Math / Calculus

QuestionQuestion 4 of 8 Barring disasters (human-made or natural), the population $P$ of humans grows at a rate proportional to its current size.
According to the U.N. World Population studies, from 2006 to 2008 the population of the more developed regions of the world (Europe, North America, Australia, New Zealand, and Japan) grew at an annual rate of $1.33 \%$ per year. (Use symbolic notation and fractions where needed.) (a) Write a differential equation that models the growth rate of population. $\frac{d P}{d t}=$ $\square$ (b) Find the general solution to the differential equation. Use $P_{0}$ as $P(0)$ if needed. $P=$ $\square$ Search

Studdy Solution

STEP 1

What is this asking? We need to write a differential equation that describes how the population changes over time, and then solve that equation to find a general formula for the population at any time. Watch out! Don't mix up the growth rate (percentage) with the actual change in population.
Also, remember that "growing proportionally to its current size" means bigger populations grow faster!

STEP 2

1. Set up the differential equation.
2. Solve the differential equation.

STEP 3

Alright, so the problem says the population grows at a rate proportional to its current size.
That's the key!
This means the rate of change of the population, written as $\frac{dP}{dt}$ , is equal to some constant, let's call it $k$ , times the current population, $P$ .

STEP 4

So, we can write this relationship as a differential equation: $\frac{dP}{dt} = k \cdot P$ Boom! There's our differential equation.
The problem tells us the annual growth rate is 1.33%, which as a decimal is 0.0133.
Since this is the growth rate, it's our value for $k$ !

STEP 5

So, our specific differential equation is: $\frac{dP}{dt} = \textbf{0.0133} \cdot P$ We did it!
Now on to solving this bad boy.

STEP 6

We've got ourselves a separable differential equation here.
Let's separate the variables by dividing both sides by $P$ and multiplying both sides by $dt$ : $\frac{1}{P} \cdot dP = \textbf{0.0133} \cdot dt$ Perfect! Now we can integrate both sides: $\int \frac{1}{P} \, dP = \int \textbf{0.0133} \, dt$

STEP 7

The integral of $\frac{1}{P}$ with respect to $P$ is $\ln|P|$ , and the integral of 0.0133 with respect to $t$ is 0.0133 $\cdot t$ .
Don't forget the constant of integration, which we'll call $C$ , on the right side: $\ln|P| = \textbf{0.0133} \cdot t + C$

STEP 8

To solve for $P$ , we'll exponentiate both sides using $e$ as the base: $e^{\ln|P|} = e^{\textbf{0.0133} \cdot t + C}$ This simplifies to: $|P| = e^C \cdot e^{\textbf{0.0133} \cdot t}$

STEP 9

Since $e^C$ is just another constant, we can replace it with a new constant, let's call it $A$ .
Also, since population is always positive, we can drop the absolute value around $P$ : $P = A \cdot e^{\textbf{0.0133} \cdot t}$

STEP 10

The problem mentions using $P_0$ for $P(0)$ .
This is the initial population when $t = 0$ .
Let's plug that in: $P_0 = A \cdot e^{\textbf{0.0133} \cdot 0}$ Since $e^0 = 1$ , we get $P_0 = A$ .

STEP 11

So, our final general solution is: $P = P_0 \cdot e^{\textbf{0.0133} \cdot t}$ And there you have it!

STEP 12

(a) $\frac{dP}{dt} = 0.0133 \cdot P$ (b) $P = P_0 \cdot e^{0.0133 \cdot t}$

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Studdy Solution

STEP 1

STEP 2

1. Set up the differential equation.
2. Solve the differential equation.

STEP 3

Alright, so the problem says the population grows at a rate proportional to its current size.
That's the key!
This means the rate of change of the population, written as $\frac{dP}{dt}$ , is equal to some constant, let's call it $k$ , times the current population, $P$ .

STEP 4

So, we can write this relationship as a differential equation: $\frac{dP}{dt} = k \cdot P$ Boom! There's our differential equation.
The problem tells us the annual growth rate is 1.33%, which as a decimal is 0.0133.
Since this is the growth rate, it's our value for $k$ !

STEP 5

So, our specific differential equation is: $\frac{dP}{dt} = \textbf{0.0133} \cdot P$ We did it!
Now on to solving this bad boy.

