Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problema 7.4.13

Capitolo 7, Problema 7.4.13

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

Guida verificata passo dopo passo

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 \( x \).

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.

Video consigliato:

06:06

Solving Separable Differential Equations

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.

Video consigliato:

Percorso guidato

06:18

Integration by Parts for Definite Integrals

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.

Video consigliato:

05:02

Intro to the Chain Rule

Pratica correlata

Domanda del libro di testo

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)

views

Domanda del libro di testo

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)

views

Domanda del libro di testo

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

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