Respuesta :
Answer:
Population after 10 years is approximately 124 persons.
Approximately 4 people dies when t = 10
Step-by-step explanation:
We are given the following information in the question:
The population in thousands at time t is given by a function:
[tex]P(t) = \displaystyle\frac{20t^2 + 100t}{240t^2 + 5t + 60}[/tex]
a) Rate of change of population
Derivative with the help of quotient rule:
[tex]\displaystyle\frac{d}{dx}\bigg(\frac{u}{v}\bigg) = \frac{u'v - uv'}{v^2}[/tex]
Differentiating, we get:
[tex]\displaystyle\frac{d}{dt}\bigg(\frac{20t^2 + 100t}{240t^2 + 5t + 60}\bigg)\\\\= \frac{(20t^2 + 100t)'(240t^2 + 5t + 60)-(20t^2 + 100t)(240t^2 + 5t + 60)'}{(240t^2 + 5t + 60)^2}[/tex]
Solving, we get:
[tex]P'(t)=\displaystyle\frac{-4(239t^2-24x-60)}{(48t^2 + t + 12)^2}[/tex]
b) P(10) =
[tex]\displaystyle\frac{20(10)^2 + 100(10)}{240(10)^2 + 5(10) + 60} = 0.12442[/tex]
Since the population is given in thousands, we multiply it by 1000.
Thus, population after 10 years is approximately 124 persons.
c) P'(10) = [tex]\displaystyle\frac{-4(239(10)^2-24(10)-60)}{(48(10)^2 + (10) + 12)^2} = -0.004059[/tex]
Since the population is given in thousands, we multiply it by 1000.
Thus, approximately 4 people dies when t = 10.
The rate at which the population be increasing when t = 10 is 0.3265.
What is differentiation?
Differentiation can be defined as the rate of a function or variable with respect to other variable.
A.) In order to find the rate at which Glen Cove's population is changing with respect to time, we need to differentiate the function of the population with respect to time, and in order to do that we will use the quotient rule, therefore,
[tex]P(t) = \dfrac{20t^2 + 100t + 240}{t^2 + 5t + 60}\\\\\dfrac{dP}{dt} = \dfrac{d(\frac{20t^2 + 100t + 240}{t^2 + 5t + 60})}{dt}[/tex]
According to the quotient rule,
[tex]\dfrac{du}{dv} = \dfrac{u'v-uv'}{v^2}[/tex]
[tex]P'(t) = \dfrac{(20t^2 + 100t + 240)' (t^2 + 5t + 60)- (20t^2 + 100t + 240)(t^2 + 5t + 60)'}{(t^2 + 5t + 60)^2}\\\\P'(t) = \dfrac{(20t + 100)(t^2 + 5t + 60)- (20t^2 + 100t + 240)(t + 5)}{t^4 + 25t^2 + 3600 + 2(t^2 \times 5t) + 2(5t \times 60) + 2(t^2 \times 60)}\\\\[/tex]
Hence, the rate at which the population of Glen Cove's population is changing with respect to time is given as P'(t).
B.) The population of Glen Cove's after 10 years will be,
The population of Glen Cove's after 10 years can be found by substituting the value of time, t=10, in the function of the population,
[tex]\begin{aligned}P(t) &= \dfrac{20t^2 + 100t + 240}{t^2 + 5t + 60}\\\\&= \dfrac{20(10^2) + 100(10) + 240}{(10^2) + 5(10) + 60}\\\\&= \dfrac{2000 + 1000 + 240}{100 + 50 + 60}\\\\&= \dfrac{3240}{210}\\\\&= 15.4285\end{aligned}[/tex]
Since population can not be in decimal, therefore,
[tex]P(10) = 15.4285 \approx 15[/tex]
Hence, the population of Glen Cove after 10 years will be 15.
C.) The rate will the population be increasing when t = 10
[tex]P'(t) = \dfrac{(20t + 100)(t^2 + 5t + 60)- (20t^2 + 100t + 240)(t + 5)}{t^4 + 25t^2 + 3600 + 2(t^2 \times 5t) + 2(5t \times 60) + 2(t^2 \times 60)}\\\\\\P'(10) = \dfrac{(200 + 100)(100 + 50 + 60)- (2000 + 1000 + 240)(10 + 5)}{10000 + 2500+ 3600 + 2(5000) + 2(3000)+ 2(6000)}\\\\P'(10) = 0.3265[/tex]
Hence, the rate at which the population be increasing when t = 10 is 0.3265.
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