STEP 6

STEP 7

The integral of $\frac{1}{P}$ with respect to $P$ is $\ln|P|$ , and the integral of 0.0133 with respect to $t$ is 0.0133 $\cdot t$ .
Don't forget the constant of integration, which we'll call $C$ , on the right side: $\ln|P| = \textbf{0.0133} \cdot t + C$

STEP 8

To solve for $P$ , we'll exponentiate both sides using $e$ as the base: $e^{\ln|P|} = e^{\textbf{0.0133} \cdot t + C}$ This simplifies to: $|P| = e^C \cdot e^{\textbf{0.0133} \cdot t}$

STEP 9

Since $e^C$ is just another constant, we can replace it with a new constant, let's call it $A$ .
Also, since population is always positive, we can drop the absolute value around $P$ : $P = A \cdot e^{\textbf{0.0133} \cdot t}$

STEP 10

The problem mentions using $P_0$ for $P(0)$ .
This is the initial population when $t = 0$ .
Let's plug that in: $P_0 = A \cdot e^{\textbf{0.0133} \cdot 0}$ Since $e^0 = 1$ , we get $P_0 = A$ .

STEP 11

So, our final general solution is: $P = P_0 \cdot e^{\textbf{0.0133} \cdot t}$ And there you have it!

STEP 12

(a) $\frac{dP}{dt} = 0.0133 \cdot P$ (b) $P = P_0 \cdot e^{0.0133 \cdot t}$

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Studdy Solution

STEP 1

STEP 2

1. Set up the differential equation.2. Solve the differential equation.

STEP 3

Alright, so the problem says the population grows at a rate *proportional* to its current size.That's the key!This means the rate of change of the population, written as dPdt\frac{dP}{dt}dtdP​, is equal to some constant, let's call it kkk, *times* the current population, PPP.

STEP 4

STEP 5

So, our specific differential equation is: dPdt=0.0133⋅P\frac{dP}{dt} = \textbf{0.0133} \cdot PdtdP​=0.0133⋅P We did it!Now on to solving this bad boy.

STEP 6

STEP 7

STEP 8

To solve for PPP, we'll **exponentiate both sides** using eee as the base: eln⁡∣P∣=e0.0133⋅t+Ce^{\ln|P|} = e^{\textbf{0.0133} \cdot t + C}eln∣P∣=e0.0133⋅t+C This simplifies to: ∣P∣=eC⋅e0.0133⋅t|P| = e^C \cdot e^{\textbf{0.0133} \cdot t}∣P∣=eC⋅e0.0133⋅t

STEP 9

Since eCe^CeC is just another constant, we can replace it with a new constant, let's call it AAA.Also, since population is always positive, we can drop the absolute value around PPP: P=A⋅e0.0133⋅tP = A \cdot e^{\textbf{0.0133} \cdot t}P=A⋅e0.0133⋅t

STEP 10

The problem mentions using P0P_0P0​ for P(0)P(0)P(0).This is the **initial population** when t=0t = 0t=0.Let's plug that in: P0=A⋅e0.0133⋅0P_0 = A \cdot e^{\textbf{0.0133} \cdot 0}P0​=A⋅e0.0133⋅0 Since e0=1e^0 = 1e0=1, we get P0=AP_0 = AP0​=A.

STEP 11

So, our final general solution is: P=P0⋅e0.0133⋅tP = P_0 \cdot e^{\textbf{0.0133} \cdot t}P=P0​⋅e0.0133⋅t And there you have it!

STEP 12

(a) dPdt=0.0133⋅P\frac{dP}{dt} = 0.0133 \cdot PdtdP​=0.0133⋅P (b) P=P0⋅e0.0133⋅tP = P_0 \cdot e^{0.0133 \cdot t}P=P0​⋅e0.0133⋅t

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1. Set up the differential equation.
2. Solve the differential equation.

Alright, so the problem says the population grows at a rate proportional to its current size.
That's the key!
This means the rate of change of the population, written as $\frac{dP}{dt}$ , is equal to some constant, let's call it $k$ , times the current population, $P$ .

So, our specific differential equation is: $\frac{dP}{dt} = \textbf{0.0133} \cdot P$ We did it!
Now on to solving this bad boy.

To solve for $P$ , we'll exponentiate both sides using $e$ as the base: $e^{\ln|P|} = e^{\textbf{0.0133} \cdot t + C}$ This simplifies to: $|P| = e^C \cdot e^{\textbf{0.0133} \cdot t}$

Since $e^C$ is just another constant, we can replace it with a new constant, let's call it $A$ .
Also, since population is always positive, we can drop the absolute value around $P$ : $P = A \cdot e^{\textbf{0.0133} \cdot t}$

The problem mentions using $P_0$ for $P(0)$ .
This is the initial population when $t = 0$ .
Let's plug that in: $P_0 = A \cdot e^{\textbf{0.0133} \cdot 0}$ Since $e^0 = 1$ , we get $P_0 = A$ .

So, our final general solution is: $P = P_0 \cdot e^{\textbf{0.0133} \cdot t}$ And there you have it!

(a) $\frac{dP}{dt} = 0.0133 \cdot P$ (b) $P = P_0 \cdot e^{0.0133 \cdot t}$