Problem 29 - Thomas Calculus

Question

You will use a CAS to explore the osculating circle at a point \(P\) on a plane curve where \(\kappa \neq 0 .\) Use a CAS to perform the following steps: a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like. b. Calculate the curvature \(\kappa\) of the curve at the given value \(t_{0}\) using the appropriate formula from Exercise 5 or \(6 .\) Use the parametrization \(x=t\) and \(y=f(t)\) if the curve is given as a function \(y=f(x) .\) c. Find the unit normal vector \(N\) at \(t_{0}\) . Notice that the signs of the components of \(N\) depend on whether the unit tangent vector \(T\) is turning clockwise or counterclockwise at \(t=t_{0}\) . (See Exercise 7\()\) d. If \(\mathbf{C}=a \mathbf{i}+b \mathbf{j}\) is the vector from the origin to the center \((a, b)\) of the osculating circle, find the center \(\mathbf{C}\) from the vector equation $$ \mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right) $$ The point \(P\left(x_{0}, y_{0}\right)\) on the curve is given by the position vector \(\mathbf{r}\left(t_{0}\right) .\) e. Plot implicitly the equation \((x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}\) of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure the axes are equally scaled. $$ \mathbf{r}(t)=(3 \cos t) \mathbf{i}+(5 \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi, \quad t_{0}=\pi / 4 $$

Step-by-Step Solution

Answer

Plot the curve, compute curvature, normal vector, find center, plot circle.

Step 1: Plot the Parametric Curve

We start by plotting the given parametric curve to visualize it. The curve is given as \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 5 \sin t \mathbf{j} \) for \( 0 \leq t \leq 2 \pi \). In your computer algebra system (CAS), use the parametric plot function to generate this curve and inspect its shape.

Step 2: Calculate the Curvature \( \kappa \) at \( t_0 \)

To calculate the curvature \( \kappa \) at the specified \( t_0 = \frac{\pi}{4} \), we use the formula \( \kappa = \frac{|x' y'' - y' x''|}{((x')^2 + (y')^2)^{3/2}} \), where \( x(t) = 3 \cos t \) and \( y(t) = 5 \sin t \). Differentiate these with respect to \( t \) and evaluate at \( t = \frac{\pi}{4} \).

Step 3: Find the Unit Normal Vector \( N \)

Calculate the unit tangent vector \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \) first by differentiating \( \mathbf{r}(t) \). Then, determine the normal vector \( \mathbf{N}(t) \) by rotating \( \mathbf{T}(t) \) ninety degrees counterclockwise, which switches its components and changes the sign of the second one. Evaluate \( \mathbf{N}(t) \) at \( t = \frac{\pi}{4} \).

Step 4: Determine the Center of the Osculating Circle

To find the center \( \mathbf{C} = a \mathbf{i} + b \mathbf{j} \) of the osculating circle, use the formula: \( \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0) \). Substitute the point \( \mathbf{r}(t_0) = (3 \sqrt{2}/2, 5 \sqrt{2}/2) \) and the normal vector \( \mathbf{N}(t_0) \), along with the curvature \( \kappa(t_0) \) computed earlier.

Step 5: Plot the Osculating Circle

Finally, plot the osculating circle using its implicit equation \( (x-a)^2 + (y-b)^2 = 1/\kappa^2 \), where \( (a, b) \) is the center found in Step 4. In the same plot, include the original parametric curve to visualize the circle touching the curve at \( t_0 = \frac{\pi}{4} \). Ensure the plot axes are equally scaled for accuracy.

Key Concepts

Curvature Calculation

Calculating the curvature of a curve at a given point is a fundamental task in understanding how the curve bends at that point. This measure is essential when analyzing curves in differential geometry. The curvature, denoted by \( \kappa \), provides insight into how sharply a curve is turning at a particular location.

To find the curvature \( \kappa \) of a curve given in parametric form, you typically use the formula:
\[\kappa = \frac{|x' y'' - y' x''|}{((x')^2 + (y')^2)^{3/2}}\]

Where:

• \( x(t) \) and \( y(t) \) describe the position of the curve parametrically.
• \( x' \) and \( y' \) are the derivatives of \( x(t) \) and \( y(t) \) respectively.
• \( x'' \) and \( y'' \) are the second derivatives of \( x(t) \) and \( y(t) \).

By evaluating \( \kappa \) at a specific parameter value like \( t_0 = \frac{\pi}{4} \), you determine how the curve bends at that precise spot. This step is crucial when determining the osculating circle, as high curvature implies a tighter fitting circle.

The osculating circle, or the "kissing circle," helps visualize the curve's sharpness by approximating the curve with a circle at a specific point.

Parametric Curves

Parametric curves are representations of paths traced by points that depend on a parameter, usually denoted by \( t \). This representation is incredibly useful in calculus and physics, offering flexibility when defining the curve's shape.

The parametric expression of a curve can often be denoted as:
\[\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}\]

Where:

• \( x(t) \) controls the horizontal component or the horizontal motion of the curve.
• \( y(t) \) dictates the vertical component or motion.
• \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors in the direction of the x-axis and y-axis, respectively.

In our specific exercise, the curve \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 5 \sin t \mathbf{j} \) illustrates an ellipse with its parameter, \( t \), ranging from \( 0 \) to \( 2\pi \).

This approach is particularly beneficial when describing complex shapes that simple functional forms \( y=f(x) \) cannot cover efficiently. Distinguished by their ability to succinctly describe curves, parametric forms are widely used in computer graphics and robotics.

Unit Normal Vector

The unit normal vector, \( \mathbf{N}(t) \), plays a pivotal role in curve analysis by indicating the orientation of the curve's concave side. To derive it, we first calculate the unit tangent vector, \( \mathbf{T}(t) \), and then perform a 90-degree counterclockwise rotation.

• First, find the derivative of the parametric equation \( \mathbf{r}(t) \), resulting in \( \mathbf{r}'(t) \).
• The unit tangent vector is given by \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \).
• Performing a 90-degree rotation of \( \mathbf{T}(t) \) provides the unit normal vector.

Consider that the rotation involves switching the vector components and altering the sign of the second component. At \( t_0 = \frac{\pi}{4} \), calculating \( \mathbf{N}(t) \) lets us establish the direction perpendicularly away from the curve.

This directional orientation is critical when determining the osculating circle's center, as it directly influences the direction in which you offset from the curve's path. Remember, the unit normal vector helps indicate how the curve is turning either clockwise or counterclockwise, affecting the curvature analysis.