Question

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 4 turtles/yr t = 6 1516 Incorrect: Your answer is incorrect. turtles/yr What will be the population 10 yr after the conservation measures are implemented? 3250 Incorrect: Your answer is incorrect. turtles Need Help? Read It Watch It

Answer 1

Part 1: Growth Rate = 48.63%

Part 2: 3260

Explanation:

Part1:

The number of turtles, t years from now, is found by the equation:

`N(t)=2t^3 +3t^2 -4t + 1000`

Rate of growth from t=2 to t = 6:

Turtle population at end of year 2:

`N(2)=2(2)^3 +3(2)^2 -4(2) + 1000\\N(2)=1020`

Turtle population at end of year 6:

`N(6)=2(6)^3 +3(6)^2 -4(6) + 1000\\N(6)=1516`

Rate of Growth is:

`(1516-1020)/(1020)=0.4863`

Growth Rate = 48.63%

Part2:

To find population after 10th year, we simply plug in 10 into the equation of N(t):

`N(t)=2t^3 +3t^2 -4t + 1000\\N(10)=2(10)^3+3(10)^2-4(10)+1000\\N(10)=3260`

The population of turtles after 10th year is 3260