Problem 20 - Essential Calculus Early Transcendentals

Question

Find the curvature of \(\mathbf{r}(t)=\left\langle t^{2}, \ln t, t \ln t\right\rangle\) at the point \((1,0,0) .\)

Step-by-Step Solution

Answer

The curvature is \( \frac{\sqrt{30}}{18} \).

Step 1: Verify the Given Point on the Curve

First, verify that the point (1, 0, 0) lies on the curve defined by \( \mathbf{r}(t) = \langle t^2, \ln t, t \ln t \rangle \). Substitute \(t = 1\) into the components of \( \mathbf{r}(t) \): \( x = 1^2 = 1 \), \( y = \ln 1 = 0 \), and \( z = 1 \cdot \ln 1 = 0 \). So, the point (1, 0, 0) indeed lies on the curve when \( t = 1 \).

Step 2: Find the Velocity Vector \(\mathbf{v}(t)\)

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \). Differentiate each component:- \( \frac{d}{dt}(t^2) = 2t \)- \( \frac{d}{dt}(\ln t) = \frac{1}{t} \)- \( \frac{d}{dt}(t \ln t) = \ln t + 1 \)Thus, \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \ln t + 1 \rangle \).

Step 3: Find the Acceleration Vector \(\mathbf{a}(t)\)

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \). Differentiate each component:- \( \frac{d}{dt}(2t) = 2 \)- \( \frac{d}{dt}(\frac{1}{t}) = -\frac{1}{t^2} \)- \( \frac{d}{dt}(\ln t + 1) = \frac{1}{t} \)Thus, \( \mathbf{a}(t) = \langle 2, -\frac{1}{t^2}, \frac{1}{t} \rangle \).

Step 4: Evaluate \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\) at \(t = 1\)

Substitute \(t = 1\) into \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \):- \( \mathbf{v}(1) = \langle 2 \cdot 1, \frac{1}{1}, \ln 1 + 1 \rangle = \langle 2, 1, 1 \rangle \)- \( \mathbf{a}(1) = \langle 2, -\frac{1}{1^2}, \frac{1}{1} \rangle = \langle 2, -1, 1 \rangle \)

Step 5: Calculate the Curvature Formula

The formula for curvature \( \kappa \) of a parametric curve is given by:\[ \kappa = \frac{\|\mathbf{v}(t) \times \mathbf{a}(t)\|}{\|\mathbf{v}(t)\|^3} \]Find \( \mathbf{v}(t) \times \mathbf{a}(t) \) and its magnitude.

Step 6: Compute the Cross Product \(\mathbf{v}(1) \times \mathbf{a}(1)\)

Calculate the cross product \( \mathbf{v}(1) = \langle 2, 1, 1 \rangle \) and \( \mathbf{a}(1) = \langle 2, -1, 1 \rangle \):\[ \mathbf{v}(1) \times \mathbf{a}(1) = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \2 & 1 & 1 \2 & -1 & 1\end{array} \right| = \langle (1 \cdot 1 - 1 \cdot (-1)), - (2 \cdot 1 - 2 \cdot 1), (2\cdot(-1) - 1 \cdot 2) \rangle = \langle 2, 0, -4 \rangle \]

Step 7: Calculate Magnitude of the Cross Product

Find the magnitude of \( \mathbf{v}(1) \times \mathbf{a}(1) = \langle 2, 0, -4 \rangle \):\[ \| \mathbf{v}(1) \times \mathbf{a}(1) \| = \sqrt{2^2 + 0^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

Step 8: Calculate Magnitude of \(\mathbf{v}(1)\) and Curvature Rate

Find \( \| \mathbf{v}(1) \| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \). Then apply the values into the curvature formula:\[ \kappa = \frac{2\sqrt{5}}{(\sqrt{6})^3} = \frac{2\sqrt{5}}{6\sqrt{6}} = \frac{\sqrt{30}}{18} \]

Step 9: Final Answer for Curvature

The curvature \( \kappa \) of the curve \( \mathbf{r}(t) \) at the point \((1, 0, 0)\) is \( \frac{\sqrt{30}}{18} \).

