The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 20% in 10 years. What will be the population in 50 years? How fast is the population growing at t = 50? Step 1 We are given that the population of a town grows at a rate proportional to the population at time t. In other words, if the population at time t is given by P(t), then for some constant k we have In general with such a proportional relationship, we know that the population function has the form P(t)= cekt. The reason for this is that PKP is a simple separable linear equation. dt [-[kat In(IPI) = kt + C P = cekt Knowing the initial population, we can solve for the coefficient c. P(0) = cek(0) =C As the initial population is 500, we know that c= 500 P(t)= 500ekt P(10) 500e10k <- 600 With this equality, we can solve for k. Step 2 We know that the initial population of the town was 500. We also know that the population increased by 20% over 10 years, or (0.2) (500) 100. In terms of the population function P, this gives the following. = In(500)+(10✔ Step 4 500e10k 600 In(500e10k) In(600) 10 (k) In(e) In(600) = Step 3 We have now found the population function P. P(t) 500e1/10 In(600/500)t = k= 1/10✔ 500 We must find the population of the town in 50 years. That is, the value of P(50). What will be the population of the town in 50 years? (Round your answer to the nearest person.) 1244✔ 1,244 persons 1244.16 x 600 1/10 In 500 We are now asked to determine how fast the population is growing at t= 50. We know that the population function is given P(t) = 500ekt, where k = 10 The rate of change of the population is the derivative of P. (Round your answer to two decimal places.) P(50) KP(50) Finally, determine how fast (in persons/yr) is the population growing at t= 50? (Round your answer to two decimal places.) 23 Xpersons/yr de dt 600 500 kP. We also know that the population after 50 years is 1,244 persons.
The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 20% in 10 years. What will be the population in 50 years? How fast is the population growing at t = 50? Step 1 We are given that the population of a town grows at a rate proportional to the population at time t. In other words, if the population at time t is given by P(t), then for some constant k we have In general with such a proportional relationship, we know that the population function has the form P(t)= cekt. The reason for this is that PKP is a simple separable linear equation. dt [-[kat In(IPI) = kt + C P = cekt Knowing the initial population, we can solve for the coefficient c. P(0) = cek(0) =C As the initial population is 500, we know that c= 500 P(t)= 500ekt P(10) 500e10k <- 600 With this equality, we can solve for k. Step 2 We know that the initial population of the town was 500. We also know that the population increased by 20% over 10 years, or (0.2) (500) 100. In terms of the population function P, this gives the following. = In(500)+(10✔ Step 4 500e10k 600 In(500e10k) In(600) 10 (k) In(e) In(600) = Step 3 We have now found the population function P. P(t) 500e1/10 In(600/500)t = k= 1/10✔ 500 We must find the population of the town in 50 years. That is, the value of P(50). What will be the population of the town in 50 years? (Round your answer to the nearest person.) 1244✔ 1,244 persons 1244.16 x 600 1/10 In 500 We are now asked to determine how fast the population is growing at t= 50. We know that the population function is given P(t) = 500ekt, where k = 10 The rate of change of the population is the derivative of P. (Round your answer to two decimal places.) P(50) KP(50) Finally, determine how fast (in persons/yr) is the population growing at t= 50? (Round your answer to two decimal places.) 23 Xpersons/yr de dt 600 500 kP. We also know that the population after 50 years is 1,244 persons.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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DUE NOW. Please answer only the step 4 correctly. Thanks!
Transcribed Image Text:Tutorial Exercise
The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 20% in 10 years. What will be the population in 50 years?
How fast is the population growing at t= 50?
Step 1
We are given that the population of a town grows at a rate proportional to the population at time t. In other words, if the population at time t is given by P(t), then for some constant k we have
dP
In general with such a proportional relationship, we know that the population function has the form P(t) = cekt. The reason for this is that = KP is a simple separable linear equation.
dt
dP
[op - Skat
dt
In(IP) kt+C
P = cekt
Knowing the initial population, we can solve for the coefficient c.
P(0) = cek(0)
= C
As the initial population is 500, we know that c = 500✔
With this equality, we can solve for k.
In(500) + (10✔
Step 2
We know that the initial population of the town was 500. We also know that the population increased by 20% over 10 years, or (0.2) (500) = 100. In terms of the population function P, this gives the following.
P(t) = 500ekt
P(10) 500e10k
= 600
Step 4
500e10k = 600
In(500e10k) In(600)
=
10 (k) In(e) In(600)
=
Step 3
We have now found the population function P.
P(t) = 500e¹/10 In(600/500)t
k= 1/10✔
500
We must find the population of the town in 50 years. That is, the value of P(50).
What will be the population of the town in 50 years? (Round your answer to the nearest person.)
1244✔
1,244 persons
=1244.16 x
1/10 In
600
500
1 600
In
We are now asked to determine how fast the population is growing at t= 50. We know that the population function is given P(t) = 500ekt, where k =
10 500
Submit Skip (you cannot come back)
The rate of change of the population is the derivative of P. (Round your answer to two decimal places.)
P(50) KP(50)
Finally, determine how fast (in persons/yr) is the population growing at t= 50? (Round your answer to two decimal places.)
23
x persons/yr
dP
dt
= KP.
We also know that the population after 50 years is 1,244 persons.
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