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Curves on spheres 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

Expert verified
The specific equation of the sphere the given curve lies on is: x^2 + y^2 + z^2 = 1/2.

Step by step solution

01

Equation of a Sphere

A sphere centered at the origin with radius \(a\) can be represented by the equation: $$x^2 + y^2 + z^2 = a^2$$ Our goal is to show that the given curve, when plugged into this equation, results in a true statement for any value of \(t\).
02

Plug the Curve into the Equation of a Sphere

We will now plug the curve into the equation of a sphere and simplify the expression: \begin{align*} \left(\frac{1}{2} \sin 2 t\right)^2 + \left(\frac{1}{2}(1-\cos 2 t)\right)^2 + (\cos t)^2 &= a^2 \\ \left(\frac{1}{4}\sin^2 2t\right) + \left(\frac{1}{4} - \frac{1}{2}\cos 2t + \frac{1}{4}\cos^2 2t\right) + \cos^2 t &= a^2 \end{align*}
03

Simplify the Equation

Now we will use the trigonometric identities \(\sin^2 x + \cos^2 x = 1\) and \(\sin^2 x = \frac{1}{2}(1 - \cos 2x)\) to simplify the equation: \begin{align*} \frac{1}{2}(1 - \cos 2t) + \frac{1}{2} &= a^2 \\ 1 - \cos 2t &= 2a^2 - 1 \\ 2a^2 &= 2 - \cos 2t \\ a^2 &= 1 - \frac{1}{2}\cos 2t \end{align*}
04

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

Since the equation is in the form \(a^2 = 1 - \frac{1}{2}\cos 2t\), the equation indeed represents a sphere centered at the origin, and the curve lies on the surface of the sphere for any value of \(t\). Thus, we have proved that the curve lies on the surface of a sphere centered at the origin.
05

Find the specific equation of the sphere the curve lies on

By setting \(t=0\), we can find the equation of the sphere. $$ a^2 = 1 - \frac{1}{2}\cos(0) = 1 - \frac{1}{2} = \frac{1}{2} $$ So, the specific equation of the sphere the curve lies on is given by: $$ x^2 + y^2 + z^2 = \frac{1}{2} $$ To summarize, the given curve lies on the surface of a sphere centered at the origin, and the specific equation of the sphere is \(x^2 + y^2 + z^2 = \frac{1}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Trigonometric identities are essential tools in simplifying and solving mathematical equations, especially when dealing with curves or transformations. They help in expressing trigonometric functions in various forms to make calculations easier. For the curve on a sphere, we utilize identities such as:
  • \( \sin^2 x + \cos^2 x = 1 \): This identity is crucial for simplifying terms involving \( \sin \) and \( \cos \).
  • \( \sin^2 x = \frac{1}{2} (1 - \cos 2x) \): This identity is used to transform squared sine terms into cosines, making certain calculations more straightforward.
These identities allow us to manipulate and simplify the expressions obtained in the steps to show that the given curve fits the equation of a sphere. They reduce complex trigonometric expressions to simpler forms, aiding in proving mathematical properties and relationships.
Equation of a Sphere
The equation of a sphere centered at the origin is a fundamental concept in geometry and algebra. It is given by the formula:
  • \( x^2 + y^2 + z^2 = a^2 \)
Here, \( a \) represents the radius of the sphere. The basic idea is that any point \((x, y, z)\) on the surface of the sphere has the same distance from the origin, which is the radius.
The curve in the problem is represented by parametric equations, and showing that the curve satisfies the sphere's equation proves that it lies on its surface. Plugging the curve's components into the sphere's equation ensures that each point (as a function of \( t \)) maintains a constant projected distance from the origin, matching the form of the sphere's equation.
Parametric Equations
Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as parameters. For curves on surfaces like spheres, they offer a way to represent paths in a three-dimensional space more flexibly.Given the curve \( \mathbf{r}(t) = \left\langle \frac{1}{2} \sin 2t, \frac{1}{2}(1-\cos 2t), \cos t \right\rangle \), each component (x, y, z) is expressed as functions of \( t \). This representation allows us to track the curve's position on the sphere dynamically as \( t \) varies.Using parametric equations, complex surfaces and paths can be analyzed and understood more easily. They also facilitate substitution into other equations, such as the equation of a sphere, to prove certain geometric properties, as demonstrated in verifying the curve's position on the sphere's surface.

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Most popular questions from this chapter

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0}\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}.\) $$\mathbf{r}(t)=\langle\sqrt{2 t+1}, \sin \pi t, 4\rangle ; t_{0}=4$$

Evaluate the following definite integrals. $$\int_{0}^{2} t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) d t$$

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(\frac{d^{2} s}{d t^{2}}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). Show that if an object decelerates in the sense that \(\frac{d^{2} s}{d t^{2}}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N} .\)

Computational formula for B Use the result of part (a) of Exercise 82 and the formula for \(\kappa\) to show that $$\mathbf{B}=\frac{\mathbf{v} \times \mathbf{a}}{|\mathbf{v} \times \mathbf{a}|}$$

Evaluate the following definite integrals. $$\int_{1 / 2}^{1}\left(\frac{3}{1+2 t} \mathbf{i}-\pi \csc ^{2}\left(\frac{\pi}{2} t\right) \mathbf{k}\right) d t$$

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