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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![### Calculus: Derivatives of Vector-Valued Functions
**Problem Statement:**
Compute the derivative for \( r(t) = (\cos(8t)) \mathbf{i} + (\sin^9(t)) \mathbf{j} + (\tan(6t)) \mathbf{k} \).
\[
\frac{d\mathbf{r}}{dt} = \boxed{\phantom{answer}}
\]
**Explanation:**
In this problem, we are asked to find the derivative of a vector-valued function \( r(t) \) with components in the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) directions. Here, the function is given as:
\[ r(t) = (\cos(8t)) \mathbf{i} + (\sin^9(t)) \mathbf{j} + (\tan(6t)) \mathbf{k} \]
To find this derivative, we need to compute the derivative of each component function with respect to \( t \). The overall derivative will be the vector formed by these component derivatives.
For the \( \mathbf{i} \)-component:
\[ \frac{d}{dt} (\cos(8t)) = -8 \sin(8t) \]
For the \( \mathbf{j} \)-component:
\[ \frac{d}{dt} (\sin^9(t)) = 9 \sin^8(t) \cos(t) \]
For the \( \mathbf{k} \)-component:
\[ \frac{d}{dt} (\tan(6t)) = 6 \sec^2(6t) \]
Combining these results, we write the derivative as:
\[ \frac{d\mathbf{r}}{dt} = (-8 \sin(8t)) \mathbf{i} + (9 \sin^8(t) \cos(t)) \mathbf{j} + (6 \sec^2(6t)) \mathbf{k} \]
So, inserting this into the boxed answer format:
\[
\frac{d\mathbf{r}}{dt} = -8 \sin(8t) \mathbf{i} + 9 \sin^8(t) \cos(t) \mathbf{j} + 6 \sec^2(6t) \mathbf{k}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5adb599f-5a8a-4e18-a8b6-f0de24f7a983%2Fc64b179a-6461-4998-af98-cf67ca86ee68%2F8e833si.png&w=3840&q=75)
Transcribed Image Text:### Calculus: Derivatives of Vector-Valued Functions
**Problem Statement:**
Compute the derivative for \( r(t) = (\cos(8t)) \mathbf{i} + (\sin^9(t)) \mathbf{j} + (\tan(6t)) \mathbf{k} \).
\[
\frac{d\mathbf{r}}{dt} = \boxed{\phantom{answer}}
\]
**Explanation:**
In this problem, we are asked to find the derivative of a vector-valued function \( r(t) \) with components in the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) directions. Here, the function is given as:
\[ r(t) = (\cos(8t)) \mathbf{i} + (\sin^9(t)) \mathbf{j} + (\tan(6t)) \mathbf{k} \]
To find this derivative, we need to compute the derivative of each component function with respect to \( t \). The overall derivative will be the vector formed by these component derivatives.
For the \( \mathbf{i} \)-component:
\[ \frac{d}{dt} (\cos(8t)) = -8 \sin(8t) \]
For the \( \mathbf{j} \)-component:
\[ \frac{d}{dt} (\sin^9(t)) = 9 \sin^8(t) \cos(t) \]
For the \( \mathbf{k} \)-component:
\[ \frac{d}{dt} (\tan(6t)) = 6 \sec^2(6t) \]
Combining these results, we write the derivative as:
\[ \frac{d\mathbf{r}}{dt} = (-8 \sin(8t)) \mathbf{i} + (9 \sin^8(t) \cos(t)) \mathbf{j} + (6 \sec^2(6t)) \mathbf{k} \]
So, inserting this into the boxed answer format:
\[
\frac{d\mathbf{r}}{dt} = -8 \sin(8t) \mathbf{i} + 9 \sin^8(t) \cos(t) \mathbf{j} + 6 \sec^2(6t) \mathbf{k}
\]
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