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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
Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.1.40b
Capitolo 6, Problema 6.1.40b

40–43. Population growth


Starting with an initial value of P(0)=55, the population of a prairie dog community grows at a rate of P′(t)=20−t/5 (prairie dogs/month), for 0≤t≤200, where t is measured in months.


b. Find the population P(t), for 0≤t≤200.

Guida verificata passo dopo passo
1
Identify the given rate of change of the population, which is the derivative of the population function: \(P\'(t) = 20 - \frac{t}{5}\).
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: \(P(t) = \int \left(20 - \frac{t}{5}\right) dt + C\).
Integrate each term separately: the integral of 20 with respect to \(t\) is \$20t$, and the integral of \(-\frac{t}{5}\) with respect to $t$ is \(-\frac{1}{5} \cdot \frac{t^2}{2} = -\frac{t^2}{10}\).
Combine the results and include the constant of integration \(C\): \(P(t) = 20t - \frac{t^2}{10} + C\). Use the initial condition \(P(0) = 55\) to solve for \(C\) by substituting \(t=0\) and \(P(0)=55\).

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Initial Value Problems

An initial value problem involves finding a function that satisfies a given differential equation and meets a specified initial condition. Here, the population function P(t) must satisfy P'(t) = 20 - t/5 and the initial value P(0) = 55, which anchors the solution to a specific starting point.
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Initial Value Problems

Antiderivatives and Integration

To find P(t) from its rate of change P'(t), we integrate P'(t) with respect to t. Integration reverses differentiation, allowing us to recover the original function up to a constant, which is then determined using the initial condition.
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Antiderivatives

Applying Initial Conditions to Determine Constants

After integrating the rate function, the solution includes an arbitrary constant. Using the initial condition P(0) = 55, we substitute t = 0 into the integrated function to solve for this constant, ensuring the population model accurately reflects the starting population.
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Initial Value Problems Example 1
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For the given regions R₁ and R₂, complete the following steps.


b. Find the area of region R₂ using geometry and the answer to part (a).


R₁is the region in the first quadrant bounded by the line x=1 and the curve y=6x(2−x^2)^2; R₂ is the region in the first quadrant bounded the curve y=6x(2−x^2)^2and the line y=6x.

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Variable gravity At Earth’s surface, the acceleration due to gravity is approximately g=9.8 m/s² (with local variations). However, the acceleration decreases with distance from the surface according to Newton’s law of gravitation. At a distance of y meters from Earth’s surface, the acceleration is given by a(y) = - g / (1+y/R)², where R=6.4×10⁶ m is the radius of Earth.


b. Use the Chain Rule to show that dv/dt = 1/2 d/dy(v²).

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Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.


b. Find the displacement of the object on the interval 0≤t≤6.

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Cycling distance A cyclist rides down a long straight road with a velocity (in m/min) given by v(t) = 400−20t, for 0≤t≤10, where t is measured in minutes.


b. How far does the cyclist travel in the first 10 min?

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Blood flow A typical human heart pumps 70 mL of blood (the stroke volume) with each beat. Assuming a heart rate of 60 beats/min (1 beat/s), a reasonable model for the outflow rate of the heart is V′(t)=70(1+sin 2πt), where V(t) is the amount of blood (in milliliters) pumped over the interval [0,t],V(0)=0 and t is measured in seconds.


b. Find the function that gives the total blood pumped between t=0 and a future time t>0.

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55–58. Marginal cost Consider the following marginal cost functions.


b. Find the additional cost incurred in dollars when production is increased from 500 units to 550 units.


C′(x) = 300+10x−0.01x²

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