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Chapter 14: Problem 32

Trajectories on circles and spheres Determine whether the following trajectories lie on either a circle in \(\mathbb{R}^{2}\) or a sphere in \(\mathbb{R}^{3}\) centered at the origin. If so, find the radius of the circle or sphere, and show that the position vector and the velocity vector are everywhere orthogonal. $$\mathbf{r}(t)=\langle 3 \sin t, 5 \cos t, 4 \sin t\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
Answer: The given trajectory lies on a sphere in \(\mathbb{R}^3\) with a radius of 5, and the position vector and the velocity vector are everywhere orthogonal.

Step by step solution

01

Calculate the position vector|

To determine whether the trajectory lies on a circle or a sphere, we first need to find the position vector, which is given by $$\mathbf{r}(t)=\langle 3 \sin t, 5 \cos t, 4 \sin t\rangle$$
02

Calculate the squared length of the position vector|

Next, we calculate the squared length of the position vector, which will help us determine if it lies on a circle or a sphere. The squared length of the vector is given by $$|\mathbf{r}(t)|^2 = (3 \sin t)^2 + (5 \cos t)^2 + (4 \sin t)^2$$
03

Simplify the squared length of the position vector|

Simplify the squared length of the position vector as follows: $$|\mathbf{r}(t)|^2 = 9 \sin^2 t + 25 \cos^2 t + 16 \sin^2 t$$ $$|\mathbf{r}(t)|^2 = 25 \cos^2 t + 25 \sin^2 t$$ We can factor out \(25\) from the expression, yielding $$|\mathbf{r}(t)|^2 = 25(\cos^2 t + \sin^2 t)$$ Since \(\cos^2 t + \sin^2 t = 1\) by the Pythagorean identity, $$|\mathbf{r}(t)|^2 = 25$$
04

Determine if the trajectory lies on a circle or a sphere|

Since the squared length of the position vector is constant, the trajectory lies on a sphere in \(\mathbb{R}^3\). The radius of the sphere is the square root of the constant squared length, which is $$r=\sqrt{25}=5$$
05

Calculate the velocity vector|

To show that the position vector and the velocity vector are everywhere orthogonal, we first need to calculate the velocity vector. The velocity vector is the derivative of the position vector with respect to time, which is given by $$\mathbf{v}(t)=\frac{d\mathbf{r}(t)}{dt}=\frac{d}{dt} \langle 3 \sin t, 5 \cos t, 4 \sin t\rangle$$ $$\mathbf{v}(t)=\langle 3 \cos t, -5 \sin t, 4 \cos t\rangle$$
06

Calculate the dot product of the position and velocity vectors|

Now, we will calculate the dot product of the position and velocity vectors to show that they are orthogonal. If their dot product is zero, then they are orthogonal. The dot product of the two vectors is given by $$\mathbf{r}(t) \cdot \mathbf{v}(t) = (3 \sin t)(3 \cos t) + (5 \cos t)(-5 \sin t) + (4 \sin t)(4 \cos t)$$
07

Simplify the dot product|

Lets simplify the dot product of the two vectors: $$\mathbf{r}(t) \cdot \mathbf{v}(t) = 9 \sin t \cos t - 25 \cos t \sin t + 16 \sin t \cos t$$ $$\mathbf{r}(t) \cdot \mathbf{v}(t) = 0$$
08

Concluding if the vectors are orthogonal|

Since the dot product of the position vector \(\mathbf{r}(t)\) and the velocity vector \(\mathbf{v}(t)\) is zero for the given trajectory, it is verified that the position vector and the velocity vector are everywhere orthogonal. In conclusion, the given trajectory lies on a sphere in \(\mathbb{R}^3\) with a radius of 5 and the position vector and the velocity vector are everywhere orthogonal.

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Key Concepts

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

Position Vector
When working with trajectories on spheres, understanding the concept of a position vector is essential. A position vector refers to a vector that originates from a fixed point, usually the origin, and ends at a point representing the current location of an object in space. In our example, the position vector is \( \mathbf{r}(t)=\langle 3 \sin t, 5 \cos t, 4 \sin t\rangle \), which indicates the object's location in 3-dimensional space at any given time \(t\).

