Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.4.13

Chapter 7, Problem 7.4.13

Solve the differential equation in Exercises 9–22.
13. (dy/dx) = √y cos²√y

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Recognize that the given differential equation is \( \frac{dy}{dx} = \sqrt{y} \cos^{2}(\sqrt{y}) \). Our goal is to separate variables to integrate both sides.

Make the substitution \( t = \sqrt{y} \), which implies \( y = t^{2} \) and \( \frac{dy}{dx} = 2t \frac{dt}{dx} \). Rewrite the differential equation in terms of \( t \): \( 2t \frac{dt}{dx} = t \cos^{2}(t) \).

Simplify the equation by dividing both sides by \( t \) (assuming \( t \neq 0 \)): \( 2 \frac{dt}{dx} = \cos^{2}(t) \). Then, solve for \( \frac{dt}{dx} \): \( \frac{dt}{dx} = \frac{1}{2} \cos^{2}(t) \).

Separate variables to isolate \( t \) and \( x \): \( \frac{dt}{\cos^{2}(t)} = \frac{1}{2} dx \). Recall that \( \frac{1}{\cos^{2}(t)} = \sec^{2}(t) \), so the equation becomes \( \sec^{2}(t) dt = \frac{1}{2} dx \).

Integrate both sides: \( \int \sec^{2}(t) dt = \int \frac{1}{2} dx \). Use the integral formula \( \int \sec^{2}(t) dt = \tan(t) + C \) to find the implicit solution, then substitute back \( t = \sqrt{y} \) to express the solution in terms of \( y \) and \( x \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the solution.

Integration Techniques

Solving the equation requires integrating functions involving square roots and trigonometric expressions. Familiarity with substitution methods and standard integral formulas is essential to handle integrals like ∫(1/√y cos²√y) dy or ∫ cos²(u) du.

Chain Rule and Substitution in Integration

Using substitution, such as setting u = √y, simplifies the integral by transforming complex expressions into manageable forms. Understanding how to apply the chain rule in reverse during integration helps in correctly changing variables and integrating composite functions.

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Related Practice

Textbook Question

For problems 49–52 use implicit differentiation to find dy/dx at the given point P.

49. 3arctan(x) + arcsin(y) = π/4; P(1, -1)

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Textbook Question

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.

sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞

cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1

tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1

sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1

csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1

coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1

Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.

65. sech⁻¹(3/5)

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Textbook Question

In Exercises 139–142, find the length of each curve.

141. y = ln(cos(x)) from x = 0 to x = π/4.