/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Path on a sphere Show that the f... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 33

Path on a sphere Show that the following trajectories lie on a sphere centered at the origin, and find the radius of the sphere. $$r(t)=\left\langle\frac{5 \sin t}{\sqrt{1+\sin ^{2} 2 t}}, \frac{5 \cos t}{\sqrt{1+\sin ^{2} 2 t}}, \frac{5 \sin 2 t}{\sqrt{1+\sin ^{2} 2 t}}\right\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
If so, what is the radius of the sphere? Answer: Yes, the trajectory lies on a sphere centered at the origin with a radius of 5.

Step by step solution

01

Write the components of the trajectory vector

The given trajectory vector $$r(t)=\left\langle\frac{5 \sin t}{\sqrt{1+\sin ^{2} 2 t}}, \frac{5 \cos t}{\sqrt{1+\sin ^{2} 2 t}}, \frac{5 \sin 2 t}{\sqrt{1+\sin ^{2} 2 t}}\right\rangle$$ has the following components: $$x = \frac{5 \sin t}{\sqrt{1+\sin ^{2} 2 t}}$$ $$y = \frac{5 \cos t}{\sqrt{1+\sin ^{2} 2 t}}$$ $$z = \frac{5 \sin 2 t}{\sqrt{1+\sin ^{2} 2 t}}$$ Now, we will substitute these components into the equation of a sphere to see if it simplifies to a constant value.
02

Substitute the components into the sphere equation

Substitute the components \(x, y, z\) into the sphere equation: $$\left(\frac{5 \sin t}{\sqrt{1+\sin ^{2} 2 t}}\right)^2 + \left(\frac{5 \cos t}{\sqrt{1+\sin ^{2} 2 t}}\right)^2 + \left(\frac{5 \sin 2 t}{\sqrt{1+\sin ^{2} 2 t}}\right)^2 = R^2$$
03

Simplify the equation

Simplify the equation to see if it represents a sphere: $$\frac{25 \sin^2 t}{1+\sin^2 2t} + \frac{25 \cos^2 t}{1+\sin^2 2t} + \frac{25 \sin^2 2t}{1+\sin^2 2t} = R^2$$ Notice that the denominators of all the fractions are the same. Combine the numerators: $$\frac{25 (\sin^2 t + \cos^2 t + \sin^2 2t)}{1+\sin^2 2t} = R^2$$ Since \(\sin^2 t + \cos^2 t = 1\), we can substitute this into the equation: $$\frac{25 (1 + \sin^2 2t)}{1+\sin^2 2t} = R^2$$ Now, divide both sides by \((1+\sin^2 2t)\): $$25 = R^2$$
04

Calculate the radius

Since we have found that \(25 = R^2\), we can determine the radius of the sphere. Take the square root of both sides: $$R = \sqrt{25} = 5$$ The radius of the sphere is 5. Conclusion: The given trajectory $$r(t)=\left\langle\frac{5 \sin t}{\sqrt{1+\sin ^{2} 2 t}}, \frac{5 \cos t}{\sqrt{1+\sin ^{2} 2 t}}, \frac{5 \sin 2 t}{\sqrt{1+\sin ^{2} 2 t}}\right\rangle$$ lies on a sphere centered at the origin with a radius of 5.

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Key Concepts

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

Trajectories
In mathematics, a **trajectory** refers to the path that a moving object follows through space as a function of time. The concept of trajectories is key within physics and mathematics when it comes to analyzing the motion of objects, especially in the context of parametric equations. In this exercise, we explore the trajectory of a point defined by the vector function \(r(t)\).
This vector function incorporates trigonometric terms, indicating that it traces a path within a three-dimensional space. The specific form of \(r(t)\) in this problem includes sine and cosine functions which are periodic, allowing the path to loop back upon itself within a specified interval \(0 \leq t \leq 2\pi\).
Through substitution and simplification, you can determine if this trajectory lies on a sphere by matching it to the general equation of a sphere. Understanding how such a path behaves helps in visualizing and validating the geometric representation it implies.
Spherical Coordinates
In three-dimensional space, **spherical coordinates** provide a way of representing points in terms of a radius and two angles. This system is akin to using latitude and longitude to specify locations on Earth. For any given point in space, spherical coordinates utilize:
  • \( r \): The radial distance from the origin.
  • \( \theta \): The angle in the \(xy\)-plane from the positive \(x\)-axis.
  • \( \phi \): The angle from the positive \(z\)-axis.

In the context of the exercise, while the representation \(r(t)\) is not explicitly in spherical coordinates, the usage of trigonometric functions to describe each component hints at a natural coordinate transformation into a spherical form. The radius of 5 derived in the solution correlates with the constant radial distance one would expect if these points lay on the surface of a sphere. Understanding spherical coordinates is essential for grasping how this trajectory fits within the three-dimensional space.
Equation of a Sphere
The **equation of a sphere** in three dimensions is a mathematical representation depicting the set of all points that maintain a fixed distance, the radius, from a central point. The general form of this equation is:
\[ x^2 + y^2 + z^2 = R^2 \] where \((x, y, z)\) are the coordinates of any point on the sphere, and \(R\) is the radius.
In the exercise, the trajectory defined by \(r(t)\) aimed to confirm if it lies on a sphere. By equating the components of \(r(t)\) to \(x\), \(y\), and \(z\), they were collectively squared and summed. Simplification led to a constant value of 25, indicating that \(x^2 + y^2 + z^2 = 25 = R^2\).
Taking the square root of 25 revealed the radius \(R = 5\), confirming that the path does indeed lie on a sphere centered at the origin, providing a key insight into the relationship between the trajectory and three-dimensional geometry.

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