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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 6.1.41b

40–43. Population growth


When records were first kept (t=0), the population of a rural town was 250 people. During the following years, the population grew at a rate of P′(t) = 30(1+√t), where t is measured in years.


b. Find the population P(t) at any time t≥0.

Guida verificata passo dopo passo
1
Identify the given rate of change of the population, which is the derivative of the population function: \(P\'(t) = 30(1 + \sqrt{t})\).
Recall that to find the population function \(P(t)\), you need to integrate the rate function \(P\'(t)\) with respect to \(t\): \(P(t) = \int P\'(t) \, dt + C\).
Set up the integral: \(P(t) = \int 30(1 + t^{1/2}) \, dt + C\).
Integrate each term separately: \(\int 30 \, dt\) and \(\int 30 t^{1/2} \, dt\).
Use the initial condition \(P(0) = 250\) to solve for the constant of integration \(C\) after performing the integration.

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Derivative as a Rate of Change

The derivative P′(t) represents the instantaneous rate of change of the population with respect to time. Understanding that P′(t) = 30(1 + √t) means the population grows faster as time increases, and this rate function is essential for finding the original population function P(t).
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04:16
Intro To Related Rates

Antiderivative and Indefinite Integration

To find the population function P(t) from its rate of change P′(t), we use antiderivatives or indefinite integrals. Integrating P′(t) with respect to t recovers P(t) up to a constant, which can be determined using initial conditions.
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Introduction to Indefinite Integrals

Initial Conditions and Constant of Integration

When integrating, an unknown constant appears because differentiation loses constant terms. Using the initial population P(0) = 250 allows us to solve for this constant, ensuring the population function accurately reflects the starting value.
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Initial Value Problems
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