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Chapter 12: Problem 34

Given an acceleration vector, initial velocity $\left\langle u_{0}, v_{0}\right\rangle,\( and initial position \)\left\langle x_{0}, y_{0}\right\rangle,\( find the velocity and position vectors, for \)t \geq 0$. $$\mathbf{a}(t)=\langle 1, t\rangle,\left\langle u_{0}, v_{0}\right\rangle=\langle 2,-1\rangle,\left\langle x_{0}, y_{0}\right\rangle=\langle 0,8\rangle$$

Short Answer

Expert verified
Answer: The velocity vector, as a function of time, is given by: $$ \mathbf{v}(t) = \langle t, \frac{1}{2}t^2 + \frac{1}{2} \rangle $$ The position vector, as a function of time, is given by: $$ \mathbf{r}(t) = \langle \frac{1}{2}t^2, \frac{1}{6}t^3 + \frac{1}{2}t + 8 \rangle $$

Step by step solution

01

1) Integration of the acceleration function to find the velocity function

First, we will integrate the acceleration function component-wise to find the general form of the velocity vector: $$ \mathbf{v}(t) = \int \mathbf{a}(t)dt = \int \langle 1, t \rangle dt $$ Integrate each component separately: $$ \int 1 dt = t + C_1 $$ $$ \int t dt = \frac{1}{2}t^2 + C_2 $$ So the general form of the velocity vector is: $$ \mathbf{v}(t) = \langle t + C_1, \frac{1}{2}t^2 + C_2 \rangle $$
02

2) Use initial velocity to find the constants in the velocity function

Now, we plug the initial velocity \(\left\langle u_{0},v_{0}\right\rangle=\langle 2,-1\rangle\) into the general form of the velocity function and solve for \(C_1\) and \(C_2\): $$ \langle 2,-1 \rangle = \langle 2+C_1, -\dfrac{1}{2} + C_2 \rangle $$ Equating the components, we find \(C_1 = 0\) and \(C_2 = \dfrac{1}{2}\). So the velocity function is: $$ \mathbf{v}(t) = \langle t, \dfrac{1}{2}t^2 + \frac{1}{2} \rangle $$
03

3) Integration of the velocity function to find the position function

Next, we will integrate the velocity function component-wise to find the general form of the position vector: $$ \mathbf{r}(t) = \int \mathbf{v}(t)dt = \int \langle t, \frac{1}{2}t^2 + \frac{1}{2} \rangle dt $$ Integrate each component separately: $$ \int t dt = \frac{1}{2}t^2 + C_3 $$ $$ \int \left(\frac{1}{2}t^2 + \frac{1}{2}\right) dt = \frac{1}{6}t^3 + \frac{1}{2}t + C_4 $$ So the general form of the position vector is: $$ \mathbf{r}(t) = \langle \frac{1}{2}t^2 + C_3, \frac{1}{6}t^3 + \frac{1}{2}t + C_4 \rangle $$
04

4) Use initial position to find the constants in the position function

Now, we plug the initial position \(\left\langle x_{0},y_{0}\right\rangle=\langle 0,8\rangle\) into the general form of the position function and solve for \(C_3\) and \(C_4\): $$ \langle 0,8 \rangle = \left\langle \frac{1}{2}(0)^2 + C_3, \frac{1}{6}(0)^3 + \frac{1}{2}(0) + C_4 \right\rangle $$ Equating the components, we find \(C_3 = 0\) and \(C_4 = 8\). So the position function is: $$ \mathbf{r}(t) = \langle \frac{1}{2}t^2, \frac{1}{6}t^3 + \frac{1}{2}t + 8 \rangle $$ Thus, the velocity and position vectors are given by: $$ \mathbf{v}(t) = \langle t, \frac{1}{2}t^2 + \frac{1}{2} \rangle $$ and $$ \mathbf{r}(t) = \langle \frac{1}{2}t^2, \frac{1}{6}t^3 + \frac{1}{2}t + 8 \rangle $$

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Find a general expression for a nonzero vector orthogonal to the plane containing the curve. $$\begin{aligned} \mathbf{r}(t)=&(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j} \\\ &+(e \cos t+f \sin t) \mathbf{k} \end{aligned}$$ where \(\langle a, c, e\rangle \times\langle b, d, f\rangle \neq \mathbf{0}.\)

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 3+4 t, 1-6 t, 4 t\rangle;\\\ &\mathbf{R}(s)=\langle-2 s, 5+3 s, 4-2 s\rangle \end{aligned}$$

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and \(60 \mathrm{ft}\) from home plate. Assume that the coordinate axes are oriented as shown in the figure. a. In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of \(\langle 130,0,-3\rangle \mathrm{ft} / \mathrm{s}\) (about \(90 \mathrm{mi} / \mathrm{hr}\) ). How far above the ground is the ball when it crosses home plate and how long does it take for the pitch to arrive? b. What vertical velocity component should the pitcher use so that the pitch crosses home plate exactly \(3 \mathrm{ft}\) above the ground? c. A simple model to describe the curve of a baseball assumes that the spin of the ball produces a constant sideways acceleration (in the \(y\) -direction) of \(c \mathrm{ft} / \mathrm{s}^{2}\). Assume a pitcher throws a curve ball with \(c=8 \mathrm{ft} / \mathrm{s}^{2}\) (one-fourth the acceleration of gravity). How far does the ball move in the \(y\) -direction by the time it reaches home plate, assuming an initial velocity of (130,0,-3) ft/s? d. In part (c), does the ball curve more in the first half of its trip to the plate or in the second half? How does this fact affect the batter? e. Suppose the pitcher releases the ball from an initial position of (0,-3,6) with initial velocity \((130,0,-3) .\) What value of the spin parameter \(c\) is needed to put the ball over home plate passing through the point (60,0,3)\(?\)

Practical formula for \(\mathbf{N}\) Show that the definition of the principal unit normal vector $\mathbf{N}=\frac{d \mathbf{T} / d s}{|d \mathbf{T} / d s|}\( implies the practical formula \)\mathbf{N}=\frac{d \mathbf{T} / d t}{|d \mathbf{T} / d t|} .\( Use the Chain Rule and Note that \)|\mathbf{v}|=d s / d t>0.$

\(\mathbb{R}^{3}\) Consider the vectors \(\mathbf{I}=\langle 1 / 2,1 / 2,1 / \sqrt{2}), \mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}, 0\rangle,\) and \(\mathbf{K}=\langle 1 / 2,1 / 2,-1 / \sqrt{2}\rangle\) a. Sketch I, J, and K and show that they are unit vectors. b. Show that \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\) are pairwise orthogonal. c. Express the vector \langle 1,0,0\rangle in terms of \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\).
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