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Theory and Examples
Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 4, Problema 4.7.87
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5
Guida verificata passo dopo passo1
Recognize that the given differential equation is a third-order ordinary differential equation: \(\frac{d^{3}y}{dx^{3}} = 6\). Our goal is to find the function \(y(x)\) that satisfies this equation along with the initial conditions.
Integrate the differential equation once with respect to \(x\) to find the second derivative \(y''(x)\). Since \(\frac{d^{3}y}{dx^{3}} = 6\), integrating gives \(y''(x) = 6x + C_1\), where \(C_1\) is an integration constant.
Use the initial condition \(y''(0) = -8\) to solve for \(C_1\). Substitute \(x=0\) into \(y''(x) = 6x + C_1\) and set it equal to \(-8\) to find \(C_1\).
Integrate \(y''(x)\) to find the first derivative \(y'(x)\). This gives \(y'(x) = \int y''(x) \, dx = \int (6x + C_1) \, dx = 3x^{2} + C_1 x + C_2\), where \(C_2\) is another integration constant.
Use the initial condition \(y'(0) = 0\) to solve for \(C_2\). Substitute \(x=0\) into \(y'(x)\) and set it equal to \(0\) to find \(C_2\). Then integrate \(y'(x)\) once more to find \(y(x)\), and use the initial condition \(y(0) = 5\) to solve for the last constant.
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Higher-Order Derivatives and Integration
This involves understanding derivatives beyond the first order, such as the third derivative given here. To solve the differential equation, you integrate the third derivative step-by-step to find the second, first derivatives, and finally the original function, adding constants of integration at each step.
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02:42
Higher Order Derivatives
Initial Conditions in Differential Equations
Initial conditions specify the values of the function and its derivatives at a particular point, allowing us to determine the constants of integration uniquely. Here, values for y(0), y′(0), and y″(0) are given, which help solve for the constants after integration.
Video consigliato:
04:00Solutions to Basic Differential Equations
Solving Initial Value Problems (IVPs)
An IVP requires finding a function that satisfies a differential equation and meets given initial conditions. The process involves integrating the differential equation and applying the initial values to find the specific solution that fits the problem.
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Percorso guidato
05:03Initial Value Problems
Pratica correlata
Domanda del libro di testo
[Technology Exercises] When solving Exercises 14–30, you may need to use appropriate technology (such as a calculator or a computer).
26. Factoring a quartic Find the approximate values of r_1 through r_4 in the factorization
8x^4-14x^3-9x^2+11x-1=8(x-r_1)(x-r_2)(x-r_3)(x-r_4)
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Domanda del libro di testo
Finding Critical Points
In Exercises 41–50, determine all critical points and all domain endpoints for each function.
y = x − 3x²ᐟ³
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Roots (Zeros)
Show that the functions in Exercises 19–26 have exactly one zero in the given interval.
f(x) = x³ + 4x² + 7, (−∞, 0)
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Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(−3csc²x)dx
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Identifying Extrema
In Exercises 19–40:
a. Find the open intervals on which the function is increasing and those on which it is decreasing.
b. Identify the function’s local extreme values, if any, saying where they occur.
g(x) = x√8 − x²
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