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Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(\theta)=2 \cos ^{3} \theta \mathbf{i}+3 \sin ^{3} \theta \mathbf{j} $$

Short Answer

Expert verified
The curve defined by the function is smooth for all real values of \(\theta\), i.e., \(-\infty<\theta<\infty\).

Step by step solution

01

Understand the Function

The function \(\mathbf{r}(\theta)=2\cos^{3}\theta\mathbf{i}+3\sin^{3}\theta \mathbf{j}\) is a curve in the plane. It's parameterized by θ, and we need to find the values of θ where the curve is smooth.
02

Compute the Derivatives

Differentiate both of the components of the function. Derivative of \(2\cos^{3}\theta\) is \(-6\cos^{2}\theta*\sin\theta\) and the derivative of \(3\sin^{3}\theta\) is \(9\sin^{2}\theta*\cos\theta\).
03

Find Undefined Derivative

The derivatives will be undefined wherever the cosine and sine functions are undefined. But the cosine and sine functions are defined for all real numbers. Thus the derivatives are always defined.
04

Conclude the Smoothness

As the derivatives of both components of the vector-valued function always exist, it can be concluded that the curve is smooth for all open intervals.

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Key Concepts

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

Calculus
Calculus lies at the heart of understanding changes and movement in mathematics, especially when it comes to functions. A fundamental component of calculus is analyzing how objects vary over time or within a space. In the case of the vector-valued function \( \mathbf{r}(\theta)=2\cos^{3}\theta\mathbf{i}+3\sin^{3}\theta\mathbf{j} \), we're particularly interested in how the function behaves as the parameter \(\theta\) changes, thus examining its smoothness.

Smoothness of a curve is an important concept in calculus since it relates to continuity and differentiability. A function is considered smooth over an interval if it is continuously differentiable over that interval, meaning we can find a derivative for every point within the interval without any interruptions in the function's behavior. This concept ties directly into the understanding of motion and change because smoothness ensures that there are no abrupt changes or corners along the path of the curve.
Derivative of Vector-Valued Functions
When dealing with vector-valued functions, like \( \mathbf{r}(\theta) \), derivatives tell us about the rate at which the function is changing in each of its components at any given point. The derivative of a vector-valued function is obtained by differentiating each component of the function separately.

For our function \( \mathbf{r}(\theta)=2\cos^{3}\theta\mathbf{i}+3\sin^{3}\theta\mathbf{j} \), we compute the derivatives of its components, resulting in \( -6\cos^{2}\theta * \sin\theta \mathbf{i} \) and \( 9\sin^{2}\theta * \cos\theta \mathbf{j} \). These derivatives represent the instantaneous rate of change of the function's \(i\) and \(j\) components with respect to \(\theta\). The existence of these derivatives across all values of \(\theta\) is a strong indicator that our curve is smooth over the real-number domain.
Parametric Equations Smoothness
Parametric equations describe a set of relationships where one or more equations depend on a separate variable, often a parameter like \(\theta\) in our example function \( \mathbf{r}(\theta) \). The smoothness of a curve represented by parametric equations can be deduced by looking at the derivatives of these equations.

The notion of smoothness in the context of parametric equations reflects the absence of sharp turns or cusps in the curve. In other words, the curve should flow in a continuous manner without any breaks. When we differentiate the components of \( \mathbf{r}(\theta) \) and find that there are no points where the derivative is undefined or discontinuous, we can conclude the curve does not have abrupt direction changes or infinite slopes, characterising it as smooth over the range of \(\theta\).

Given that the derivatives of both components of \( \mathbf{r}(\theta) \) are well-defined for all real numbers, we can conclude that the curve is smooth across all open intervals in its domain. Thus, parametrically defined curves like this vector-valued function give us elegant ways to describe complex shapes and motions, and calculus helps us to ensure these descriptions represent continuous and smooth movements.

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Most popular questions from this chapter

Find the curvature \(K\) of the curve. \(\mathbf{r}(t)=4 t \mathbf{i}+3 \cos t \mathbf{j}+3 \sin t \mathbf{k}\)

Find the curvature \(K\) of the curve, where \(s\) is the arc length parameter. Helix in Exercise 19: \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t, t\rangle\)

Consider the graph of the vector-valued function \(\mathbf{r}(t)=t \mathbf{i}+\left(4-t^{2}\right) \mathbf{j}+t^{3} \mathbf{k}\) on the interval \([0,2]\). (a) Approximate the length of the curve by finding the length of the line segment connecting its endpoints. (b) Approximate the length of the curve by summing the lengths of the line segments connecting the terminal points of the vectors \(\mathbf{r}(0), \mathbf{r}(0.5), \mathbf{r}(1), \mathbf{r}(1.5)\), and \(\mathbf{r}(2)\) (c) Describe how you could obtain a more accurate approximation by continuing the processes in parts (a) and (b). (d) Use the integration capabilities of a graphing utility to approximate the length of the curve. Compare this result with the answers in parts (a) and (b).

You are asked to verify Kepler's Laws of Planetary Motion. For these exercises, assume that each planet moves in an orbit given by the vector- valued function \(\mathbf{r}\). Let \(\boldsymbol{r}=\|\mathbf{r}\|\), let \(\boldsymbol{G}\) represent the universal gravitational constant, let \(M\) represent the mass of the sun, and let \(m\) represent the mass of the planet. Prove Kepler's Third Law: The square of the period of a planet's orbit is proportional to the cube of the mean distance between the planet and the sun.

Centripetal Force An object of mass \(m\) moves at a constant speed \(v\) in a circular path of radius \(r .\) The force required to produce the centripetal component of acceleration is called the centripetal force and is given by \(F=m v^{2} / r .\) Newton's Law of Universal Gravitation is given by \(F=G M m / d^{2}\), where \(d\) is the distance between the centers of the two bodies of masses \(M\) and \(m\), and \(G\) is a gravitational constant. Use this law to show that the speed required for circular motion is \(v=\sqrt{G M / r}\)

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