Next we substitute 'x s(x) = √² f* VI √ 1 + (f'(t))² dt 4 = dy dx - [~ √₂ + ( ₁6(+) = 1 3x J = 3x¹/2 1 + into the equation for the arc length function and simplify. X t dt 2 +²/2) ² dt
Next we substitute 'x s(x) = √² f* VI √ 1 + (f'(t))² dt 4 = dy dx - [~ √₂ + ( ₁6(+) = 1 3x J = 3x¹/2 1 + into the equation for the arc length function and simplify. X t dt 2 +²/2) ² dt
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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![### Arc Length Function Simplification
To find the arc length of a function, we start with the formula:
\[ s(x) = \int_{4}^{x} \sqrt{1 + (f'(t))^2} \, dt \]
Given that the derivative \( \frac{dy}{dx} = 3x^{1/2} \), we'll substitute this into our arc length formula and simplify the integral.
1. Substitute \( \frac{dy}{dx} = 3x^{1/2} \) into the arc length formula:
\[ s(x) = \int_{4}^{x} \sqrt{1 + (f'(t))^2} \, dt \]
2. Simplify by substituting \( f'(t) = 3t^{1/2} \):
\[ s(x) = \int_{4}^{x} \sqrt{1 + (3t^{1/2})^2} \, dt \]
3. Further simplification:
\[ s(x) = \int_{4}^{x} \sqrt{1 + 9t} \, dt \]
4. Finally, we have:
\[ s(x) = \int_{4}^{x} \sqrt{1 + 9t} \, dt \]
This is the simplified form of the arc length integral that can be further evaluated through appropriate integration techniques.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4971f23-5570-4fed-8cd6-5e1a07d18a42%2F5ebea96d-10d2-4c6c-bb5e-b54c9fdf8587%2Fwlzf9t_processed.png&w=3840&q=75)
Transcribed Image Text:### Arc Length Function Simplification
To find the arc length of a function, we start with the formula:
\[ s(x) = \int_{4}^{x} \sqrt{1 + (f'(t))^2} \, dt \]
Given that the derivative \( \frac{dy}{dx} = 3x^{1/2} \), we'll substitute this into our arc length formula and simplify the integral.
1. Substitute \( \frac{dy}{dx} = 3x^{1/2} \) into the arc length formula:
\[ s(x) = \int_{4}^{x} \sqrt{1 + (f'(t))^2} \, dt \]
2. Simplify by substituting \( f'(t) = 3t^{1/2} \):
\[ s(x) = \int_{4}^{x} \sqrt{1 + (3t^{1/2})^2} \, dt \]
3. Further simplification:
\[ s(x) = \int_{4}^{x} \sqrt{1 + 9t} \, dt \]
4. Finally, we have:
\[ s(x) = \int_{4}^{x} \sqrt{1 + 9t} \, dt \]
This is the simplified form of the arc length integral that can be further evaluated through appropriate integration techniques.
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