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
![### Problem 3
**Objective:** Find, using a transformation, the area of the ellipse given by the equation:
\[ 9X^2 + 16Y^2 = 25 \]
**Explanation:**
To find the area of the ellipse described by the equation, we first convert the equation into a standard form. The equation of an ellipse in standard form is:
\[\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1\]
where \(a\) and \(b\) are the semi-major and semi-minor axes.
**Steps:**
1. *Transform the Equation:*
- Divide the entire equation by 25 to get it in the standard form:
\[
\frac{9X^2}{25} + \frac{16Y^2}{25} = 1
\]
2. *Identify the Semi-Axes:*
- Compare with the standard form to identify \(a\) and \(b\):
\[
\frac{X^2}{\left(\frac{5}{3}\right)^2} + \frac{Y^2}{\left(\frac{5}{4}\right)^2} = 1
\]
- Here, \(a = \frac{5}{3}\) and \(b = \frac{5}{4}\).
3. *Calculate the Area:*
- The area \(A\) of an ellipse is given by the formula:
\[
A = \pi \cdot a \cdot b
\]
- Substitute the values of \(a\) and \(b\):
\[
A = \pi \cdot \frac{5}{3} \cdot \frac{5}{4} = \pi \cdot \frac{25}{12}
\]
- Therefore, the area of the ellipse is \(\frac{25\pi}{12}\).
In summary, the transformation simplifies computation, aligning the equation to standard form for identifying axes, and ultimately aids in calculating the area of an ellipse efficiently.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F10580cf2-5676-4d37-9162-c9dfcc15f202%2Ffb091517-3a08-4d46-92f4-ac757ea951cd%2Fpc3b1il_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 3
**Objective:** Find, using a transformation, the area of the ellipse given by the equation:
\[ 9X^2 + 16Y^2 = 25 \]
**Explanation:**
To find the area of the ellipse described by the equation, we first convert the equation into a standard form. The equation of an ellipse in standard form is:
\[\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1\]
where \(a\) and \(b\) are the semi-major and semi-minor axes.
**Steps:**
1. *Transform the Equation:*
- Divide the entire equation by 25 to get it in the standard form:
\[
\frac{9X^2}{25} + \frac{16Y^2}{25} = 1
\]
2. *Identify the Semi-Axes:*
- Compare with the standard form to identify \(a\) and \(b\):
\[
\frac{X^2}{\left(\frac{5}{3}\right)^2} + \frac{Y^2}{\left(\frac{5}{4}\right)^2} = 1
\]
- Here, \(a = \frac{5}{3}\) and \(b = \frac{5}{4}\).
3. *Calculate the Area:*
- The area \(A\) of an ellipse is given by the formula:
\[
A = \pi \cdot a \cdot b
\]
- Substitute the values of \(a\) and \(b\):
\[
A = \pi \cdot \frac{5}{3} \cdot \frac{5}{4} = \pi \cdot \frac{25}{12}
\]
- Therefore, the area of the ellipse is \(\frac{25\pi}{12}\).
In summary, the transformation simplifies computation, aligning the equation to standard form for identifying axes, and ultimately aids in calculating the area of an ellipse efficiently.
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