Question

Consider the curve \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. Assuming the curve lies in a plane, prove that the curve is an ellipse in that plane.

Short answer

Question: Prove that given the curve with the parametric equation \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle,\) for \(0 \leq t \leq 2 \pi\) and \(c\) is a real number, the curve is an ellipse in the plane it lies in. Answer: By removing the third coordinate, expressing \(t\) in terms of \(z\), and simplifying the equations, we obtain the Cartesian equation \(x^2 + \left(\frac{z}{c}\right)^2 = 1\), which is in the standard form of an ellipse \((\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1)\) with \(a^2 = 1\) and \(b^2 = c^2\). Therefore, we have proved that the curve is an ellipse in the plane it lies in.

Step-by-step solution

Remove the third coordinate

To show that the curve lies on a plane, we need to eliminate the third coordinate (i.e., \(z = c\sin{t}\)). In other words, we want to express the value of \(t\) in terms of \(z\) and substitute that into the \(x\) and \(y\) coordinates to obtain a relationship between \(x\) and \(y\). From the third coordinate, we have \(z = c\sin{t}\). Thus, we can express \(t\) as \(t = \sin^{-1}{\frac{z}{c}}\).

Substitute and make a relationship between \(x\) and \(y\)

Now we can express the \(x\) and \(y\) coordinates in terms of \(z\) using the obtained expression. Recall that \(x = \cos{t}\) and \(y = \sin{t}\). By plugging in the expression for \(t\), we have - \(x = \cos{(\sin^{-1}{\frac{z}{c}})}\) - \(y = \sin{(\sin^{-1}{\frac{z}{c}})}\) To simplify things, let \(w = \sin^{-1}{\frac{z}{c}}\). Then, we have \(x = \cos{w}\) and \(y = \sin{w}\). Then, we can square both sides \(x^2 = \cos^2{w}\) \(y^2 = \sin^2{w}\)

Simplify and rearrange the equation

Now, we should simplify the equations and rearrange them to find a Cartesian equation. Using the Pythagorean identity, we know that \(\cos^2{w} + \sin^2{w} = 1\). Thus, we can set their sum equal to 1: \(x^2 + y^2 = 1\) However, from earlier, we have \(w = \sin^{-1}{\frac{z}{c}}\), which implies that \(y = \sin{w} = \frac{z}{c}\). Thus, substituting for y, we have: \(x^2 + \left(\frac{z}{c}\right)^2 = 1\)

Check if the equation is an ellipse

We have obtained a Cartesian equation for our curve as follows: \(x^2 + \left(\frac{z}{c}\right)^2 = 1\) Now, we notice that this equation is in the standard form of an ellipse \((\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1)\) with \(a^2 = 1\) and \(b^2 = c^2\). Therefore, we have proved that the curve is an ellipse in the plane it lies in.

Key concepts

Vector Functions

Vector functions play a crucial role in various branches of calculus. They are used to describe curves, surfaces, and flows in multidimensional spaces. A vector function gives a vector as output for a given scalar input, which is often represented by a parameter, such as time. In our scenario, the input parameter is \( t \), and the vector function is \( \mathbf{r}(t) = \langle \cos t, \sin t, c \sin t \rangle \).

• Components of Vector Functions: A vector function consists of components that can change with respect to the parameter \( t \).
• Representation of Curves: In three dimensions, the components \( \cos t, \sin t, \) and \( c \sin t \) collectively designate points along a curve in space.

In this exercise, the task is to demonstrate that this vector function describes an ellipse in a plane. This involves removing the third component and finding a relationship between the first two components.

Parametric Equations

Parametric equations are used to represent curves by expressing the coordinates of the points on the curve as functions of a parameter. This approach can simplify the analysis of complex curves by freeing us from solely relying on traditional Cartesian coordinates.

• Form: Coordinates \( (x, y, z) \) are given as functions of \( t \), such as \( x = \cos t \), \( y = \sin t \), and \( z = c \sin t \) in this example.
• Flexibility: Parametric equations provide flexibility in representing both simple and complex curves, making analysis and transformations more manageable.

By parameterizing these equations, we can analyze the behavior of each component individually or in combination with others, making it easier to see how changes in the parameter affect the curve. In this case, the goal is to show how these parametric equations trace out an ellipse in the plane by simplifying and rearranging them into a standard Cartesian form.

Ellipses

Ellipses are a special type of conic section, representing stretched circles. They have distinct geometric properties and are defined by specific equations in mathematics.

• Standard Equation: An ellipse can be represented by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes respectively.
• Plane Intersection: An ellipse in three-dimensional space often represents the intersection of a plane with a cylinder or a cone.

In this exercise, we started with a three-dimensional function \( \mathbf{r}(t) \) which we needed to show describes an ellipse. By eliminating the third dimension \( z = c \sin t \) and rearranging, we arrived at an equation in the form \( x^2 + \left( \frac{z}{c} \right)^2 = 1 \). This corresponds to the standard form of an ellipse, proving that the function indeed traces out an elliptical path in the plane.