Question

Consider the curve \(\mathbf{r}(t)=(\cos t, \sin t, c \sin t),\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. It can be shown that the curve lies in a plane. Prove that the curve is an ellipse in that plane.

Short answer

Based on the given curve \(\mathbf{r}(t)=(\cos t, \sin t, c \sin t)\), prove that it lies in a plane and is an ellipse. Solution: 1. We found a normal vector to the plane containing the curve, which was \(\mathbf{n}=(0,0,-c)\). 2. We expressed the curve in the plane coordinates and found that the curve lies in the plane \(z=0\). 3. Upon checking if the curve follows the standard ellipse equation, we found that it does with semi-major axis \(a = 1\) and semi-minor axis \(b = \frac{1}{c}\). Thus, the given curve lies in a plane and is an ellipse.

Step-by-step solution

Find a normal vector to the plane containing the curve

We know that the curve \(\mathbf{r}(t)=(\cos t, \sin t, c \sin t)\) lies in a plane. To find a normal vector to this plane, first, let's find any two vectors that are tangent to the curve at any point. To do this, we can compute the first derivative of \(\mathbf{r}(t)\): $$\frac{d\mathbf{r}(t)}{dt} = (-\sin t, \cos t, c \cos t)$$ Since this derivative vector is tangent to the curve at any point for \(0 \leq t \leq 2 \pi\), we can use this to our advantage. Let's pick two points with different values of \(t\) and find the corresponding tangent vectors: $$\frac{d\mathbf{r}(0)}{dt} = (-\sin 0, \cos 0, c \cos 0) = (0, 1, c)$$ $$\frac{d\mathbf{r}(\pi/2)}{dt} = (-\sin (\pi/2), \cos (\pi/2), c \cos (\pi/2)) = (-1, 0, 0)$$ Now, we can find a normal vector to the plane by taking the cross product of these two tangent vectors: $$\mathbf{n} = \frac{d\mathbf{r}(0)}{dt} \times \frac{d\mathbf{r}(\pi/2)}{dt} = (0, 1, c) \times (-1, 0, 0) = (0, 0, -c)$$

Express the curve in the plane coordinates

Now that we have a normal vector to the plane, we can express the curve \(\mathbf{r}(t)\) using coordinates of the plane containing it. To do this, we first find the equation of the plane: $$\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0$$ where \(\mathbf{n}\) is the normal vector, \(\mathbf{r}\) is a point on the curve, and \(\mathbf{r_0}\) is a point on the plane. Since both \(\mathbf{r_0}\) and \(\mathbf{r}\) are on the curve, we can pick a convenient point on the curve and plug the normal vector. For example, we can use the point \(\mathbf{r}(0) = (1, 0, 0)\): $$\mathbf{n} \cdot (\mathbf{r} - (1, 0, 0)) = (0, 0, -c) \cdot ((\cos t, \sin t, c \sin t) - (1, 0, 0)) = 0$$ Now, let's simplify the equation: $$0 \cdot (\cos t - 1) + 0 \cdot (\sin t) - c \cdot (c \sin t) = 0$$ $$- c^2 \sin t = 0$$ Since \(c\) is a real number, this equation indicates that the curve lies in a plane \(z = 0\).

Prove that the curve is an ellipse

Now that we have the curve expressed in the coordinates of the plane, let's check if it follows the standard ellipse equation: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{(\cos t)^2}{a^2} + \frac{(\sin t)^2}{b^2} = 1$$ We notice that the equation becomes: $$\frac{(\cos t)^2}{a^2} + \frac{(\sin t)^2}{b^2} = \frac{(\cos t)^2}{1^2} + \frac{(\sin t)^2}{(1/c)^2} = 1$$ Since we know that \((\cos t)^2 + (\sin t)^2 = 1\), this equation holds true, and our curve is indeed an ellipse in the plane with semi-major axis \(a = 1\) and semi-minor axis \(b = \frac{1}{c}\).

Key concepts

Ellipses

Ellipses are fascinating shapes in geometry, often characterized as stretched circles. They are defined as the locus of all points where the sum of the distances from two fixed points, called foci, is constant. In Cartesian coordinates, the standard form of an ellipse's equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.
In the context of parametric curves, an ellipse can be represented by parametric equations, which essentially express the coordinates \(x\) and \(y\) in terms of a parameter, usually \(t\). These equations take the form \(x = a\cos(t)\) and \(y = b\sin(t)\).
The beauty of ellipses lies in their symmetry, made evident in both algebraic form and graphical representation. In problems involving curves and their classification, showing that a parametric curve matches the ellipse's equation is a common task.

Plane Geometry

Plane geometry deals with shapes and figures that exist in a two-dimensional plane. It encompasses basic shapes like lines, circles, and polygons, and their properties and relationships. Understanding plane geometry is crucial for analyzing parametric curves since it involves recognizing and working within a certain plane.
In our original exercise, the curve \(\mathbf{r}(t)=(\cos t, \sin t, c \sin t)\) needed to be shown to lie within a plane. To do this, various strategies involving the cross product and finding the plane's normal vector were employed—a testament to the deep interconnection between curves and plane geometry.
Knowing how to determine if a curve resides within a plane is a fundamental skill that helps in the simplification and understanding of complex geometric and algebraic problems, such as identifying the type of curve and its properties.

Cross Product

The cross product is a powerful tool in three-dimensional mathematics, used primarily in vector analysis. It involves two vectors and results in a third vector that is perpendicular to both. For two vectors \(\mathbf{a}\) and \(\mathbf{b}\), the cross product is defined as \(\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\).
This operation is essential for finding the normal vector to a plane. In the exercise, the cross product of two tangent vectors along the curve \(\mathbf{r}(t)\) was calculated to help establish the normal vector \((0, 0, -c)\).
Once you have the normal vector, you can easily derive the equation of the plane, knowing a point through which the plane passes. Understanding and using the cross product is vital for tasks like proving that a curve lies in a plane or determining the nature of intersections between planes and lines.

Parametric Equations

Parametric equations are a versatile mathematical tool used to define curves and surfaces in various dimensions. They express the coordinates of the points on the curve as functions of one or more parameters. For example, a circle can be represented parametrically by \(x = \cos(t)\) and \(y = \sin(t)\), where \(t\) is the parameter that varies, typically over \([0, 2\pi]\).
In the original exercise, the curve \(\mathbf{r}(t)=(\cos t, \sin t, c\sin t)\) is specified in terms of \(t\), clearly showing the role of parametric equations in simplifying complex curves into manageable forms. This approach makes it easier to calculate derivatives, integrals, and other geometric properties, like length and tangent vectors.
Parametric equations are incredibly useful in advanced mathematics and applications, aiding in the transition from standard geometric shapes to more intricate forms like ellipses and spirals that can't be easily represented using simple Cartesian equations.