Problem 101 - Calculus Early Transcendentals

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. a. What is the equation of the plane \(P\) in which the curve lies? b. What is the angle between \(P\) and the \(x y\) -plane? c. Prove that the curve is an ellipse in \(P\).

Step-by-Step Solution

Answer

The equation of the plane P containing the parametric curve is cy = 0. The angle between the plane P and the xy-plane is π/2 radians, which means that the planes are perpendicular.

Step 1: Finding a normal vector and a point on the plane

We can find the normal vector to the plane \(P\) by taking the gradient of the function, \(z = c \sin t\). The gradient of the function is given by the vector \(\nabla z = \langle \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \rangle = \langle 0, c \rangle\). Pick a point on the curve, for example, when \(t=0\), we have \(\mathbf{r}(0) = \langle 1, 0, 0 \rangle\). This point also lies on the plane \(P\).

Step 2: Equation of the plane \(P\)

To find the equation of the plane \(P\), we use the point-normal form of the plane. Given a normal vector \(\mathbf{n} = \langle A, B, C \rangle\) and a point \(P_0 = \langle x_0, y_0, z_0 \rangle\) on the plane, the point-normal form of the plane is given by \(A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\). From the point and normal vector found in Step 1, we have the equation of the plane \(P\) as: \(0(x-1) + c(y-0) + 0(z-0) = 0\), which simplifies to \(cy = 0\).

Step 3: Finding the angle between plane \(P\) and the \(xy\)-plane

To find the angle between plane \(P\) and the \(xy\)-plane, we need to find the angle between their normal vectors. The normal vector to the \(xy\)-plane is \(\mathbf{n}_{xy} = \langle 0, 0, 1 \rangle\). The normal vector to the plane \(P\), found in step 1, is \(\mathbf{n}_P = \langle 0, c, 0 \rangle\). To find the angle between these planes, we calculate the cosine of the angle, \(\theta\), which is given by: $$\cos{\theta} = \frac{\mathbf{n}_{xy} \cdot \mathbf{n}_P}{\Vert \mathbf{n}_{xy}\Vert \Vert \mathbf{n}_P \Vert} = \frac{0 \cdot 0 + 0 \cdot c + 1 \cdot 0}{\sqrt{0^2+0^2+1^2}\sqrt{0^2+c^2+0^2}} = 0$$ This implies that the angle \(\theta\) between the two planes is \(\cos^{-1}{(0)} = \frac{\pi}{2}\). In other words, the two planes are perpendicular.

Step 4: Proving the curve is an ellipse in plane \(P\)

To prove the curve is an ellipse in plane \(P\), we first write the formula of the curve in terms of functions of x and y. Recall the parametric representation of the curve: \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle\). Now, express the curve in terms of \(x\) and \(y\): $$x = \cos t, \qquad y = \sin t, \qquad z = c \sin t$$ Notice that from \(x=\cos t\) and \(y=\sin t\), we have: $$x^2+y^2=\left(\cos t\right)^2+\left(\sin t\right)^2=1$$ This equation is the standard form of an ellipse in \(x\) and \(y\) coordinates, when in plane \(P\), as \(\frac{x^2}{1^2}+\frac{y^2}{1^2}=1\). It confirms the curve is an ellipse in plane \(P\).

Key Concepts

Ellipse

An ellipse is a geometric shape that forms when slicing an inclined plane across a cone. It appears as an elongated circle, characterized by its two distinct axes: the major and the minor axis. In the context of parametric curves and this exercise, the ellipse is described using trigonometric functions.

To understand how the given parametric curve describes an ellipse, we look at the equations involved. The curve \(\mathbf{r}(t) = \langle \cos t, \sin t, c \sin t \rangle \) represents a 3D path. The parameters \(x = \cos t \) and \(y = \sin t \) form a circular path in the projected 2D plane when isolated, satisfying \(x^2 + y^2 = 1\). This is the equation of a circle but also applies to an ellipse when interpreted within another plane. This is because an ellipse in a 2D plane can still be a circle depending on its plane's orientation.

Ellipses possess several properties that include smoothness and symmetry about two axes. Recognizing elliptical forms in various equations helps when dealing with higher-dimensional problems, especially when determining whether curves are restricted to certain planes or surfaces.

Plane Equation

A plane in three-dimensional space can be defined through the point-normal form, which is useful for various geometric purposes. When finding the equation of a plane in this exercise, we use the plane's normal vector and a point on the plane.

The general form of a plane equation is understood as \( Ax + By + Cz = D \), where \( A, B, \text{ and } C \) represent coefficients forming a normal vector \((A, B, C)\), and \(D\) is related to a point through which the plane passes. For a plane to be uniquely defined, one needs both elements: a point and a direction non-parallel to the surface.

In our exercise, the calculation led to the plane equation \(cy = 0\), implying that all points lie on this plane as long as \(y = 0\). This setup showcases how simplifying expressions provides insight into a plane's orientation and position in space, especially when concerning fixed coordinates or axes.

Angle Between Planes

Understanding the angle between two planes is essential in geometry, especially when dealing with multiple surfaces or analyzing spatial arrangements. The angle calculation involves utilizing normal vectors of the respective planes.

The formula for the angle \( \theta \) between two planes involves their normal vectors \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \), computed as:\[\cos{\theta} = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\Vert \mathbf{n}_1 \Vert \Vert \mathbf{n}_2 \Vert}\]The dot product reflects the magnitude of projection between two vectors, while the norms relate to vector lengths. In our exercise, the dot product between \( \langle 0, c, 0 \rangle \) of the plane \(P\), and \( \langle 0, 0, 1 \rangle \) of the \(xy\)-plane turns out to be zero, resulting in \( \theta = \frac{\pi}{2} \) radians, or 90 degrees.

This perpendicularity indicates the planes intersect at a right angle, explaining the spatial interaction within the given setup. Calculating angles between planes assists in determining how surfaces relate within broader structures.

Normal Vector

A normal vector is a crucial concept in geometry and vector calculus, often utilized in defining surfaces and angles between different elements. It represents a direction perpendicular to a plane, acting as an anchor for distinguishing orientation.

For plane equations, the normal vector \( \mathbf{n} = \langle A, B, C \rangle \) is derived directly from the coefficients in the plane's formula \(Ax + By + Cz = D\). It allows the identification of the plane's inclination relative to coordinate axes. Additionally, normal vectors aid in understanding and computing intersections, distances, and angles of planes within the space.

The step-by-step solution revealed the normal vector of plane \(P\) as \( \langle 0, c, 0 \rangle \). This vector indicates the plane's orientation by being parallel to the \(y\)-axis and devoid of components in the \(x\)- and \(z\)-directions. Comprehending normal vectors helps in making geometric reasoning precise, particularly when aligning various surfaces or curves in higher dimensions.