Question

Conservation of Species A certain species of turtle faces extinction because dealers collect truckloads of turtle eggs to be sold as aphrodisiacs. After severe conservation measures are implemented, it is hoped that the turtle population will grow according to the rule N(t) = 2t3 + 3t2 − 4t + 1,000 (0 ≤ t ≤ 10) where N(t) denotes the population at the end of year t. Find the rate of growth of the turtle population when t = 2 and t = 6. t = 2 turtles/yr t = 6 turtles/yr What will be the population 10 yr after the conservation measures are implemented?

Answer 1

Answer:

The turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

Ten years after the conservation measures are implemented the population will be 3260 turtles.

Step-by-step explanation:

To find the rate of growth of the turtle population at any time t you need to find $N'(t)$

$\frac{d}{dt}N(t)=\frac{d}{dt}(2t^3+3t^2-4t+1000)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dt}(2t^3)+\frac{d}{dt}(3t^2)-\frac{d}{dt}(4t)+\frac{d}{dt}(1000)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\N'(t)=6t^2+6t-4$

In particular, when t = 2 and t = 6, we have

$N'(2)=6(2)^2+6(2)-4=32\\\\N'(6)=6(6)^2+6(6)-4=248$

so the turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

The turtle population at the end of the tenth year will be

$N(10)=2(10)^3+3(10)^2-4(10)+1000\\N(10)=3260 \:turtles$