(\mathbf{T}(t)\) and the curvature \(\kappa(t)\) at the point where \(t=t_{1} .\) For calculating \(\kappa\), we suggest using Theorem \(A\), as in Example \(5 .\) $$ x(t)=1-t^{2}, y(t)=1-t^{3} ; t_{1}=1 $$

Step-by-Step Solution

Answer

The unit tangent vector \(\mathbf{T}(1)\) is \(\langle -\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \rangle\) and the curvature \(\kappa(1)\) is \(\frac{6}{13\sqrt{13}}\).

Step 1: Differentiate the position vector

First, we need to find the derivative of the position vector with respect to time to get the velocity vector. The position vector is \( \mathbf{r}(t) = \langle x(t), y(t) \rangle = \langle 1 - t^2, 1 - t^3 \rangle \). Differentiate each component:\[ \mathbf{r}'(t) = \langle \frac{d}{dt}(1-t^2), \frac{d}{dt}(1-t^3) \rangle = \langle -2t, -3t^2 \rangle \].

Step 2: Evaluate the velocity vector at \(t_1\)

Calculate the velocity vector at \(t = t_1 = 1\).\[ \mathbf{r}'(1) = \langle -2(1), -3(1)^2 \rangle = \langle -2, -3 \rangle \].

Step 3: Compute the magnitude of the velocity vector

Find the magnitude of the velocity vector \( \mathbf{r}'(1) = \langle -2, -3 \rangle \):\[ \| \mathbf{r}'(1) \| = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \].

Step 4: Find the unit tangent vector \(\mathbf{T}(t)\)

The unit tangent vector is the normalized velocity vector. Divide each component of \( \mathbf{r}'(1) \) by its magnitude:\[ \mathbf{T}(1) = \frac{1}{\sqrt{13}} \langle -2, -3 \rangle = \langle -\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \rangle \].

Step 5: Differentiate the unit tangent vector

To find the curvature, we need the derivative of the unit tangent vector \( \mathbf{T}(t) \). Differentiate \( \mathbf{r}''(t) = \langle \frac{d}{dt}(-2t), \frac{d}{dt}(-3t^2) \rangle = \langle -2, -6t \rangle \). Evaluate at \( t = 1 \):\[ \mathbf{r}''(1) = \langle -2, -6 \rangle \].

Step 6: Calculate the curvature \(\kappa(t)\)

Use the formula \( \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \) where \( \times \) denotes the determinant (or cross product in 2D). For 2D, this can be simplified:\[ \kappa(1) = \frac{| -2(-6) - (-3)(-2) |}{(\sqrt{13})^3} = \frac{|12 - 6|}{13\sqrt{13}} = \frac{6}{13\sqrt{13}} \].

Step 7: Simplify the expression for \(\kappa(t)\)

Finally, simplify this expression:\( \kappa(1) = \frac{6}{13\sqrt{13}} \) can be simplified further if desired, for instance through rationalization, but is typically left in this form for curvature calculations.

Key Concepts

Unit Tangent Vector

The unit tangent vector is a crucial concept in calculus, especially when exploring curves in space. It provides the direction of the curve at a specific point, helping us understand its orientation. Given a parametrized curve defined by a position vector \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \), the unit tangent vector \( \mathbf{T}(t) \) is derived from the velocity vector, which is the derivative of the position vector with respect to time:

• First, find the velocity vector \( \mathbf{r}'(t) = \langle \frac{d}{dt}x(t), \frac{d}{dt}y(t) \rangle \)
• Then, compute the magnitude of the velocity vector \( \| \mathbf{r}'(t) \| \)
• The unit tangent vector \( \mathbf{T}(t) \) is then \( \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \)

This normalization process ensures that \( \mathbf{T}(t) \) has a length of one, only indicating direction. For our exercise, the unit tangent vector at \( t = 1 \) is \( \mathbf{T}(1) = \langle -\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}} \rangle \). It tells us how the curve is oriented in space at that instant.

Curvature

Curvature is a measure of how sharply a curve bends at a given point. Higher curvature means a sharper turn. For a curve defined by a position vector \( \mathbf{r}(t) \), the curvature \( \kappa(t) \) is computed using the relationship between the velocity and acceleration vectors:

• First, differentiate the unit tangent vector to obtain the acceleration vector \( \mathbf{r}''(t) \)
• Curvature is found using the formula \( \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \)

In two dimensions, the cross product simplifies to a determinant-like operation. For instance, at \( t = 1 \) for our exercise, \( \kappa(1) \) is calculated as \( \frac{6}{13\sqrt{13}} \), showing how tight the curve is at that specific point.

Differentiation

In calculus, differentiation is the process of finding a derivative, which measures how a function changes as its input changes. Specifically, it helps identify the rate of change or the slope of a function at any given point.

• The first derivative, such as \( \mathbf{r}'(t) \), provides the velocity vector, indicating the direction and speed of the curve's movement.
• Further differentiation yields higher-order derivatives. For instance, \( \mathbf{r}''(t) \) provides the acceleration vector, which is crucial for computing curvature.

Differentiation is essential for analyzing curves, as it helps understand their dynamic properties. In our example, differentiation is repeatedly used to find not just the velocity vector but also the expressions required for calculating curvature.

Velocity Vector

The velocity vector is the first derivative of the position vector with respect to time. It represents the rate of change of position and is fundamental in analyzing the motion along a curve.

• For a curve \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \), the velocity vector is \( \mathbf{r}'(t) = \langle \frac{d}{dt}x(t), \frac{d}{dt}y(t) \rangle \)
• It shows both the speed and direction of motion at any point \( t \).

Calculating the velocity vector is a crucial first step in many calculus problems involving curves. This vector is then normalized to get the unit tangent vector or integrated into further calculations like curvature, as seen in our exercise where \( \mathbf{r}'(1) = \langle -2, -3 \rangle \).