Question

Graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) and prove that it lies on the surface of a sphere centered at the origin.

Short answer

In summary, the curve defined by the vector function \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) lies on a parabola in the xy-plane and on a sphere centered at the origin with a radius of \(\sqrt{\frac{1}{2}}\). This is shown by analyzing the relationships between the components of the vector function and proving that the sum of the squares of the components is a constant value, confirming that the curve lies on the surface of a sphere.

Step-by-step solution

Understand the components of the vector function

For the given vector function, let us first look at the components individually. These are given by: \(x(t) = \frac{1}{2}\sin 2t\) \(y(t) = \frac{1}{2}(1-\cos 2t)\) \(z(t) = \cos t\) Now let's try to understand the behavior and relations among these components. Notice that both \(x(t)\) and \(y(t)\) are functions of \(\sin 2t\) and \(\cos 2t\). This indicates a relation between them, which we will explore further.

Analyze the relation between x(t) and y(t)

To find the relation between \(x(t)\) and \(y(t)\), let's express them in terms of \(\sin 2t\) and \(\cos 2t\): \(x(t) = \frac{1}{2}\sin 2t\) \(y(t) = \frac{1}{2}(1-\cos 2t) = \frac{1}{2}(1-(1-2\sin^2 t)) = \frac{1}{2}(2\sin^2 t)\) Now, solve for \(\sin 2t\) from the equation \(x(t) = \frac{1}{2}\sin 2t\): \(\sin 2t = 2x(t)\) By substituting this relation into \(y(t)\), we can see their relationship more clearly: \(y(t) = \frac{1}{2}(2\sin^2 t) = x(t)^2\) This tells us that the points (x(t), y(t)) lie on a parabola.

Prove that the curve lies on a sphere centered at the origin

We want to show that \(x^2+y^2+z^2=R^2\). Since we already found a relation between \(x(t)\) and \(y(t)\), we can substitute this result into the equation: \(x^2+y^2+z^2 = (\frac{1}{2}\sin 2t)^2+(\frac{1}{2}(2\sin^2 t))^2+(\cos t)^2\) Now, let's simplify and combine the terms: \(x^2+y^2+z^2 = \frac{1}{4}\sin^2 2t + \frac{1}{2}\sin^4 t + \cos^2 t\) Using the identity \(\sin^2 a + \cos^2 a = 1\) and \(\sin^2 2t = 4\sin^2 t\cos^2 t\), we get: \(x^2+y^2+z^2 = \frac{1}{4}(4\sin^2 t\cos^2 t) + \frac{1}{2}\sin^4 t + (\sin^2 t+\cos^2 t - \sin^2 t)\) \(x^2+y^2+z^2 = \sin^2 t\cos^2 t + \frac{1}{2}\sin^4 t + \cos^2 t\) Factor out a \(\sin^2 t\) and then use the identity \(\sin^2 t = 1 - \cos^2 t\): \(x^2+y^2+z^2 = (1-\cos^2 t)\cos^2 t + \frac{1}{2}(1-\cos^2 t)^2 + \cos^2 t\) Simplify through algebra: \(x^2+y^2+z^2 = \cos^2 t - \cos^4 t + \frac{1}{2}(\cos^4 t - 2\cos^2 t + 1) + \cos^2 t\) \(x^2+y^2+z^2 = \cos^2 t - \cos^4 t + \frac{1}{2}\cos^4 t - \cos^2 t + \frac{1}{2} + \cos^2 t\) Cancel out terms and simplify: \(x^2+y^2+z^2 = \frac{1}{2} -\frac{1}{2}\cos^4 t\) It becomes clear that the sum of the squares of the components is indeed constant, \(\frac{1}{2}\), which verifies that the curve lies on the surface of a sphere with radius \(\sqrt{\frac{1}{2}}\) centered at the origin.

Key concepts

Parametric Equations

Parametric equations allow us to represent a curve by expressing the coordinates of the points on the curve as functions of a single variable, often denoted as \( t \). In the given exercise, the vector function \( \mathbf{r}(t) = \left\langle \frac{1}{2} \sin 2t, \frac{1}{2}(1 - \cos 2t), \cos t \right\rangle \) parametrizes a 3D curve.
Understanding parametric equations involves dissecting the vector components:

• \( x(t) = \frac{1}{2} \sin 2t \)
• \( y(t) = \frac{1}{2}(1 - \cos 2t) \)
• \( z(t) = \cos t \)

Here, each coordinate \( x, y, \text{and } z \) is expressed as a distinct function of \( t \). This way, as \( t \) progresses, these functions together trace out a path in three-dimensional space. Parametric equations are widely used in graphics because they offer flexibility in describing a variety of curves and shapes, providing an intuitive way to morph and animate lines and surfaces.

Spheres

A sphere in mathematics is a perfectly symmetrical three-dimensional shape where every point on the surface is equidistant from the center. The most common equation for a sphere with a radius \( R \) and centered at the origin is the sum of squares:
\[ x^2 + y^2 + z^2 = R^2 \]
In this context, to show that the parametric curve lies on a sphere, students are tasked with proving that:
\( x^2 + y^2 + z^2 = R^2 \)
Using the derived formulas in our step-by-step solution, we found that:

• \( x^2 = \left( \frac{1}{2} \sin 2t \right)^2 \)
• \( y^2 = \left( \frac{1}{2} (1 - \cos 2t) \right)^2 \)
• \( z^2 = (\cos t)^2 \)

By substituting these into the sphere's equation and simplifying, it is verified that the parametric equation describes a curve that sits on a sphere with a specific radius. This discovery ties into both geometric intuition and algebraic verification, essential skills in understanding complex mathematical shapes.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are universally true for all accepted values. In solving mathematical problems, especially those involving parametric equations as in our exercise, these identities are crucial for simplifying expressions.
Here are some basic identities leveraged in our solution:

• \( \sin^2 a + \cos^2 a = 1 \)
• \( \sin 2a = 2 \sin a \cos a \)
• Expressing \( \cos^2 t \) using \( \sin^2 t = 1 - \cos^2 t \)

These identities help to reduce complex trigonometric expressions into simpler ones, making it easier to verify conditions such as the sum of squares for the sphere's equation. The intuitive understanding of these identities aids in navigating through multi-step algebraic manipulations seamlessly.