/*! 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 55 Verify that the curve \(\mathbf{... [FREE SOLUTION] | 91Ó°ÊÓ

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Verify that the curve \(\mathbf{r}(t)\) lies on the given surface. Give the name of the surface. $$r(t)=\langle 0,2 \cos t, 3 \sin t\rangle ; x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}=1$$

Short Answer

Expert verified
Answer: Yes, the curve lies on the surface, and the name of the surface is an elliptical cone.

Step by step solution

01

Identify the components of the curve

The given vector is \(\mathbf{r}(t) = \langle 0, 2 \cos t, 3 \sin t\rangle\). We see that the individual components of the vector are: \(x(t) = 0\) \(y(t) = 2 \cos t\) \(z(t) = 3 \sin t\)
02

Plug the components into the surface equation

Now, let's substitute the above components into the given surface equation: \(x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1\) Plugging in the values, we get: \((0)^2 + \frac{(2 \cos t)^2}{4} + \frac{(3 \sin t)^2}{9} = 1\)
03

Simplify the equation

Let's simplify the equation: \(0 + \frac{4 \cos^2 t}{4} + \frac{9 \sin^2 t}{9} = 1\) Canceling out the constants, we get: \(\cos^2 t + \sin^2 t = 1\)
04

Verify if the equation holds true

We know from trigonometry that \(\cos^2 t + \sin^2 t = 1\) for all values of \(t\). Since this equation holds true, it means that the given curve, \( \mathbf{r}(t) = \langle 0, 2 \cos t, 3 \sin t \rangle\), lies on the surface \(x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1\).
05

Identify the name of the surface

The given surface equation, \(x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1\), represents an elliptical cone. This can be seen by comparing the equation to the standard form of an elliptical cone: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\), where \(a = 1, b = 2\), and \(c = 3\).

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Key Concepts

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

Vector Functions
When we talk about vector functions, we refer to functions that have vectors as their outputs. In calculus, these functions play a significant role in representing curves in a multidimensional space. Vector functions are often notated as \( \mathbf{r}(t) \), where \( t \) is the parameter and \( \mathbf{r} \) specifies the position vector of points on the curve.

For instance, in our exercise, the vector function is \( \mathbf{r}(t) = \langle 0, 2 \cos t, 3 \sin t\rangle \). This function describes a curve by giving the x, y, and z coordinates of each point as functions of the parameter \( t \). The simplicity and elegance of vector functions allow us to analyze curves' behavior by examining their components, which are simple trigonometric functions in this case.
Trigonometry in Calculus
Integrating trigonometry in calculus is essential for understanding the behavior of curves and surfaces. Key trigonometric identities, such as \( \cos^2 t + \sin^2 t = 1 \), are frequently used for simplifying problems. This fundamental identity expresses the Pythagorean theorem in the unit circle, and it is indispensable in calculus when solving equations involving trigonometric functions.

As we saw in the solution, after substituting the parametric equations into the surface equation, this identity helped us simplify the expression and verify that the curve indeed lies on the elliptical cone. In many calculus problems, identifying and applying the right trigonometric identity can be the difference between an intractable expression and a solvable equation.
Parametric Equations
The concept of parametric equations is especially crucial when studying curves and surfaces. Instead of expressing a curve as a function of x and y directly, parametric equations introduce an independent parameter that defines both x and y separately. In three-dimensional space, this extends to x, y, and z.

In our example, the parametric equations are \( x(t) = 0 \), \( y(t) = 2 \cos t \), and \( z(t) = 3 \sin t \). These equations pinpoint the exact location on the curve for each value of \( t \). Parametric equations provide a powerful way to describe motion and change over time, which is essential in many fields of science and engineering. They allow for a more flexible and detailed representation of the curve compared to traditional Cartesian equations.

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Most popular questions from this chapter

A circular trajectory An object moves clockwise around a circle centered at the origin with radius 5 m beginning at the point (0,5). a. Find a position function \(\mathbf{r}\) that describes the motion if the object moves with a constant speed, completing 1 lap every \(12 \mathrm{s}\). b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{-t}\).

Nonuniform straight-line motion Consider the motion of an object given by the position function $$\mathbf{r}(t)=f(t)\langle a, b, c\rangle+\left\langle x_{0}, y_{0}, z_{0}\right\rangle, \quad \text { for } t \geq 0$$ where \(a, b, c, x_{0}, y_{0},\) and \(z_{0}\) are constants, and \(f\) is a differentiable scalar function, for \(t \geq 0\) a. Explain why \(r\) describes motion along a line. b. Find the velocity function. In general, is the velocity constant in magnitude or direction along the path?

Solving equations of motion Given an acceleration vector, initial velocity \(\left\langle u_{0}, v_{0}, w_{0}\right\rangle,\) and initial position \(\left\langle x_{0}, y_{0}, z_{0}\right\rangle,\) find the velocity and position vectors, for \(t \geq \mathbf{0}\). $$\mathbf{a}(t)=\langle 1, t, 4 t\rangle,\left\langle u_{0}, v_{0}, w_{0}\right\rangle=\langle 20,0,0\rangle, \left\langle x_{0}, y_{0}, z_{0}\right\rangle=\langle 0,0,0\rangle$$

Evaluate the following definite integrals. $$\int_{0}^{2} t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) d t$$

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(\frac{d^{2} s}{d t^{2}}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). Show that if an object decelerates in the sense that \(\frac{d^{2} s}{d t^{2}}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N} .\)

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