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Evaluate the derivative, where r1(t) = (t°, tº, t4), r(2) = (0, 4, 5), and r'(2) = (3, 1, 2). r;(t)) t = 2 dt

Evaluate the derivative, where r1(t) = (t°, tº, t4), r(2) = (0, 4, 5), and r'(2) = (3, 1, 2). r;(t)) t = 2 dt

Calculus: Early Transcendentals
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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### Evaluating the Derivative of a Vector Function Product In this problem, we are tasked with evaluating the derivative of the product of two vector functions at \( t = 2 \). Specifically, given the vector function \( \mathbf{r}_1(t) \) and its components, as well as the values of \( \mathbf{r}(2) \) and \( \mathbf{r}'(2) \), we need to find: \[ \frac{d}{dt} \left( \mathbf{r}(t) \cdot \mathbf{r}_1(t) \right) \bigg|_{t=2} \] Given: - \( \mathbf{r}_1(t) = \langle t^5, t^3, t^4 \rangle \) - \( \mathbf{r}(2) = \langle 0, 4, 5 \rangle \) - \( \mathbf{r}'(2) = \langle 3, 1, 2 \rangle \) To begin, recall that the derivative of a dot product \( \mathbf{r}(t) \cdot \mathbf{r}_1(t) \) is given by: \[ \frac{d}{dt} \left( \mathbf{r}(t) \cdot \mathbf{r}_1(t) \right) = \mathbf{r}'(t) \cdot \mathbf{r}_1(t) + \mathbf{r}(t) \cdot \mathbf{r}_1'(t) \] We will evaluate each term separately at \( t = 2 \). ### Steps: 1. **Find \( \mathbf{r}_1'(t) \)** \[ \mathbf{r}_1(t) = \langle t^5, t^3, t^4 \rangle \\ \mathbf{r}_1'(t) = \left\langle \frac{d}{dt}(t^5), \frac{d}{dt}(t^3), \frac{d}{dt}(t^4) \right\rangle = \langle 5t^4, 3t^2, 4t^3 \rangle \] At \( t = 2 \): \[ \mathbf{r}_1'(2
Transcribed Image Text:### Evaluating the Derivative of a Vector Function Product In this problem, we are tasked with evaluating the derivative of the product of two vector functions at \( t = 2 \). Specifically, given the vector function \( \mathbf{r}_1(t) \) and its components, as well as the values of \( \mathbf{r}(2) \) and \( \mathbf{r}'(2) \), we need to find: \[ \frac{d}{dt} \left( \mathbf{r}(t) \cdot \mathbf{r}_1(t) \right) \bigg|_{t=2} \] Given: - \( \mathbf{r}_1(t) = \langle t^5, t^3, t^4 \rangle \) - \( \mathbf{r}(2) = \langle 0, 4, 5 \rangle \) - \( \mathbf{r}'(2) = \langle 3, 1, 2 \rangle \) To begin, recall that the derivative of a dot product \( \mathbf{r}(t) \cdot \mathbf{r}_1(t) \) is given by: \[ \frac{d}{dt} \left( \mathbf{r}(t) \cdot \mathbf{r}_1(t) \right) = \mathbf{r}'(t) \cdot \mathbf{r}_1(t) + \mathbf{r}(t) \cdot \mathbf{r}_1'(t) \] We will evaluate each term separately at \( t = 2 \). ### Steps: 1. **Find \( \mathbf{r}_1'(t) \)** \[ \mathbf{r}_1(t) = \langle t^5, t^3, t^4 \rangle \\ \mathbf{r}_1'(t) = \left\langle \frac{d}{dt}(t^5), \frac{d}{dt}(t^3), \frac{d}{dt}(t^4) \right\rangle = \langle 5t^4, 3t^2, 4t^3 \rangle \] At \( t = 2 \): \[ \mathbf{r}_1'(2
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