Question

Tilted ellipse 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 the curve parametrized by the position vector \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle\) for \(0 \leq t \leq 2 \pi\) is an ellipse in the plane it lies in. Answer: To prove that the given curve is an ellipse in the plane it lies in, we found the equation of the plane as \(z = cy\). Then, eliminating the parameter \(t\), we obtained the equation of the curve as \(x^2 + \left(\frac{z}{c}\right)^2 = 1\). This equation represents an ellipse in the plane \(z=cy\).

Step-by-step solution

Verify the position vector is parametrization of a curve

Given the position vector function \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle\) for \(0 \leq t \leq 2 \pi\). We have three components of the position vector: \(x(t) = \cos t\), \(y(t) = \sin t\) and \(z(t) = c \sin t\).

Find the equation of the plane

Since the curve lies on a plane, we have a relationship between x, y, and z components of \(\mathbf{r}(t)\). We know that all the points on the curve will satisfy the equation of the plane. To find the equation for this plane, let's first differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\): \[\frac{dx}{dt} = -\sin t\] \[\frac{dy}{dt} = \cos t\] \[\frac{dz}{dt} = c \cos t\] Now, let's eliminate \(t\) from these three equations. Divide the third equation by the second one: \[\frac{dz/dt}{dy/dt} = \frac{c\cos t}{\cos t}\] Since \(\cos{t} \neq 0\), we can safely cancel it out, getting: \[z = c y\] This is the equation of the plane that the curve lies in.

Eliminate parameter \(t\)

Now, we want to eliminate the parameter \(t\) to express the curve as an equation in \(x\), \(y\), and \(z\). We know the first two components of the curve to be \(x(t) = \cos t\) and \(y(t) = \sin t\). We can express \(t\) in terms of \(x\) and \(y\) as: \[x^2 + y^2 = \cos^2 t + \sin^2 t = 1 \] Now, substitute \(z = c y\) so the curve is now in the form of an equation with \(x\), \(y\), and \(z\): \[x^2 + \left(\frac{z}{c}\right)^2 = 1\]

Conclusion

We have now expressed the given curve as a function of \(x\), \(y\), and \(z\): \[x^2 + \left(\frac{z}{c}\right)^2 = 1\] This is the equation of an ellipse in the plane \(z = c y\). Thus, we have proven that the given curve is an ellipse in the plane it lies in.

Key concepts

Parametric Equations

A parametric equation represents a set of equations in which one or more variables depend on an independent parameter. In our case, the parameter is typically represented by \(t\). In the context of the ellipse, parametric equations offer flexibility to describe the trajectory of points in a plane or space. For the tilted ellipse in the exercise, the position vector \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle\) is given with \(0 \leq t \leq 2\pi\).

• \(x(t) = \cos t\) represents the horizontal coordinate.
• \(y(t) = \sin t\) gives the vertical coordinate.
• \(z(t) = c \sin t\) provides the third dimension in space related to \(c\), a constant.

In parametric form, the ellipse is intuitively described as the pair of \(x(t)\) and \(y(t)\) trace points in the plane according to the parameter \(t\). This approach is powerful because it allows to express complex curves such as ellipses in simpler mathematical terms.

Plane Geometry

Plane geometry concerns shapes like circles, lines, and ellipses that lie in a flat surface. In this exercise, we demonstrate the curve of the position vector forms an ellipse on a plane. The statement begins with assuming the curve given by \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle\) is indeed on some plane. Thus, the critical task is to find an equation connecting \(x\), \(y\), and \(z\) that confirms the shape.

By differentiating the components, and using the parametric constraint \(x^2 + y^2 = 1\), we derive:

• Plane equation: \(z = c y\)
• Ellipse equation: \(x^2 + \left(\frac{z}{c}\right)^2 = 1\)

This final equation indeed represents an ellipse since it matches the general form \(\frac{x^2}{a^2} + \frac{z^2}{b^2} = 1\), confirming that the figure described by points \(x\), \(y\), and \(z\) is an ellipse on the plane \(z = c y\). Understanding the role of each term in this equation is pivotal in grasping its representation in plane geometry.

Differential Calculus

Differential calculus involves calculating the derivatives or rates of change. It is a vital part of proving the ellipse lies on a plane. By differentiating each part of the position vector \(\mathbf{r}(t)\) for the components \(x(t)\), \(y(t)\), and \(z(t)\), critical insights about the relationship between these values are gained.

• The derivative \(\frac{dx}{dt} = -\sin t\) provides the rate of change for \(x(t)\).
• \(\frac{dy}{dt} = \cos t\) shows how \(y(t)\) changes with \(t\).
• \(\frac{dz}{dt} = c \cos t\) indicates how the height \(z(t)\) evolves.

By analyzing \(\frac{dz}{dt}\) and \(\frac{dy}{dt}\), setting up a fraction such as \(\frac{dz/dt}{dy/dt} = \frac{c \cos t}{\cos t} = c\), eliminates \(\cos t\) and presents a direct link \(z = c y\). Differential calculus transforms the parameters into a useful plane equation. This mixture of derivatives and algebra results in simplifying the task to uncovering the plane equation. Understanding these concepts also aids in solving more intricate problems involving curves and multidimensional spaces.