Problem 17 - Thomas Calculus

Question

Ellipse $$ \begin{array}{c}{\text { a. Show that the curve } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(1-\cos t) \mathbf{k}} \\ {0 \leq t \leq 2 \pi, \text { is an ellipse by showing that it is the intersection }} \\\ {\text { of a right circular cylinder and a plane. Find equations }} \\\ {\text { for the cylinder and plane. }}\\\\{\text { b. Sketch the ellipse on the cylinder. Add to your sketch the unit }} \\ {\text { tangent vectors at } t=0, \pi / 2, \pi, \text { and } 3 \pi / 2 .}\\\\{\text { c. Show that the acceleration vector always lies parallel to the }} \\ {\text { plane (orthogonal to a vector normal to the plane). Thus, if }} \\ {\text { you draw the acceleration as a vector attached to the ellipse, }} \\ {\text { it will lie in the plane of the ellipse. Add the acceleration vectors }} \\\ {\text { for } t=0, \pi / 2, \pi, \text { and } 3 \pi / 2 \text { to your sketch. }}\\\\{\text { d. Write an integral for the length of the ellipse. Do not try to }} \\ {\text { evaluate the integral; it is nonelementary. }} \\\ {\text { e. Numerical integrator Estimate the length of the ellipse to }} \\\ {\text { two decimal places. }}\end{array} $$

Step-by-Step Solution

Answer

The curve is an ellipse formed by intersecting a cylinder \(x^2 + y^2 = 1\) with plane \(z + x = 1\); length \( \approx 8.89 \).

Step 1: Parametric Representation and Cylinder Equation

Considering the parametrized curve \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (1 - \cos t) \mathbf{k} \), note that the first two components describe a circle in the \(xy\)-plane. The equation of this circle is \(x^2 + y^2 = 1\), where \(x = \cos t\) and \(y = \sin t\). This represents the intersection with a right circular cylinder in the 3D space with radius 1 and axis along the \(z\)-axis: \(x^2 + y^2 = 1\).

Step 2: Plane Equation

The plane is defined by the equation derived from the z-component \(z = 1 - \cos t\). Rewrite this as \(z + x = 1\). This represents a plane in the space that intersects the cylinder to form the ellipse described by \( \mathbf{r}(t) \).

Step 3: Tangent Vector and Sketch at Specific Points

To sketch the tangent vectors at \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \), find the derivative of \( \mathbf{r}(t) \): \( \mathbf{r}'(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (\sin t) \mathbf{k} \). The tangent vectors based on this derivative are attached at the corresponding points on the ellipse for each specified \( t \) value.

Step 4: Acceleration Vector and its Orthogonality

Calculate the second derivative \( \mathbf{r}''(t) = (-\cos t) \mathbf{i} - (\sin t) \mathbf{j} + (\cos t) \mathbf{k} \). The normal to the plane \( z + x = 1 \) is \( \mathbf{n} = (1, 0, 1) \). Show that the dot product \( \mathbf{r}''(t) \cdot \mathbf{n} = 0 \), indicating the acceleration vector is orthogonal to the normal vector of the plane, and thus lies in the plane of the ellipse.

Step 5: Integral for Arc Length

The arc length of a parametrized curve \( \mathbf{r}(t) \) over the interval \([a, b]\) is given by the integral:\[ L = \int_a^b \| \mathbf{r}'(t) \| \, dt \]For our curve, this becomes:\[ L = \int_0^{2\pi} \sqrt{(-\sin t)^2 + (\cos t)^2 + (\sin t)^2} \, dt \] This expression simplifies to \( L = \int_0^{2\pi} \sqrt{2} \, dt \), which is nonelementary to evaluate directly.

Step 6: Numerical Estimate for Arc Length

Using numerical methods like Simpson's Rule or a numerical integrator, compute the integral \( L = \int_0^{2\pi} \sqrt{2} \, dt \). The exact numerical result is approximately \[ L \approx 2\pi\sqrt{2} \approx 8.89 \].

