Textbook Question
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
27. y=arccsc(x²+1)
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.2.82
82. Find a curve through the point (1, 0) whose length from x=1 to x=2 is
L = ∫(from 1 to 2)√(1 + 1/x²)dx.
Verified step by step guidance1
Recognize that the problem asks for a curve \( y = f(x) \) passing through the point \( (1, 0) \) such that the length of the curve from \( x=1 \) to \( x=2 \) is given by the integral \( L = \int_1^2 \sqrt{1 + \frac{1}{x^2}} \, dx \). This integral represents the arc length formula \( L = \int_a^b \sqrt{1 + (y')^2} \, dx \).
From the arc length formula, identify that \( \sqrt{1 + (y')^2} = \sqrt{1 + \frac{1}{x^2}} \). Equate the expressions inside the square roots to find \( (y')^2 = \frac{1}{x^2} \).
Solve for \( y' \) by taking the square root of both sides, giving \( y' = \pm \frac{1}{x} \). This is a separable differential equation.
Integrate both sides with respect to \( x \): \( dy = \pm \frac{1}{x} dx \) leads to \( y = \pm \ln|x| + C \), where \( C \) is the constant of integration.
Use the initial condition \( y(1) = 0 \) to find \( C \). Substitute \( x=1 \) and \( y=0 \) into the equation to solve for \( C \), which will give the specific curve satisfying the problem.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Arc Length Formula
The arc length of a curve y = f(x) from x = a to x = b is given by the integral L = ∫ from a to b √(1 + (dy/dx)²) dx. This formula calculates the length of the curve by summing infinitesimal line segments along the curve.
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Arc Length of Parametric Curves
Differential Equation from Arc Length Integrand
Given the integrand √(1 + (dy/dx)²) = √(1 + 1/x²), we can equate (dy/dx)² = 1/x², leading to a differential equation dy/dx = ±1/x. Solving this differential equation helps find the explicit form of the curve.
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06:29Arc Length of Parametric Curves
Initial Condition and Integration
Using the initial point (1, 0), we integrate dy/dx = 1/x to find y = ln|x| + C. Applying the initial condition y(1) = 0 determines the constant C, yielding the specific curve that satisfies the given length integral.
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05:03Initial Value Problems
Related Practice
Textbook Question
Rewrite the expressions in Exercises 5–10 in terms of exponentials and simplify the results as much as you can.
8. cosh(3x) - sinh(3x)
Textbook Question
In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
27. y = θ(sin(lnθ) + cos(lnθ))
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Textbook Question
In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
23. y = (x²+1)sech(ln x)
(Hint: Before differentiating, express in terms of exponentials and simplify.)
Textbook Question
77. The region in the first quadrant bounded by the coordinate axes, the line y=3, and the curve x=2/√(y+1) is revolved about the y-axis to generate a solid. Find the volume of the solid.
Textbook Question
Verify the integration formulas in Exercises 37–40.
39. ∫x coth⁻¹(x)dx = ((x²-1)/2)coth⁻¹(x) + x/2 + C
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