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 What will be the population 10 yr after the conservation measures are implemented? t= 6 I need to know what t=6 is

Answer 1

The rate of growth of the turtle population at t = 2 is 32 turtles per year, and at t = 6 it is 248 turtles per year. Ten years after conservation, the turtle population is projected to be 3,260 turtles.

Step-by-step explanation:

To find the rate of growth of the turtle population at t = 2 and t = 6, we need to calculate the first derivative of the given polynomial function N(t) = 2t3 + 3t2 − 4t + 1,000. The derivative N'(t) represents the rate of growth at any given time t.

First, we calculate the derivative:

N'(t) = d/dt(2t3 + 3t2 − 4t + 1,000)

N'(t) = 6t2 + 6t − 4

Then we plug in t = 2 to find the rate of growth at that time:

N'(2) = 6(2)2 + 6(2) − 4 = 24 + 12 − 4 = 32

So, the rate of growth when t = 2 is 32 turtles per year.

Similarly, we plug in t = 6:

N'(6) = 6(6)2 + 6(6) − 4 = 216 + 36 − 4 = 248

Therefore, the rate of growth when t = 6 is 248 turtles per year.

Lastly, to find the turtle population 10 years after conservation measures are implemented, we evaluate N(t) at t = 10:

N(10) = 2(10)3 + 3(10)2 − 4(10) + 1,000

N(10) = 2,000 + 300 − 40 + 1,000 = 3,260

The population is projected to be 3,260 turtles after 10 years.