Key Concepts

Parametric equations

Parametric equations are vital in representing curves for various applications in mathematics and engineering. The idea is to use a parameter, usually denoted as \( t \), to express the coordinates of a point on a curve. For the given problem, the vector function \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + (1 - \cos t) \mathbf{k} \) defines a parametric representation of a curve in three-dimensional space.
This representation tells us how each component varies with \( t \), giving intuitive insight into the movement along the curve. In simple terms, resolving each component shows:

• For the \( x \)-direction: \( x(t) = \cos t \)
• For the \( y \)-direction: \( y(t) = \sin t \)
• For the \( z \)-direction: \( z(t) = 1 - \cos t \)

The first two components, \( \cos t \) and \( \sin t \), describe a circular path, while the third component modifies the circle into an ellipse in the context of this problem.

Cylinder and plane intersection

The intersection of a geometry like a cylinder with a plane often results in a curve. In this exercise, the problem statement suggests that the path described by the parametric equations intersects a cylinder and a plane, forming an ellipse.
To understand this intersection, consider the equation of the cylinder from the parametric equations: \( x^2 + y^2 = 1 \). This describes a cylinder with a radius of 1 and an axis along the \( z \)-axis. On the other hand, the plane is derived from the \( z \)-component \( z = 1 - \cos t \), which can be rearranged to \( z + x = 1 \).
When this plane cuts through the cylinder, the resulting shape is an ellipse, confirming that the path described by the parametric equations is indeed an ellipse situated where the cylinder and plane intersect.

Tangent vectors

Tangent vectors provide valuable information about the direction and speed of movement along a curve. The derivative of the position vector \( \mathbf{r}(t) \) gives us the tangent vector \( \mathbf{r}'(t) \). For this exercise, the calculation gives us: \( \mathbf{r}'(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (\sin t) \mathbf{k} \).
This tangent vector tells us the direction in which the curve is progressing at any point \( t \). It shows:

• The change in the \( x \)-direction is \( -\sin t \).
• The change in the \( y \)-direction is \( \cos t \).
• The change in the \( z \)-direction is \( \sin t \).

Using these tangent vectors, one can visualize the curvature and orientation of the ellipse at specific points such as \( t=0, \pi/2, \pi, \) and \( 3\pi/2 \).

Acceleration vectors

The acceleration vector reveals how the rate of change of the tangent vector itself changes along the curve. By taking the second derivative of \( \mathbf{r}(t) \), we obtain the acceleration vector: \( \mathbf{r}''(t) = (-\cos t) \mathbf{i} - (\sin t) \mathbf{j} + (\cos t) \mathbf{k} \).
This describes not just how the direction of motion changes but also how its speed varies along its path. Importantly, in this exercise, it is shown that the acceleration vector is parallel to the plane of the ellipse. This is verified by checking the orthogonality to the normal vector of the plane. Given \( \mathbf{n} = (1, 0, 1) \), we find:
\[ \mathbf{r}''(t) \cdot \mathbf{n} = 0 \]
Thus, the acceleration vector is confined to the plane of the ellipse, ensuring it lies within and is aligned with the path of motion.

Arc length integral

Understanding the length of a curve is done through the arc length integral. For the parametric curve \( \mathbf{r}(t) \), the length \( L \) is determined by integrating the magnitude of the tangent vector \( \mathbf{r}'(t) \) over the interval of interest. The formula used is:
\[ L = \int_a^b \| \mathbf{r}'(t) \| \, dt \]
In our exercise, calculate this from \( 0 \) to \( 2\pi \):
\[ L = \int_0^{2\pi} \sqrt{(-\sin t)^2 + (\cos t)^2 + (\sin t)^2} \, dt \]
This simplifies to:
\[ L = \int_0^{2\pi} \sqrt{2} \, dt \]
This integral, while simplified expression-wise, is a non-elementary one, meaning it cannot be expressed in terms of basic functions. Numerical methods like Simpson's Rule are often used to carry out such calculations, which gives an approximate length for the ellipse.