Problem 17 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The ellipse is formed by intersecting a cylinder \( x^2 + y^2 = 1 \) with the plane \( z = 1-x \); estimated length is approximately 8.89 units.

Step 1: Parametrization of the Curve

The curve is given by \( \mathbf{r}(t) = (\cos t)\mathbf{i} + (\sin t)\mathbf{j} + (1-\cos t)\mathbf{k} \) for \( 0 \leq t \leq 2\pi \). The functions represent coordinates in 3D space. Our goal is to demonstrate that this path is indeed an ellipse.

Step 2: Equation for the Cylinder

Express \( (\cos t)^2 + (\sin t)^2 = 1 \), corresponding to unit radius components in \( i \) and \( j \). Recognizing this as the equation of a right circular cylinder \( x^2 + y^2 = 1 \), where \( x = \cos t \) and \( y = \sin t \). Thus, \( x^2 + y^2 = 1 \) describes a cylinder centered around the z-axis.

Step 3: Equation for the Plane

The parameter \( z = 1 - \cos t \) describes a transformation on the \( z \)-axis. This defines the plane \( z = 1 - x \), using the relation \( x = \cos t \). The plane intersects the cylinder to form the ellipse.

Step 4: Sketch the Ellipse with Tangent Vectors

Draw the cylinder \( x^2 + y^2 = 1 \) with the intersecting plane \( z = 1 - x \). The ellipse lies at the intersection. Introduce tangent vectors by differentiating \( \mathbf{r}(t) \). Evaluate the tangent vector \( \mathbf{r}'(t) = (-\sin t)\mathbf{i} + (\cos t)\mathbf{j} + (\sin t)\mathbf{k} \) at \( t = 0, \pi/2, \pi, 3\pi/2 \).

Step 5: Confirm the Acceleration Vector Alignment

Calculate the acceleration vector \( \mathbf{r}''(t) = (-\cos t)\mathbf{i} - (\sin t)\mathbf{j} + (\cos t)\mathbf{k} \). Show orthogonality to the normal of the plane \( \mathbf{n} = \langle 1, 0, 1 \rangle \) by confirming \( \mathbf{r}''(t) \cdot \mathbf{n} = 0 \). Add acceleration vectors for \( t = 0, \pi/2, \pi, 3\pi/2 \) to the sketch.

Step 6: Write Integral for Ellipse Length

Use the formula for arc length in parametric form: \[ L = \int_{0}^{2\pi} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \ dt \]. Evaluate each derivative: \( \frac{dx}{dt} = -\sin t \), \( \frac{dy}{dt} = \cos t \), \( \frac{dz}{dt} = \sin t \). Substitute and form \( L = \int_{0}^{2\pi} \sqrt{(-\sin t)^2 + (\cos t)^2 + (\sin t)^2} \ dt \).

Step 7: Numerical Estimate of Ellipse Length

The integral for the arc length simplifies to \( L = \int_{0}^{2\pi} \sqrt{2} \ dt \). Evaluate numerically: \[ L = \sqrt{2} \cdot \int_{0}^{2\pi} dt = \sqrt{2} \cdot 2\pi \]. Approximate \( \sqrt{2} \approx 1.41 \), thus \( L \approx 1.41 \times 2\pi \approx 8.89 \).

Key Concepts

Parametric Equations

Parametric equations are a way to express a set of related quantities as functions of one or more independent variables, called parameters. In the context of our ellipse, the parametric equations for the curve \( \mathbf{r}(t) = (\cos t)\mathbf{i} + (\sin t)\mathbf{j} + (1-\cos t)\mathbf{k} \) describe a trajectory in three-dimensional space. Here, \( t \) is the parameter that ranges from \( 0 \) to \( 2\pi \).

This method of using parameters allows us to break down complex motion into simpler, more manageable components. Note how each coordinate—\( x, y, \) and \( z \)—is represented as a function of \( t \). This method is particularly useful for modeling curves and surfaces in multidimensional spaces.

The curve described by these parametric equations traces out an ellipse, as we will show, formed by the intersection of a cylinder and a slanted plane.

Right Circular Cylinder

A right circular cylinder is a three-dimensional surface consisting of parallel lines, each of which is tangent to a circle of a fixed radius. The lines (also called rulings) are perpendicular to the base of the cylinder.

In our problem, the equation \( x^2 + y^2 = 1 \) defines such a cylinder, where the circle's radius is 1 and it extends infinitely along the \( z \)-axis. This is because the coordinates \( x \) and \( y \) satisfy the Pythagorean identity for a unit circle: \( (\cos t)^2 + (\sin t)^2 = 1 \).

This cylinder is crucial because it provides one of the boundaries needed to define the ellipse. The cylinder and the slanted plane together create the elliptical intersection within the bounds defined by the parameter \( t \).

Arc Length of Ellipse

The arc length of a curve in parametric form can be found using the integral formula for arc length, which requires computing derivatives of each component function of the curve. Specifically, for a curve \( \mathbf{r}(t) = (x(t), y(t), z(t)) \), the arc length \( L \) from \( t = a \) to \( t = b \) is given by:
\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \ dt \]

Here, the derivatives indicate the rates of change of each component with respect to the parameter \( t \). For our ellipse, these derivatives are: \( \frac{dx}{dt} = -\sin t \), \( \frac{dy}{dt} = \cos t \), \( \frac{dz}{dt} = \sin t \).

Substituting them back into the arc length formula results in \( L = \int_{0}^{2\pi} \sqrt{2} \ dt \), ultimately simplifying to \( L = \sqrt{2} \cdot 2\pi \). This expression shows that even though evaluating the integral directly is complex, it can be approximated numerically, resulting in a length of approximately 8.89 units.

Acceleration Vector

The acceleration vector is found by differentiating the velocity vector, itself the derivative of the position vector \( \mathbf{r}(t) \). For our ellipse, differentiation yields the acceleration vector \( \mathbf{r}''(t) = (-\cos t)\mathbf{i} - (\sin t)\mathbf{j} + (\cos t)\mathbf{k} \).

What's special here is the alignment of this vector with the plane of the ellipse. The vector normal to the plane is \( \mathbf{n} = \langle 1, 0, 1 \rangle \), and the dot product \( \mathbf{r}''(t) \cdot \mathbf{n} = 0 \) confirms orthogonality, meaning the acceleration vector is always parallel to the plane of the ellipse. This is why, when plotted, the acceleration vectors appear to lie on the ellipse, illustrating their confinement to the plane.

This understanding is key in verifying how forces acting on the body would maintain its motion purely within the plane, as orthogonality to a normal is indicative of maintaining a direction within that plane itself.