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

Yes, the curve $\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle$ lies on the surface of a sphere centered at the origin. This is proven by showing that the square of the distance between any point on the curve and the origin is constant and by graphing the curve, which visually demonstrates that it wraps around a sphere's surface.

Step-by-step solution

To graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\), you can either use a graphing calculator or a software like GeoGebra. Set the parameter \(t\) to vary from \(0\) to \(2\pi\) (since trigonometric functions' periods usually occur within this range). The resulting curve should look like a looping curve that wraps around the surface of a sphere. #Step 2: Prove the Curve Lies on a Sphere Centered at the Origin#

To prove that the curve lies on the surface of a sphere centered at the origin, we need to show that the distance between any point on the curve (\(\mathbf{r}(t)\)) and the origin is constant. The sphere's equation is given by \(x^2+y^2+z^2=R^2\), where \(R\) is the radius of the sphere. First, let's compute the square of the distance from a point on the curve to the origin. The coordinates of a point on the curve are given by \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\), so we have: \(x^2+y^2+z^2 = \left(\frac{1}{2} \sin 2 t\right)^2 + \left(\frac{1}{2}(1 - \cos 2 t)\right)^2 + (\cos t)^2\) Simplify the expression: \(x^2+y^2+z^2 = \frac{1}{4} (\sin^2 2t + 1 - 2\cos 2t +\cos^2 2t) + \cos^2 t\) Now, we need to check if this expression results in a constant value, which would indicate that the curve lies on the surface of the sphere. Using the trigonometric identities \(\sin^2a+\cos^2a=1\) and \(2\sin^2a=1-\cos 2a\), we can simplify the expression further: \(x^2+y^2+z^2 = \frac{1}{4} (4 - (1 + \cos 2t)) + \cos^2 t = 1 - \frac{1}{4} (1 + \cos 2t) + \cos^2 t\) Since \(2\sin^2t=1-\cos 2t\), we can express the aforementioned expression as: \(x^2+y^2+z^2 = 1 -\frac{1}{2}\cos 2t + \cos^2 t = 1 + \left(\cos t - \frac{\cos 2t}{2}\right)^2\) The expression given above is constant, as it depends only on \(\cos t\), and its variations lie between \(0\) and \(1\). Therefore, the curve \(\mathbf{r}(t)\) lies on the surface of a sphere centered at the origin.

Key concepts

Trigonometric Functions

Trigonometric functions are essential tools for describing phenomena involving periodicity and angles. They play a crucial role in defining curves and surfaces, especially in parametric equations. In this exercise, we see how trigonometric functions help create a three-dimensional path or curve using a parameter, often denoted as \( t \).

• Functions like \( \sin \) (sine) and \( \cos \) (cosine) repeat their values in cycles, usually achieving one full cycle between \( 0 \) and \( 2\pi \).
• These functions are used to describe oscillations, such as waves or rotational movements.
• In parametric equations, trigonometric functions allow us to control the behavior of each coordinate separately by adjusting the parameter \( t \).

For example, in our curve \( \mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t,\frac{1}{2}(1-\cos 2 t), \cos t\right\rangle \), the parameters are divided among \( \sin \) and \( \cos \) functions to distribute the path in three separate, but related dimensions. The amplitude, frequency and phase shifts these functions provide are instrumental in tracing unique paths, like the loop around a sphere seen in this problem.

Sphere

A sphere is a perfect symmetrical three-dimensional object with all points on its surface equidistant from its center. Understanding spheres is key to solving problems that deal with spatial relationships.

• The equation of a sphere centered at the origin is \( x^2 + y^2 + z^2 = R^2 \), where \( R \) is the radius.
• This equation denotes all points \((x, y, z)\) that maintain a constant distance \( R \) from the center \((0,0,0)\).
• In parametric terms, verifying whether a curve lies on a sphere involves checking if the expressions for \( x, y, \) and \( z \) satisfy the sphere's equation.

In this exercise, we showed how to use trigonometric identities to simplify and confirm that the curve maintains a constant distance from the origin, effectively proving it lies on a sphere. This highlights the harmony between algebraic representations and geometric interpretations, illustrating the precision and beauty of mathematical descriptions of shapes.

Curve Sketching

Curve sketching is the art of drawing a curve that represents a mathematical function or parametric equation. It involves visualizing a curve's trajectory in space, which can greatly aid in understanding complex equations.

• The main goal of curve sketching with parametric equations is to visualize what the movement or path looks like over a specific range of the parameter \( t \).
• Tools like graphing calculators or software such as GeoGebra can be helpful, allowing us to input equations and see the curves they create.
• Sketching helps in identifying key features such as where the curve loops, crosses itself, or how it is oriented in three-dimensional space.

For the curve \( \mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\), sketching shows us how this loop behaves around the origin, wrapping around to form sections of a sphere. The periodic nature of trigonometric functions results in a striking, symmetric trajectory. By visualizing these curves, students can gain a deeper intuitive understanding of the transformation from algebraic expressions to geometric forms in space.