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Sketch the circle \(\mathbf{r}(t)=\cos t \mathbf{i}-\sin t \mathbf{j}\), and in each part draw the vector with its correct length. (a) \(\mathbf{r}^{\prime}(\pi / 4)\) (b) \(\mathbf{r}^{\prime \prime}(\pi)\) (c) \(\mathbf{r}(2 \pi)-\mathbf{r}(3 \pi / 2)\)

Short Answer

Expert verified
(a) The vector is \((-\sqrt{2}/2, -\sqrt{2}/2)\) with length 1. (b) The vector is \(\mathbf{i}\) with length 1. (c) The vector is \((1, -1)\) with length \(\sqrt{2}\).

Step by step solution

01

Understanding the Given Vector Function

The vector function \( \mathbf{r}(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \) represents a circle in the xy-plane. As \( t \) varies, the point \( (\cos t, -\sin t) \) traces a circle of radius 1 centered at the origin.
02

Find \( \mathbf{r}^{\prime}(t) \)

Calculate the derivative of \( \mathbf{r}(t) \): \( \mathbf{r}^{\prime}(t) = \frac{d}{dt}[\cos t \mathbf{i} - \sin t \mathbf{j}] = -\sin t \mathbf{i} - \cos t \mathbf{j} \).
03

Evaluate \( \mathbf{r}^{\prime}(\pi/4) \)

Substitute \( t = \pi/4 \) into \( \mathbf{r}^{\prime}(t) \): \( \mathbf{r}^{\prime}(\pi/4) = -\sin(\pi/4) \mathbf{i} - \cos(\pi/4) \mathbf{j} = -\frac{\sqrt{2}}{2} \mathbf{i} - \frac{\sqrt{2}}{2} \mathbf{j} \).
04

Determine the Length of \( \mathbf{r}^{\prime}(\pi/4) \)

The length (magnitude) of vector \( \mathbf{r}^{\prime}(\pi/4) \) is \( \sqrt{(-\frac{\sqrt{2}}{2})^2 + (-\frac{\sqrt{2}}{2})^2} = 1 \).
05

Find \( \mathbf{r}^{\prime\prime}(t) \)

Calculate the second derivative of \( \mathbf{r}(t) \): \( \mathbf{r}^{\prime\prime}(t) = -\cos t \mathbf{i} + \sin t \mathbf{j} \).
06

Evaluate \( \mathbf{r}^{\prime\prime}(\pi) \)

Substitute \( t = \pi \) into \( \mathbf{r}^{\prime\prime}(t) \): \( \mathbf{r}^{\prime\prime}(\pi) = -\cos(\pi) \mathbf{i} + \sin(\pi) \mathbf{j} = \mathbf{i} \).
07

Determine the Length of \( \mathbf{r}^{\prime\prime}(\pi) \)

The length of vector \( \mathbf{r}^{\prime\prime}(\pi) \) is \( \sqrt{1^2 + 0^2} = 1 \).
08

Evaluate \( \mathbf{r}(2\pi) \) and \( \mathbf{r}(3\pi/2) \)

Calculate \( \mathbf{r}(2\pi) \): \( (\cos(2\pi), -\sin(2\pi)) = (1, 0) \). \( \mathbf{r}(3\pi/2) \): \( (\cos(3\pi/2), -\sin(3\pi/2)) = (0, 1) \).
09

Compute \( \mathbf{r}(2\pi) - \mathbf{r}(3\pi/2) \)

Subtract the vectors: \( \mathbf{r}(2\pi) - \mathbf{r}(3\pi/2) = (1, 0) - (0, 1) = (1, -1) \).
10

Determine the Length of \( \mathbf{r}(2\pi) - \mathbf{r}(3\pi/2) \)

The length of the vector \( \mathbf{r}(2\pi) - \mathbf{r}(3\pi/2) \) is \( \sqrt{1^2 + (-1)^2} = \sqrt{2} \).

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Key Concepts

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

Parametric Equations
Parametric equations are a way to describe a curve by defining coordinates as functions of a parameter, usually denoted as \( t \). Instead of describing a path in the form of \( y = f(x) \), parametric equations provide both \( x(t) \) and \( y(t) \), which describe how each component changes with respect to \( t \).

For instance, in our exercise, the vector function \( \mathbf{r}(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \) uses \( t \) as the parameter. As \( t \) changes, \( x(t) = \cos t \) and \( y(t) = -\sin t \) generate a circle centered at the origin with a radius of 1.

