Answer:
(a)
[tex]\frac{dP(t)}{dt}=\frac{800(2t+5)}{(t^2+5t+40)^2}[/tex]
(b)
[tex]P(10)\approx20789[/tex]
[tex]\frac{dP(t)}{dt} \left \{ {{} \atop {t=10}} \right. \approx0.55[/tex]
Step-by-step explanation:
(a) In order to find the rate at which Glen Cove's population is changing with respect to time we need to calculate the derivative of P(t):
Use the quotient rule:
[tex]\frac{d}{dt} (\frac{u}{v} )=\frac{v\frac{du}{dt} -u\frac{dv}{dt} }{v^2}[/tex] (1)
Where:
[tex]u=25t^2+125t+200\\v=t^2+5t+40[/tex]
[tex]\frac{du}{dt} =50t+125\\\\\frac{dv}{dt} =2t+5[/tex]
Replacing into the equation (1) and symplifying:
[tex]\frac{dP(t)}{dt}=\frac{800(2t+5)}{(t^2+5t+40)^2}[/tex]
(b)
In order to find what's the population after 10 years, just evaluate P(t) at t=10
[tex]P(10)=\frac{25(10)^2+125(10)+200}{(10)^2+5(10)+40} =\frac{2500+1250+200}{100+50+40} =\frac{3950}{190} =20.78947368[/tex]
Glen Cove's population is in thousands so:
[tex]P(10)=20.78947368\times10^3=20789.47368\approx 20789[/tex]
Finally, to find the rate at which the population is increasing, you need to evaluate the derivative of P(t) at t=10
[tex]\frac{dP(t)}{dt} \left \{ {{} \atop {t=10}} \right. \approx0.55 =\frac{800(2(100)+5)}{((10)^2+5(10)+40)^2} =\frac{20000}{36100}=0.5540166205\approx0.55[/tex]
