2018 international practice exam ab mcq, part b - calculator required\nthe number of insects in a certain…

2018 international practice exam ab mcq, part b - calculator required\nthe number of insects in a certain population at time t days is modeled by the function p with first derivative p(t)=0.3t² + 12t + 210. at time t = 0, the number of insects in the population is 40.\ni. at time t = 10, the number of insects in the population is 2840.\nii. at time t = 10, the number of insects in the population is increasing at a rate of 360 insects per day.\niii. at time t = 10, the rate of change of the number of insects in the population is increasing at a rate of 18 insects per day per day.\n14 mark for review\nwhich of the following statements are true?\na i only\nb ii only\nc iii only\nd i, ii, and iii

2018 international practice exam ab mcq, part b - calculator required\nthe number of insects in a certain population at time t days is modeled by the function p with first derivative p(t)=0.3t² + 12t + 210. at time t = 0, the number of insects in the population is 40.\ni. at time t = 10, the number of insects in the population is 2840.\nii. at time t = 10, the number of insects in the population is increasing at a rate of 360 insects per day.\niii. at time t = 10, the rate of change of the number of insects in the population is increasing at a rate of 18 insects per day per day.\n14 mark for review\nwhich of the following statements are true?\na i only\nb ii only\nc iii only\nd i, ii, and iii

Answer

Explanation:

Step1: Find the rate of change of population at $t = 10$

We are given $P'(t)=0.3t^{2}+12t + 210$. Substitute $t = 10$ into $P'(t)$: $P'(10)=0.3\times(10)^{2}+12\times10 + 210=0.3\times100+120 + 210=30+120+210 = 360$. So the population is increasing at a rate of 360 insects per day at $t = 10$, and statement II is true.

Step2: Find the second - derivative of $P(t)$

Differentiate $P'(t)=0.3t^{2}+12t + 210$ with respect to $t$ to get $P''(t)$. Using the power rule, $P''(t)=0.6t+12$.

Step3: Evaluate the second - derivative at $t = 10$

Substitute $t = 10$ into $P''(t)$: $P''(10)=0.6\times10+12=6 + 12=18$. So the rate of change of the number of insects in the population is increasing at a rate of 18 insects per day per day at $t = 10$, and statement III is true. We are not given enough information to find the number of insects in the population at $t = 10$ (we only have the derivative and the initial - condition $P(0)=40$, and we would need to integrate $P'(t)$ to find $P(t)$), so statement I is not necessarily true.

Answer:

B. II only