Since these components are trigonometric functions, they smoothly describe the circular motion of a point. Parametric equations are especially useful in physics and engineering when analyzing paths and trajectories.
Derivatives of Vector Functions
The derivative of a vector function relates to how the vector changes as its parameter changes. If you have a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \), finding its derivative requires calculating the derivative of each component with respect to the parameter \( t \).

In our problem, the derivative \( \mathbf{r}^{\prime}(t) = -\sin t \mathbf{i} - \cos t \mathbf{j} \) was computed from \( \mathbf{r}(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \). This derivative vector represents the "velocity vector" of the moving point on the circle, showing its instant rate of change at any given time \( t \).

The second derivative \( \mathbf{r}^{\prime\prime}(t) = -\cos t \mathbf{i} + \sin t \mathbf{j} \) provides even deeper insight. This "acceleration vector" indicates how the "velocity vector" changes. In this context, it describes the circular motion's centripetal acceleration.
Unit Circle
The unit circle is a fundamental concept in mathematics representing a circle with a radius of 1, centered at the origin of the coordinate system. It serves as a model for understanding angles, trigonometric functions, and many properties of circles.

For our scenario, the parametric equations \( x(t) = \cos t \) and \( y(t) = -\sin t \) describe motion precisely along the unit circle. As \( t \) travels from 0 to \( 2\pi \), the point traverses the entire circumference of the circle once.

The coordinates on the unit circle directly relate to sine and cosine values, making it invaluable for conversions between radians and coordinates. When addressing problems involving circular paths or oscillations, the unit circle often plays a crucial role in simplifying calculations and visualizing movements.
Motion in the Plane
Motion in the plane deals with how an object or particle moves along a flat, two-dimensional surface. By using vectors, we can encapsulate both the direction and magnitude of motion.

In the context of our vector function \( \mathbf{r}(t) = \cos t \mathbf{i} - \sin t \mathbf{j} \), this represents a specific case of uniform circular motion in the xy-plane. Here, the point steadily moves around the circle. The velocity vector \( \mathbf{r}^{\prime}(t) \) and acceleration vector \( \mathbf{r}^{\prime\prime}(t) \) describe the dynamics of rotation and position change.

Such analyses allow for understanding more intricate movements like projectile paths, rotations, or even simple oscillating systems like pendulums.
  • Velocity vectors tell us the speed and direction of motion at any instant.
  • Acceleration vectors provide insights into changes in motion, helping predict the future trajectory of the moving body.
Overall, studying motion in the plane gives a meaningful look into the mechanics of everyday phenomena, from machinery to natural occurrences.

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Most popular questions from this chapter

Determine whether the statement is true or false. Explain your answer. The binormal vector \(\mathbf{B}(t)\) to the graph of a vector-valued function \(\mathbf{r}(t)\) in 3 -space is the dot product of unit tangent and unit normal vectors, \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\).

As illustrated in the accompanying figure, copper cable with a diameter of \(\frac{1}{2}\) inch is to be wrapped in a circular helix around a cylinder that has a 12 -inch diameter. What length of cable (measured along its centerline) will make one complete turn around the cylinder in a distance of 20 inches (between centerlines) measured parallel to the axis of the cylinder?

(a) Sketch the graph of \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}\). Show that \(\mathbf{r}(t)\) is a smooth vector-valued function but the change of parameter \(t=\tau^{3}\) produces a vector-valued function that is not smooth, yet has the same graph as \(\mathbf{r}(t)\) (b) Examine how the two vector-valued functions are traced, and see if you can explain what causes the problem.

The evolute of a smooth parametric curve \(C\) in 2 -space is the curve formed from the centers of curvature of \(C\). The accompanying figure shows the ellipse \(x=3 \cos t\) \(y=2 \sin t(0 \leq t \leq 2 \pi)\) and its evolute graphed together. (a) Which points on the evolute correspond to \(t=0\) and \(t=\pi / 2 ?\) (b) In what direction is the evolute traced as \(t\) increases from 0 to \(2 \pi\) ? (c) What does the evolute of a circle look like? Explain your reasoning.

Find the velocity, speed, and acceleration at the given time \(t\) of a particle moving along the given curve. I" $$ x=2 \cos t, y=2 \sin t, z=t ; t=\pi / 4 $$

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