This vector changes over time as the object moves, tracing out its trajectory. To find out if this path lies on a sphere centered at the origin, we look for constancy in the vector's length, a clear indication that every position maintains the same distance from the center鈥攖hus defining a sphere.
Velocity Vector Orthogonal
A velocity vector represents the rate of change of the position vector with respect to time 鈥� essentially, how the position is changing at any given moment. It is calculated by taking the derivative of the position vector. In the context of our spherical trajectory, knowing whether the velocity vector remains orthogonal (at a right angle) to the position vector is key.

When a velocity vector is orthogonal to the position vector at all points on a trajectory, it means that the movement of the object is always tangent to the sphere, and the object is neither moving towards nor away from the center. This condition is critical for confirming that the path is indeed confined to the surface of the sphere.
Dot Product
An effective way to determine if two vectors are orthogonal is to compute their dot product. The dot product is a mathematical operation that multiplies two vectors to yield a scalar quantity. It is given by multiplying the corresponding components of two vectors and then summing those products. Mathematically, \( \mathbf{a}\cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

For vectors to be orthogonal, this dot product must equal zero. In our example, calculating the dot product of the position and velocity vectors results in a zero value, confirming that they are indeed orthogonal at every point of the trajectory on the sphere.
Pythagorean Identity
The Pythagorean identity is a fundamental trigonometric identity that expresses a relationship between the sine and cosine functions: \( \sin^2 t + \cos^2 t = 1 \). This property is derived from the Pythagorean theorem applied to the unit circle and is crucial in analyzing the trajectory on a sphere.

In our trajectory problem, applying the Pythagorean identity simplifies the expression for the square of the position vector's length, leading us to find that it is constant (\(|\mathbf{r}(t)|^2 = 25\)) for all \(t\). This constant magnitude indicates that the object is moving along a sphere with a fixed radius, mathematically tying together the spatial geometry and trigonometry at play.

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Most popular questions from this chapter

Consider the parameterized curves \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) and \(\mathbf{R}(t)=\langle f(u(t)), g(u(t)), h(u(t))\rangle\) where \(f, g, h,\) and \(u\) are continuously differentiable functions and \(u\) has an inverse on \([a, b]\) a. Show that the curve generated by \(\mathbf{r}\) on the interval \(a \leq t \leq b\) is the same as the curve generated by \(\mathbf{R}\) on \(u^{-1}(a) \leq t \leq u^{-1}(b)\left(\text { or } u^{-1}(b) \leq t \leq u^{-1}(a)\right)\) b. Show that the lengths of the two curves are equal. (Hint: Use the Chain Rule and a change of variables in the are length integral for the curve generated by \(\mathbf{R} .\) )

Curvature of \(e^{x}\) Find the curvature of \(f(x)=e^{x}\) and find the point at which it is a maximum. What is the value of the maximum curvature?

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\sqrt{t+4} \mathbf{i}+\frac{t}{t+1} \mathbf{j}-e^{-t^{2}} \mathbf{k}$$

The position functions of objects \(A\) and \(B\) describe different motion along the same path for \(t \geq 0\). a. Sketch the path followed by both \(A\) and \(B\). b. Find the velocity and acceleration of \(A\) and \(B\) and discuss the differences. c. Express the acceleration of A and \(B\) in terms of the tangential and normal components and discuss the differences. $$A: \mathbf{r}(t)=\langle\cos t, \sin t\rangle, B: \mathbf{r}(t)=\langle\cos 3 t, \sin 3 t\rangle$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the speed of an object is constant, then its velocity components are constant. b. The functions \(\mathbf{r}(t)=\langle\cos t, \sin t\rangle\) and \(\mathbf{R}(t)=\left\langle\sin t^{2}, \cos t^{2}\right\rangle\) generate the same set of points, for \(t \geq 0\). c. A velocity vector of variable magnitude cannot have a constant direction. d. If the acceleration of an object is a( \(t\) ) \(=0,\) for all \(t \geq 0,\) then the velocity of the object is constant. e. If you double the initial speed of a projectile, its range also doubles (assume no forces other than gravity act on the projectile). If If you double the initial speed of a projectile, its time of flight also doubles (assume no forces other than gravity). g. A trajectory with \(v(t)=a(t) \neq 0,\) for all \(t,\) is possible.
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