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)

Verified step by step guidance

1. Identify the given derivative of the function, which is r'(θ) = 8 - csc²θ. Our goal is to find the original function r(θ).
2. To find r(θ), we need to integrate the derivative r'(θ). Set up the integral: ∫(8 - csc²θ) dθ.
3. 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.
4. 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.
5. 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.

Key Concepts

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.

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.

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.