Key Concepts

Velocity Vector

In the context of a parametric curve like \( \mathbf{r}(t) = \langle t^2, \ln t, t \ln t \rangle \), the **velocity vector** describes the rate of change of the curve with respect to its parameter, \( t \). To find the velocity vector, \( \mathbf{v}(t) \), you differentiate each component of \( \mathbf{r}(t) \):

• The derivative of \( t^2 \) is \( 2t \).
• The derivative of \( \ln t \) is \( \frac{1}{t} \).
• The derivative of \( t \ln t \) is \( \ln t + 1 \).

Thus, the velocity vector \( \mathbf{v}(t) = \langle 2t, \frac{1}{t}, \ln t + 1 \rangle \). The velocity vector is critical because it tells us the direction and speed of the curve's path at any point \( t \).
By evaluating the velocity vector at a specific point, such as \( t = 1 \), we find \( \mathbf{v}(1) = \langle 2, 1, 1 \rangle \). This means that at the point \((1, 0, 0)\), the curve moves in the direction of \( \langle 2, 1, 1 \rangle \).

Acceleration Vector

The **acceleration vector** is the derivative of the velocity vector, \( \mathbf{v}(t) \). It measures how the velocity itself changes over time. This is equivalent to looking at the second derivative of the original parametric equation. To find the acceleration vector \( \mathbf{a}(t) \), calculate the derivative of each component of \( \mathbf{v}(t) \):

• \( \frac{d}{dt}(2t) = 2 \)
• \( \frac{d}{dt}\left(\frac{1}{t}\right) = -\frac{1}{t^2} \)
• \( \frac{d}{dt}(\ln t + 1) = \frac{1}{t} \)

Thus, \( \mathbf{a}(t) = \langle 2, -\frac{1}{t^2}, \frac{1}{t} \rangle \). Evaluating \( \mathbf{a}(t) \) at \( t = 1 \) gives us \( \mathbf{a}(1) = \langle 2, -1, 1 \rangle \), showing how the velocity is changing at the point \((1, 0, 0)\).
In a practical sense, understanding the acceleration vector helps us determine how the speed and direction of the curve's motion are changing.

Cross Product

The **cross product** is a mathematical operation that takes two vectors and produces another vector that is perpendicular to both of them. For vectors \( \mathbf{v}(1) = \langle 2, 1, 1 \rangle \) and \( \mathbf{a}(1) = \langle 2, -1, 1 \rangle \), the cross product \( \mathbf{v}(1) \times \mathbf{a}(1) \) is computed using the determinant of a matrix formulated with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):\[\mathbf{v}(1) \times \mathbf{a}(1) = \left( \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \2 & 1 & 1 \2 & -1 & 1\end{array} \right)\]Which results in \( \langle 2, 0, -4 \rangle \).
The resulting vector represents the area of the parallelogram spanned by \( \mathbf{v}(1) \) and \( \mathbf{a}(1) \), as well as the orientation out of the plane formed by the two vectors.
In finding curvature, we use the magnitude of this cross product to determine the rate at which the direction of the curve changes.

Magnitude of a Vector

The **magnitude of a vector** represents its length. For a vector \( \mathbf{v} = \langle x, y, z \rangle \), the magnitude \( \| \mathbf{v} \| \) is found using the Pythagorean theorem in three dimensions:\[ \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \]In our exercise, consider the cross product vector \( \langle 2, 0, -4 \rangle \). Its magnitude is calculated as:

• \( \| \langle 2, 0, -4 \rangle \| = \sqrt{2^2 + 0^2 + (-4)^2} = \sqrt{20} = 2\sqrt{5} \)

Another key vector is \( \mathbf{v}(1) = \langle 2, 1, 1 \rangle \) with magnitude:

• \( \| \mathbf{v}(1) \| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6} \)

These magnitudes are crucial in calculating the curvature of the parametric curve since the curvature formula relies on both the magnitude of the cross product and the velocity vector.