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Chapter 11: Problem 1

In Problems \(1-8\), find the required limit or indicate that it does not exist. $$ \lim _{t \rightarrow 1}\left[2 t \mathbf{i}-t^{2} \mathbf{j}\right] $$

Short Answer

Expert verified
The limit is \( 2\mathbf{i} - \mathbf{j} \).

Step by step solution

01

Recognize the Vector Function Components

The given problem involves finding the limit of a vector function as \( t \) approaches 1. The vector function is \( 2t \mathbf{i} - t^2 \mathbf{j} \), where \( 2t \) is the \( i \)-component and \( -t^2 \) is the \( j \)-component.
02

Find the Limit of Each Component Separately

To find the limit of the vector, we must find the limit of each of its components as \( t \rightarrow 1 \). The limit of the \( i \)-component is \( \lim_{t \to 1} 2t \) and the limit of the \( j \)-component is \( \lim_{t \to 1} (-t^2) \).
03

Calculate the Limit of the i-Component

Evaluate \( \lim_{t \to 1} 2t \). Substituting \( t = 1 \) into \( 2t \), we get \( 2 \times 1 = 2 \). Thus, the limit of the \( i \)-component is 2.
04

Calculate the Limit of the j-Component

Evaluate \( \lim_{t \to 1} (-t^2) \). Substituting \( t = 1 \) into \( -t^2 \), we get \(-1^2 = -1\). Thus, the limit of the \( j \)-component is -1.
05

Combine the Limits to Find the Vector Limit

Combine the results of the limits of the components into a vector form: \( 2\mathbf{i} - 1\mathbf{j} \). Therefore, the limit of the vector function is \( 2\mathbf{i} - \mathbf{j} \).

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Key Concepts

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

Vector Functions
A vector function is an essential concept in calculus and vector calculus. It involves functions that return vectors instead of scalar values. When dealing with vector functions, each output at a given point is a vector composed of multiple components. For instance, the vector function in our exercise is given as \( 2t \mathbf{i} - t^2 \mathbf{j} \). Each part of this vector function, \( 2t \) and \( -t^2 \), represents different components of the vector, specifically along the \( \mathbf{i} \) and \( \mathbf{j} \) directions respectively.
Vector functions are useful because they can describe movements and forces in multiple dimensions, such as a curve in 3D space. When learning about vector functions, it's important to understand how they relate to their component parts, providing both magnitude and direction for each unit vector involved.
Component-Wise Limits
A helpful approach when evaluating the limits of vector functions is the component-wise method. This means you can break down a vector into its component functions and consider each one separately. For example, in order to find the limit of the vector function \( 2t \mathbf{i} - t^2 \mathbf{j} \) as \( t \) approaches 1, we analyze the individual limits:
  • \( \lim_{t \to 1} 2t \)
  • \( \lim_{t \to 1} (-t^2) \)
By evaluating each component separately, the calculation becomes more straightforward. You determine the behavior of each part as \( t \) approaches a certain value and then recombine them into a final vector form.
This method highlights the simplicity of analyzing vector functions by treating them as collections of simpler scalar functions.
Calculus Limits
Limits play a central role in calculus, providing a way to understand the behavior of functions as inputs approach certain values. The concept of a limit helps in investigating function continuity, derivatives, and integrals. Limits allow us to examine how functions behave near a point without necessarily having to compute the function's value exactly at that point.
When working with limits, it is crucial to substitute the approaching value into the function whenever possible, to observe the tendency. For the vector function in question, once you substitute \( t = 1 \) into the respective components \( 2t \) and \( -t^2 \), you get specific concrete values without ambiguity. This step ensures the calculation of the individual component limits to be merged into the overall limit of the vector.
Vector Calculus
Vector calculus extends regular calculus into multiple dimensions, dealing with vector fields and operations like gradients, curls, and divergences. Understanding vector limits forms the foundational knowledge needed to explore advanced topics within vector calculus. In the realm of vector calculus, analyzing how vector functions behave as they approach certain points is essential for many applications, such as fluid dynamics, electromagnetism, and other fields involving vector quantities.
By practicing limits of vector functions, you develop an intuition for how components of vectors interact and change. These exercises lay the groundwork for tackling more complex vector calculus problems where analyzing directional derivatives and multi-dimensional surfaces is necessary.
Mastering concepts like vector functions, component-wise operations, and calculus limits in the context of vector calculus paves the way for successful applications in physics and engineering.

