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Chapter 12: Problem 4

Determine whether \(\mathbf{r}(t)\) is a smooth function of the parameter \(t .\) $$ \mathbf{r}(t)=\sin \pi t \mathbf{i}+(2 t-\ln t) \mathbf{j}+\left(t^{2}-t\right) \mathbf{k} $$

Short Answer

Expert verified
The function \( \mathbf{r}(t) \) is smooth for \( t > 0 \).

Step by step solution

01

Understand Smoothness Criteria

A function \( \mathbf{r}(t) \) is considered smooth if it is continuously differentiable, which means its first derivative \( \mathbf{r}'(t) \) exists and is continuous over its domain. Determine the domain of \( \mathbf{r}(t) \) first. Check the smoothness of each component of \( \mathbf{r}(t) \).
02

Domain Determination

Examine the expression \( 2t - \ln t \) inside the \( \mathbf{j} \) component to determine the domain. The function \( \ln t \) only exists for \( t > 0 \). Hence, the domain of \( \mathbf{r}(t) \) is \( t > 0 \).
03

Differentiate Each Component

Differentiate each component of \( \mathbf{r}(t) \): 1. \( \mathbf{i} \) component: Differentiate \( \sin \pi t \) to get \( \pi \cos \pi t \).2. \( \mathbf{j} \) component: Differentiate \( 2t - \ln t \) to get \( 2 - \frac{1}{t} \).3. \( \mathbf{k} \) component: Differentiate \( t^2 - t \) to get \( 2t - 1 \).
04

Analyze the Continuity of Derivatives

Check the continuity of each derivative:1. \( \pi \cos \pi t \) is continuous for all real \( t \).2. \( 2 - \frac{1}{t} \) is continuous for \( t > 0 \) as long as \( t eq 0 \).3. \( 2t - 1 \) is continuous for all real \( t \).
05

Combine Results

From the derivative analysis, all components have continuous derivatives on the domain \( t > 0 \). Therefore, \( \mathbf{r}(t) \) is continuously differentiable for \( t > 0 \). As such, \( \mathbf{r}(t) \) is a smooth function on its domain.

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Key Concepts

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

Domain Determination
Determining the domain of a function is the first crucial step in verifying its smoothness. The domain of a function refers to the set of input values (in this case, values of the parameter \( t \)) for which the function is defined and real. For the vector function \( \mathbf{r}(t) = \sin \pi t \mathbf{i} + (2t - \ln t) \mathbf{j} + (t^2 - t) \mathbf{k} \), we need to consider the restrictions on each term:
  • The \( \sin \pi t \) function is well defined for all real \( t \).
  • The \( 2t - \ln t \) component requires \( t > 0 \) because the natural logarithm \( \ln t \) is only defined for positive values of \( t \).
  • The \( t^2 - t \) is a polynomial, which is defined for all real \( t \).
This analysis reveals the domain of \( \mathbf{r}(t) \) to be \( t > 0 \). This domain ensures that all components of the function are real and defined, which is a prerequisite for smoothness.
Continuously Differentiable
To understand whether a function is smooth on its domain, it needs to be continuously differentiable. In simple terms, this means that you should be able to differentiate the function and obtain a derivative for every single component. Moreover, each of these derivatives should be continuous over the domain.
Let's differentiate each component of \( \mathbf{r}(t) \):
  • The derivative of the \( \sin \pi t \mathbf{i} \) component is \( \pi \cos \pi t \).
  • The \( 2t - \ln t \mathbf{j} \) component differentiates to \( 2 - \frac{1}{t} \).
  • The \( t^2 - t \mathbf{k} \) results in a derivative of \( 2t - 1 \).
This step confirms that the function's components are differentiable within the domain \( t > 0 \), as each component's derivative results in a real expression. Hence, \( \mathbf{r}(t) \) meets the first criterion of smoothness.
Smoothness Criteria
Smoothness of a vector function like \( \mathbf{r}(t) \) implies that all its components are continuously differentiable over the domain. After establishing that \( \mathbf{r}(t) \) is defined for \( t > 0 \), we also checked if all derivatives are continuous:
  • \( \pi \cos \pi t \) holds continuity for all \( t \).
  • \( 2 - \frac{1}{t} \) is continuous as long as \( t > 0 \) since \( t \) is never zero within the domain.
  • \( 2t - 1 \) is a simple linear expression, continuous for all \( t \).
With all components continuously differentiable and their derivatives continuous across the specified domain, \( \mathbf{r}(t) \) adheres to the smoothness criteria. Therefore, we conclude that \( \mathbf{r}(t) \) is indeed smooth for \( t > 0 \).

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Most popular questions from this chapter

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 $$ \alpha_{0}=\tan ^{-1} \frac{v}{\sqrt{v^{2}+2 g h}} $$

(a) Find the arc length parametrization of the line $$ x=1+t, \quad y=3-2 t, \quad z=4+2 t $$ that has the same direction as the given line and has reference point \((1,3,4) .\) (b) Use the parametric equations obtained in part (a) to find the point on the line that is 25 units from the reference point in the direction of increasing parameter.

Find two elevation angles that will enable a shell, fired from ground level with a muzzle speed of \(800 \mathrm{ft} / \mathrm{s}\), to hit a ground- level target \(10,000 \mathrm{ft}\) away.

Find the arc length of the parametric curve. $$ x=e^{t}, y=e^{-t}, z=\sqrt{2} t ; 0 \leq t \leq 1 $$

Suppose that the position vector of a particle moving in the plane is \(\mathbf{r}=12 \sqrt{t} \mathbf{i}+t^{3 / 2} \mathbf{j}, t>0 .\) Find the minimum speed of the particle and its location when it has this speed.
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