Problem 22 - Thomas Calculus

Question

Find an equation for the circle of curvature of the curve \(\mathbf{r}(t)=\) \((2 \ln t) \mathbf{i}-[t+(1 / t)] \mathbf{j}, e^{-2} \leq t \leq e^{2},\) at the point \((0,-2)\) where \(t=1 .\)

Step-by-Step Solution

Answer

\(x^2 + (y+4)^2 = 4\)

Step 1: Find Velocity and Acceleration at t=1

\(\mathbf{r}(t) = (2\ln t)\mathbf{i} - [t + 1/t]\mathbf{j}\)
\(\mathbf{r}'(t) = \frac{2}{t}\mathbf{i} - [1 - 1/t^2]\mathbf{j}\)
At \(t=1\): \(\mathbf{r}'(1) = 2\mathbf{i} - 0\mathbf{j} = 2\mathbf{i}\), \(|\mathbf{r}'(1)| = 2\)
\(\mathbf{r}''(t) = -\frac{2}{t^2}\mathbf{i} - \frac{2}{t^3}\mathbf{j}\)
At \(t=1\): \(\mathbf{r}''(1) = -2\mathbf{i} - 2\mathbf{j}\)

Step 2: Find Curvature

\(\kappa = \frac{|\mathbf{r}' \times \mathbf{r}''|}{|\mathbf{r}'|^3}\)
In 2D: \(|\mathbf{r}' \times \mathbf{r}''| = |2(-2) - 0(-2)| = |-4| = 4\)
\(\kappa = \frac{4}{8} = \frac{1}{2}\), so radius \(\rho = 2\).

Step 3: Find Center and Equation

The unit normal points in the \(-\mathbf{j}\) direction (toward the concave side).
\(\mathbf{N} = -\mathbf{j}\) (since \(\mathbf{T} = \mathbf{i}\) and the acceleration component perpendicular to \(\mathbf{T}\) points in \(-\mathbf{j}\)).
Center = \((0, -2) + 2(-\mathbf{j}) = (0, -4)\)
Equation: \(x^2 + (y+4)^2 = 4\)

Key Concepts

Circle of Curvature

The circle of curvature provides a geometrical understanding of how sharply a curve bends at a given point. It is also known as the osculating circle, meaning it "kisses" the curve at one point. This circle has the same tangent and curvature at the specific point on the curve. To find the circle of curvature, you need to calculate the radius of curvature, denoted by \( R \), which is the reciprocal of the curvature \( \kappa \) at that point. To compute the curvature, one typically uses the formula \( \kappa = \frac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{|\mathbf{v}(t)|^3} \) where \( \mathbf{v}(t) \) is the velocity vector and \( \mathbf{a}(t) \) is the acceleration vector. Once the curvature is known, the center of the circle of curvature can be determined by shifting from the point of tangency along the normal in the direction the curve turns. Here's the vital part: when you've computed all these elements, the circle offers a powerful way to visualize the motion or configuration of the curve at that immediate point.

Velocity Vector

The velocity vector is crucial in determining the rate and direction of movement of a point along a curve. In mathematical terms, the velocity vector \( \mathbf{v}(t) \) is derived by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \). For the curve given by \( \mathbf{r}(t) = (2 \ln t) \mathbf{i} - [t + (1/t)] \mathbf{j} \), the velocity vector comes out to be \( \mathbf{v}(t) = \frac{2}{t} \mathbf{i} - \left(1 - \frac{1}{t^2}\right) \mathbf{j} \). This expression tells us the tangent direction of the curve at any time \( t \).

• It has both magnitude and direction, providing complete motion information at any instant.
• The components \( \frac{2}{t} \) and \( \left(1 - \frac{1}{t^2}\right) \) highlight how the curve navigates through its domain.

Acceleration Vector

Acceleration vector provides insights into how the velocity of an object changes over time. Structurally, it is the derivative of the velocity vector. When you take the derivative of \( \mathbf{v}(t) \), you get the acceleration vector \( \mathbf{a}(t) \). So, for our function \( \mathbf{v}(t) = \frac{2}{t} \mathbf{i} - \left(1 - \frac{1}{t^2}\right) \mathbf{j} \), the acceleration vector turns out to be \( \mathbf{a}(t) = -\frac{2}{t^2} \mathbf{i} - \frac{2}{t^3} \mathbf{j} \).

• It helps determine how the object speeding along the curve is behaving -- whether it's accelerating or decelerating, and in which direction.
• The negative values indicate a decrease in the velocity components over time.

Understanding acceleration can provide crucial insights into the dynamics of motion and curvature behavior at any given point.

Derivatives

Derivatives are fundamental mathematical tools that measure how a function changes as its input changes. They're the cornerstone in calculus for understanding motion and change. In the context of curves, derivatives are used to find both velocity and acceleration vectors.

• The first derivative of a position function gives the velocity vector, indicating direction and speed.
• The second derivative provides the acceleration vector, showing how the velocity changes over time.

To differentiate the function \( \mathbf{r}(t) = (2 \ln t) \mathbf{i} - [t + (1/t)] \mathbf{j} \), you find \( \mathbf{v}(t) = \frac{2}{t} \mathbf{i} - \left(1 - \frac{1}{t^2}\right) \mathbf{j} \) for the first derivative and \( \mathbf{a}(t) = -\frac{2}{t^2} \mathbf{i} - \frac{2}{t^3} \mathbf{j} \) for the second derivative. These calculations are important to describe how the motion along the curve evolves with respect to time. Employing derivatives provides a precise and efficient way to determine instantaneous rates of change, critical in both theoretical and applied mathematics.