A particle has position function r(t) and velocity function v(t) = { 2e“, sin(−t), ln(t+1) ) (a) Find its acceleration function a(t). (b) At time t=0, its position is r(0)=i+k. Find r(t).
A particle has position function r(t) and velocity function v(t) = { 2e“, sin(−t), ln(t+1) ) (a) Find its acceleration function a(t). (b) At time t=0, its position is r(0)=i+k. Find r(t).
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement
A particle has position function \( \mathbf{r}(t) \) and velocity function \( \mathbf{v}(t) = \left\langle 2e^{-t}, \sin(-t), \ln(t+1) \right\rangle \).
#### (a) Find its **acceleration function** \( \mathbf{a}(t) \).
#### (b) At time \( t = 0 \), its position is \( \mathbf{r}(0) = \mathbf{i} + \mathbf{k} \). Find \( \mathbf{r}(t) \).
#### Solution Explanation
- The velocity function given is \( \mathbf{v}(t) = \left\langle 2e^{-t}, \sin(-t), \ln(t+1) \right\rangle \).
**Part (a): Acceleration Function**
To find the acceleration function \( \mathbf{a}(t) \), we need to take the derivative of the velocity function \( \mathbf{v}(t) \) with respect to time \( t \).
1. **Component-wise Differentiation**:
- The first component of \( \mathbf{v}(t) \) is \( 2e^{-t} \). Its derivative is \( -2e^{-t} \).
- The second component of \( \mathbf{v}(t) \) is \( \sin(-t) \). Its derivative is \( -\cos(-t) \), which simplifies to \( -\cos(t) \) (since cosine is an even function).
- The third component of \( \mathbf{v}(t) \) is \( \ln(t+1) \). Its derivative is \( \frac{1}{t+1} \).
Thus, the acceleration function \( \mathbf{a}(t) \) is:
\[
\mathbf{a}(t) = \left\langle -2e^{-t}, -\cos(t), \frac{1}{t+1} \right\rangle
\]
**Part (b): Position Function**
To find the position function \( \mathbf{r}(t) \), we must integrate the velocity function \( \mathbf{v}(t) \) and use the given initial condition \( \mathbf{r}(0) = \mathbf{i} + \mathbf{k} \).
1. **Component-wise](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc74d85e6-643d-432d-8ed6-9fb95f3e5fbd%2F31a9d8da-e1d8-4c67-a83f-88b5fb23cb70%2Fgab3a9e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A particle has position function \( \mathbf{r}(t) \) and velocity function \( \mathbf{v}(t) = \left\langle 2e^{-t}, \sin(-t), \ln(t+1) \right\rangle \).
#### (a) Find its **acceleration function** \( \mathbf{a}(t) \).
#### (b) At time \( t = 0 \), its position is \( \mathbf{r}(0) = \mathbf{i} + \mathbf{k} \). Find \( \mathbf{r}(t) \).
#### Solution Explanation
- The velocity function given is \( \mathbf{v}(t) = \left\langle 2e^{-t}, \sin(-t), \ln(t+1) \right\rangle \).
**Part (a): Acceleration Function**
To find the acceleration function \( \mathbf{a}(t) \), we need to take the derivative of the velocity function \( \mathbf{v}(t) \) with respect to time \( t \).
1. **Component-wise Differentiation**:
- The first component of \( \mathbf{v}(t) \) is \( 2e^{-t} \). Its derivative is \( -2e^{-t} \).
- The second component of \( \mathbf{v}(t) \) is \( \sin(-t) \). Its derivative is \( -\cos(-t) \), which simplifies to \( -\cos(t) \) (since cosine is an even function).
- The third component of \( \mathbf{v}(t) \) is \( \ln(t+1) \). Its derivative is \( \frac{1}{t+1} \).
Thus, the acceleration function \( \mathbf{a}(t) \) is:
\[
\mathbf{a}(t) = \left\langle -2e^{-t}, -\cos(t), \frac{1}{t+1} \right\rangle
\]
**Part (b): Position Function**
To find the position function \( \mathbf{r}(t) \), we must integrate the velocity function \( \mathbf{v}(t) \) and use the given initial condition \( \mathbf{r}(0) = \mathbf{i} + \mathbf{k} \).
1. **Component-wise
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