A major corporation is building a 4325-acre complex of homes, offices, stores, schools, and churches in the rural community of Glen Cove. As a result of this development, the planners have estimated that Glen Cove's population (in thousands) t years from now will be given by(25t2 + 150t + 100)/(t2 + 6t + 24)(a) Find the rate at which Glen Cove's population is changing with respect to time.(b) What will be the population after 10 years? (Round your answer to the nearest person.)(c)At what rate will the population be increasing when t = 10? (Round your answer to the nearest integer.)

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]