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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 4 - Applications of DerivativesProblema 39
Capitolo 4, Problema 39

Finding Functions from Derivatives


In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.


r'(θ) = 8 − csc²θ, P(π/4, 0)

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Identify the given derivative of the function, which is r'(θ) = 8 - csc²θ. Our goal is to find the original function r(θ).
To find r(θ), we need to integrate the derivative r'(θ). Set up the integral: ∫(8 - csc²θ) dθ.
Integrate each term separately. The integral of 8 with respect to θ is 8θ. The integral of -csc²θ is a standard integral, which is -cotθ.
Combine the results of the integration to get the general form of the function: r(θ) = 8θ + cotθ + C, where C is the constant of integration.
Use the given point P(π/4, 0) to find the constant C. Substitute θ = π/4 and r(θ) = 0 into the equation: 0 = 8(π/4) + cot(π/4) + C. Solve for C to find the specific function.

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Antiderivatives

Antiderivatives, or indefinite integrals, are functions that reverse the process of differentiation. To find a function from its derivative, we integrate the derivative. In this case, integrating r'(θ) = 8 − csc²θ will yield the original function r(θ) plus a constant of integration, C.
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Antiderivatives

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometric terms. For r'(θ) = 8 − csc²θ, we need to integrate both 8 and −csc²θ separately. The integral of csc²θ is known to be −cotθ, which helps in finding the antiderivative of the given function.
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Introduction to Trigonometric Functions

Initial Conditions

Initial conditions are used to determine the constant of integration when finding an antiderivative. Given the point P(π/4, 0), we substitute θ = π/4 and r(θ) = 0 into the antiderivative to solve for the constant C. This ensures the function satisfies the condition of passing through the specified point.
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Initial Value Problems
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Domanda del libro di testo

Finding Functions from Derivatives


In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.


r'(t) = sec t tan t − 1, P(0, 0)

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Domanda del libro di testo

Graphs and Graphing


Graph the curves in Exercises 33–42.


y = 𝓍³ (8―𝓍 )

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Domanda del libro di testo

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

36. y = (x³ + x - 2) / (x - x²)

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Domanda del libro di testo

Finding Functions from Derivatives


In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.


g'(x) = 1 / x² + 2x, P(−1, 1)

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Domanda del libro di testo

Graphs and Graphing


Graph the curves in Exercises 33–42.

______

y = 𝓍√4 ― 𝓍²

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Domanda del libro di testo

Finding Position from Velocity or Acceleration


Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.


v = 9.8t + 5, s(0) = 10

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