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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Question
20
![**Problem 20:**
Find \(\Delta y\) and \(dy\).
Given:
\[ y = \arctan x, \quad x = 1, \quad \Delta x = dx = 0.05 \]
**Explanation:**
This is a calculus problem where we're asked to find the actual change in \(y\), denoted as \(\Delta y\), and the differential \(dy\) for the function \(y = \arctan x\).
- **\(\Delta y\)** represents the actual change in the value of the function as \(x\) changes from its initial value.
- **\(dy\)** is the approximate change based on the differential calculus, calculated using the derivative of the function.
**Steps to Solve the Problem:**
1. **Find the Derivative of \(y = \arctan x\):**
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]
2. **Calculate \(dy\):**
- Use the derivative to find \(dy\):
\[ dy = \frac{1}{1 + x^2} \cdot dx \]
- Substitute \(x = 1\) and \(dx = 0.05\):
\[ dy = \frac{1}{1 + 1^2} \cdot 0.05 = \frac{1}{2} \cdot 0.05 = 0.025 \]
3. **Calculate \(\Delta y\):**
- Calculate the actual change in \(y\) for the given change \(\Delta x\):
\[ \Delta y = \arctan(1.05) - \arctan(1) \]
- This requires evaluating the arctan values and computing the difference.
**Conclusion:**
Through these steps, you determine both the approximate and actual changes in \(y\) as \(x\) changes from 1 by 0.05 units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbc57344-aee0-4c7a-83a5-ce0e8c0e16d3%2F9c2dc643-ec77-45dd-a752-9bd9b4bcec20%2Fa5ecxb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 20:**
Find \(\Delta y\) and \(dy\).
Given:
\[ y = \arctan x, \quad x = 1, \quad \Delta x = dx = 0.05 \]
**Explanation:**
This is a calculus problem where we're asked to find the actual change in \(y\), denoted as \(\Delta y\), and the differential \(dy\) for the function \(y = \arctan x\).
- **\(\Delta y\)** represents the actual change in the value of the function as \(x\) changes from its initial value.
- **\(dy\)** is the approximate change based on the differential calculus, calculated using the derivative of the function.
**Steps to Solve the Problem:**
1. **Find the Derivative of \(y = \arctan x\):**
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]
2. **Calculate \(dy\):**
- Use the derivative to find \(dy\):
\[ dy = \frac{1}{1 + x^2} \cdot dx \]
- Substitute \(x = 1\) and \(dx = 0.05\):
\[ dy = \frac{1}{1 + 1^2} \cdot 0.05 = \frac{1}{2} \cdot 0.05 = 0.025 \]
3. **Calculate \(\Delta y\):**
- Calculate the actual change in \(y\) for the given change \(\Delta x\):
\[ \Delta y = \arctan(1.05) - \arctan(1) \]
- This requires evaluating the arctan values and computing the difference.
**Conclusion:**
Through these steps, you determine both the approximate and actual changes in \(y\) as \(x\) changes from 1 by 0.05 units.
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