/*! 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 10 Describe the graph of the equati... [FREE SOLUTION] | 91Ó°ÊÓ

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Describe the graph of the equation. $$ \mathbf{r}=2 \sin 3 t \mathbf{i}-2 \cos 3 t \mathbf{j} $$

Short Answer

Expert verified
It's a circle with radius 2 centered at the origin.

Step by step solution

01

Identify the Components

The vector function is given as \( \mathbf{r}=2 \sin 3t \mathbf{i}-2 \cos 3t \mathbf{j} \). This can be broken down into its components: \( x(t) = 2 \sin 3t \) and \( y(t) = -2 \cos 3t \).
02

Recognize the Parametric Form

This vector equation describes a parametric form with \( x = 2 \sin 3t \) and \( y = -2 \cos 3t \). These are typical forms for parametric equations of an ellipse or a circle.
03

Locate the Standard Circle Equation

To identify whether it's a circle, we need the form \( x^2 + y^2 = r^2 \). Substitute the parametric equations: \( (2 \sin 3t)^2 + (-2 \cos 3t)^2 = (2)^2 \).
04

Simplify the Equation

Calculate \( (2 \sin 3t)^2 + (-2 \cos 3t)^2 \): - \( 4 \sin^2 3t + 4 \cos^2 3t \). Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find: - \( 4(\sin^2 3t + \cos^2 3t ) = 4 \).
05

Conclude the Graph Type

Since \( x^2 + y^2 = 4 \) is consistent with \( r^2 = 4 \), the graph represents a circle with radius 2, centered at the origin.

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Key Concepts

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

Vector Functions
Vector functions are mathematical expressions that describe vector-valued quantities depending on one or more parameters. They are often used in space and plane representations. A vector function can be written in terms of its individual components. For example, the vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \) specifies a path in the plane defined by the changes in \( x(t) \) and \( y(t) \).
The given vector \( \mathbf{r} = 2 \sin 3t \mathbf{i} - 2 \cos 3t \mathbf{j} \) consists of two functions: \( x(t) = 2 \sin 3t \) and \( y(t) = -2 \cos 3t \). This representation indicates both the direction in which the point moves over time as \( t \) changes and its position at any given \( t \).
Understanding vector functions helps us visualize how quantities linked with a given parameter evolve, an essential skill in physics and engineering.
Circle Equation
The equation of a circle is a fundamental concept in geometry. It is given by the formula \( x^2 + y^2 = r^2 \), where \( r \) is the radius of the circle, translating the geometric property into an algebraic one. Circles are centered at the origin (0,0) unless stated otherwise.
In the provided solution, the parametric equations \( x(t) = 2 \sin 3t \) and \( y(t) = -2 \cos 3t \) describe a circle when plugged into this standard form. The expression \( (2 \sin 3t)^2 + (-2 \cos 3t)^2 \) simplifies to \( 4(\sin^2 3t + \cos^2 3t) \). Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we find that the equation becomes \( 4 = 4 \), indicating a circle with \( r = 2 \).
This parametric circle is centered at the origin, and traversed as the parameter \( t \) varies, demonstrating how the circle is formed dynamically by the vector function.
Trigonometric Identities
Trigonometric identities are crucial mathematical tools that simplify complex trigonometric expressions. One of the most pivotal identities in mathematics is \( \sin^2 \theta + \cos^2 \theta = 1 \). This identity helps in transforming parametric forms like those in vector functions into recognizable geometric shapes.
In our exercise, to transition from the parametric to the standard circle equation form, this identity was utilized. By applying it, the expression \( 4(\sin^2 3t + \cos^2 3t) \) reduces to 4, echoing the equality \( x^2 + y^2 = 4 \) or \( r^2 = 4 \).
This straightforward yet profound identity shows a seamless bridge from trigonometry to geometry, unraveling the ways in which parametric equations describe simple geometrical forms like circles. Understanding and applying these identities allow for the conversion of paths or loci described through vector functions into familiar and simplified geometric terms.

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Most popular questions from this chapter

Let \(\left.\mathbf{r}(t)=t^{2} \mathbf{i}+t^{3} \mathbf{j} \text { (see Figure } 12.3 .1\right) .\) Let \(\theta(t)\) be the angle between \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{i} .\) Show that $$ \theta(t) \rightarrow \pi \text { as } t \rightarrow 0^{-} \quad \text { and } \quad \theta(t) \rightarrow 0 \text { as } t \rightarrow 0^{+} $$

These exercises are concerned with the problem of creating a single smooth curve by piecing together two separate smooth curves. If two smooth curves \(C_{1}\) and \(C_{2}\) are joined at a point \(P\) to form a curve \(C,\) then we will say that \(C_{1}\) and \(C_{2}\) make a smooth transition at \(P\) if the curvature of \(C\) is continuous at \(P\). (a) Sketch the graph of the curve defined piecewise by \(y=x^{2}\) for \(x<0, y=x^{4}\) for \(x \geq 0\) (b) Show that for the curve in part (a) the transition at \(x=0\) is not smooth.

At time \(t=0\) a projectile is fired from a height \(h\) above level ground at an elevation angle of \(\alpha\) with a speed \(v .\) Let \(R\) be the horizontal distance to the point where the projectile hits the ground. (a) Show that \(\alpha\) and \(R\) must satisfy the equation $$ g\left(\sec ^{2} \alpha\right) R^{2}-2 v^{2}(\tan \alpha) R-2 v^{2} h=0 $$ (b) If \(g, h,\) and \(v\) are constant, then the equation in part (a) defines \(R\) implicitly as a function of \(\alpha\). Let \(R_{0}\) be the maximum value of \(R\) and \(\alpha_{0}\) the value of \(\alpha\) when \(R=R_{0} .\) Use implicit differentiation to find \(d R / d \alpha\) and show that $$ \tan \alpha_{0}=\frac{v^{2}}{g R_{0}} $$ [Hint: Assume that \(d R / d \alpha=0\) when \(R\) attains a maximum. (c) Use the results in parts (a) and (b) to show that $$ R_{0}=\frac{v}{g} \sqrt{v^{2}+2 g h} $$ and $$ \alpha_{0}=\tan ^{-1} \frac{v}{\sqrt{v^{2}+2 g h}} $$

Determine whether the statement is true or false. Explain your answer. If the line \(y=x\) is parametrized by the vector-valued function \(\mathbf{r}(t),\) then \(\mathbf{r}(t)\) is smooth.

In these exercises \(\mathbf{r}(t)\) is the position vector of a particle moving in the plane. Find the velocity, acceleration, and speed at an arbitrary time \(t .\) Then sketch the path of the particle together with the velocity and acceleration vectors at the indicated time \(t .\) $$ \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} ; t=2 $$

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