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Most popular questions from this chapter

Name and sketch the graph of each of the following equations in three-space. $$ y=\cos x $$

Find the length of the curve with the given vector equation. $$ \mathbf{r}(t)=t^{2} \mathbf{i}-2 t^{3} \mathbf{j}+6 t^{3} \mathbf{k} ; 0 \leq t \leq 1 $$

Let \(\left(\rho_{1}, \theta_{1}, \phi_{1}\right)\) and \(\left(\rho_{2}, \theta_{2}, \phi_{2}\right)\) be the spherical coordinates of two points, and let \(d\) be the straight-line distance between them. Show that $$ \begin{aligned} d=\left\\{\left(\rho_{1}-\rho_{2}\right)^{2}+2 \rho_{1} \rho_{2}\left[1-\cos \left(\theta_{1}-\theta_{2}\right)\right.\right.& \sin \phi_{1} \sin \phi_{2} \\\ -&\left.\left.\cos \phi_{1} \cos \phi_{2}\right]\right\\}^{1 / 2} \end{aligned} $$

EXPL 48. In this exercise you will derive Kepler's First Law, that planets travel in elliptical orbits. We begin with the notation. Place the coordinate system so that the sun is at the origin and the planet's closest approach to the sun (the perihelion) is on the positive \(x\) -axis and occurs at time \(t=0\). Let \(\mathbf{r}(t)\) denote the position vector and let \(r(t)=\|\mathbf{r}(t)\|\) denote the distance from the sun at time \(t\). Also, let \(\theta(t)\) denote the angle that the vector \(\mathbf{r}(t)\) makes with the positive \(x\) -axis at time \(t\). Thus, \((r(t), \theta(t))\) is the polar coordinate representation of the planet's position. Let \(\mathbf{u}_{1}=\mathbf{r} / r=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j} \quad\) and \(\quad \mathbf{u}_{2}=(-\sin \theta) \mathbf{i}+(\cos \theta) \mathbf{j}\) Vectors \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal unit vectors pointing in the directions of increasing \(r\) and increasing \(\theta\), respectively. Figure 12 summarizes this notation. We will often omit the argument \(t\), but keep in mind that \(\mathbf{r}, \theta, \mathbf{u}_{1}\), and \(\mathbf{u}_{2}\) are all functions of \(t .\) A prime in. dicates differentiation with respect to time \(t\). (a) Show that \(\mathbf{u}_{1}^{\prime}=\theta^{\prime} \mathbf{u}_{2}\) and \(\mathbf{u}_{2}^{\prime}=-\theta^{\prime} \mathbf{u}_{1}\). (b) Show that the velocity and acceleration vectors satisfy $$ \begin{array}{l} \mathbf{v}=r^{\prime} \mathbf{u}_{1}+r \theta^{\prime} \mathbf{u}_{2} \\ \mathbf{a}=\left(r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2}\right) \mathbf{u}_{1}+\left(2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime}\right) \mathbf{u}_{2} \end{array} $$ (c) Use the fact that the only force acting on the planet is the gravity of the sun to express a as a multiple of \(\mathbf{u}_{1}\), then explain how we can conclude that $$ \begin{aligned} r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2} &=\frac{-G M}{r^{2}} \\ 2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime} &=0 \end{aligned} $$ (d) Consider \(\mathbf{r} \times \mathbf{r}^{\prime}\), which we showed in Example 8 was a constant vector, say D. Use the result from (b) to show that \(\mathbf{D}=r^{2} \theta^{\prime} \mathbf{k} .\) (e) Substitute \(t=0\) to get \(\mathbf{D}=r_{0} v_{0} \mathbf{k}\), where \(r_{0}=r(0)\) and \(v_{0}=\|\mathbf{v}(0)\|\). Then argue that \(r^{2} \theta^{\prime}=r_{0} v_{0}\) for all \(t\). (f) Make the substitution \(q=r^{\prime}\) and use the result from (e) to obtain the first-order (nonlinear) differential equation in \(q\) : $$ q \frac{d q}{d r}=\frac{r_{0}^{2} v_{0}^{2}}{r^{3}}-\frac{G M}{r^{2}} $$ (g) Integrate with respect to \(r\) on both sides of the above equation and use an initial condition to obtain $$ q^{2}=2 G M\left(\frac{1}{r}-\frac{1}{r_{0}}\right)+v_{0}^{2}\left(1-\frac{r_{0}^{2}}{r^{2}}\right) $$ (h) Substitute \(p=1 / r\) into the above equation to obtain $$ \frac{r_{0}^{2} v_{0}^{2}}{\left(\theta^{\prime}\right)^{2}}\left(\frac{d p}{d t}\right)^{2}=2 G M\left(p-p_{0}\right)+v_{0}^{2}\left(1-\frac{p^{2}}{p_{0}^{2}}\right) $$

If the curve \(z=x^{2}\) in the \(x z\) -plane is revolved about the \(z\) -axis, the resulting surface has equation \(z=x^{2}+y^{2}\), obtained as a result of replacing \(x\) by \(\sqrt{x^{2}+y^{2}}\). If \(y=2 x^{2}\) in the \(x y\) -plane is revolved about the \(y\) -axis, what is the equation of the resulting surface?
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