/*! 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 7 Sketch the circle \(\mathbf{r}(t... [FREE SOLUTION] | 91Ó°ÊÓ

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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) \(-\frac{\sqrt{2}}{2} \mathbf{i} + \frac{\sqrt{2}}{2} \mathbf{j}\); (b) \(\mathbf{i}\); (c) \(\mathbf{i} + \mathbf{j}\).

Step by step solution

01

Understand the Circle

The vector function \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \) describes a circle in the Cartesian plane with a radius of 1. The \( \cos t \) and \( \sin t \) terms represent the coordinates \( x \) and \( y \), respectively, as functions of the parameter \( t \).
02

First Derivative for Part (a)

To find \( \mathbf{r}^{\prime}(t) \), differentiate \( \mathbf{r}(t) \) with respect to \( t \). The derivative is \( \mathbf{r}^{\prime}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \).
03

Evaluate First Derivative at \( t = \pi / 4 \)

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

Second Derivative for Part (b)

Differentiate \( \mathbf{r}^{\prime}(t) \) to find \( \mathbf{r}^{\prime\prime}(t) \), yielding \( \mathbf{r}^{\prime\prime}(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \).
05

Evaluate Second Derivative at \( t = \pi \)

Substitute \( t = \pi \) into \( \mathbf{r}^{\prime\prime}(t) \): \( \mathbf{r}^{\prime\prime}(\pi) = -\cos(\pi) \mathbf{i} - \sin(\pi) \mathbf{j} = \mathbf{i} \).
06

Calculate \( \mathbf{r}(2\pi) \) and \( \mathbf{r}(3\pi/2) \) for Part (c)

Find \( \mathbf{r}(2\pi) = \cos(2\pi) \mathbf{i} + \sin(2\pi) \mathbf{j} = \mathbf{i} \) and \( \mathbf{r}(3\pi/2) = \cos(3\pi/2) \mathbf{i} + \sin(3\pi/2) \mathbf{j} = -\mathbf{j} \).
07

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

Subtract to find the difference: \( \mathbf{r}(2\pi) - \mathbf{r}(3\pi/2) = \mathbf{i} - (-\mathbf{j}) = \mathbf{i} + \mathbf{j} \).

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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 powerful way to represent shapes and paths in mathematics. Instead of having one equation relating two variables, parametric equations describe each variable independently with respect to a third variable, called a parameter.
In our exercise, the vector function \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \) describes a circle in the Cartesian plane. Here, the parameter is \( t \), which can be thought of as an angle or a time variable. This parameter allows us to model the coordinates of points along the circle. The x-coordinate is given by \( \cos t \), and the y-coordinate by \( \sin t \).
These parametric equations make it easy to understand how objects move through space along a path. When \( t \) varies from 0 to \( 2\pi \), the vector traces a complete circle with radius 1 centered at the origin.
Differentiation
Differentiation is a mathematical technique used to find rates of change. In vector calculus, we can differentiate vector functions with respect to a parameter to understand the changes in motion.
For our circle, differentiating \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \) with respect to \( t \) gives us the velocity vector \( \mathbf{r}^{\prime}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \).

  • This velocity vector \( \mathbf{r}^{\prime}(t) \) tells us the direction and speed with which a point moves around the circle.
  • Further differentiating, we find \( \mathbf{r}^{\prime\prime}(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \), which is the acceleration vector.


These results demonstrate key aspects of circular motion. The velocity is perpendicular to the radius, and the acceleration points toward the center, illustrating centripetal acceleration.
Vector Functions
Vector functions are essential for describing complex paths and geometric shapes in space. A vector function assigns a vector to each value of a parameter. These vectors can represent quantities that have both a magnitude and a direction, such as position, velocity, and acceleration.
In this exercise, we look at the vector function \( \mathbf{r}(t) \) and its derivatives.
  • Each derivative helps us understand different elements of the system: the first derivative gives us the velocity, and the second gives us the acceleration.
  • This multi-faceted representation allows modeling of a range of phenomena, from simple geometric shapes to the motion of objects in physics.
Through such an approach, one can analyze how changes in time or other parameters affect the motion or position in a spatial domain.

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Most popular questions from this chapter

At what point(s) does \(4 x^{2}+9 y^{2}=36\) have minimum radius of curvature?

Discuss some of the advantages of parametrizing a curve by arc length.

Determine whether the statement is true or false. Explain your answer. If the smooth vector-valued function \(\mathbf{r}(s)\) is parametrized by arc length and \(\mathbf{r}^{\prime \prime}(s)\) is defined, then \(\mathbf{r}^{\prime}(s)\) and \(\mathbf{r}^{\prime \prime}(s)\) are orthogonal vectors.

Consider the various forces that a passenger in a car would sense while traveling over the crest of a hill or around a curve. Relate these sensations to the tangential and normal vector components of the acceleration vector for the car's motion. Discuss how speeding up or slowing down (e.g., doubling or halving the car's speed) affects these components.

Assume that \(s\) is an arc length parameter for a smooth vector-valued function \(\mathbf{r}(s)\) in 3 -space and that \(d \mathbf{T} / d s\) and \(d \mathbf{N} / d s\) exist at each point on the curve. (This implies that \(d \mathbf{B} / d s\) exists as well, since \(\mathbf{B}=\mathbf{T} \times \mathbf{N}\).) (a) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{B}(s)\). (b) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{T}(s)\). [Hint: Use the fact that \(\mathbf{B}(s)\) is perpendicular to both \(\mathbf{T}(s)\) and \(\mathbf{N}(s)\) and differentiate \(\mathbf{B} \cdot \mathbf{T}\) with respect to \(s .]\) (c) Use the results in parts (a) and (b) to show that \(d \mathbf{B} / d s\) is a scalar multiple of \(\mathbf{N}(s)\). The negative of this scalar is called the torsion of \(\mathbf{r}(s)\) and is denoted by \(\tau(s)\). Thus, $$ \frac{d \mathbf{B}}{d s}=-\tau(s) \mathbf{N}(s) $$ (d) Show that \(\tau(s)=0\) for all \(s\) if the graph of \(\mathbf{r}(s)\) lies in a plane. [Note: For reasons that we cannot discuss here, the torsion is related to the "twisting" properties of the curve, and \(\tau(s)\) is regarded as a numerical measure of the tendency for the curve to twist out of the osculating plane.]

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