question 16: 8 pts solve the problem. a population grows from an initial size of 10 people to an amount…

question 16: 8 pts solve the problem. a population grows from an initial size of 10 people to an amount p(t), given by p(t)=10(1 + 0.5t + t³) where t is measured in years from 1996. find the acceleration in the population t years from 1996. edit view insert format tools table 12pt paragraph b i u a t²

question 16: 8 pts solve the problem. a population grows from an initial size of 10 people to an amount p(t), given by p(t)=10(1 + 0.5t + t³) where t is measured in years from 1996. find the acceleration in the population t years from 1996. edit view insert format tools table 12pt paragraph b i u a t²

Answer

Explanation:

Step1: Recall acceleration - derivative relation

Acceleration of population is the second - derivative of population function. First, find the first - derivative of $P(t)=10(1 + 0.5t+t^{3})=10 + 5t+10t^{3}$.

Step2: Differentiate $P(t)$ to get $P^\prime(t)$

Using the power rule $\frac{d}{dt}(x^{n})=nx^{n - 1}$, we have $P^\prime(t)=\frac{d}{dt}(10)+\frac{d}{dt}(5t)+\frac{d}{dt}(10t^{3})=0 + 5+30t^{2}=5 + 30t^{2}$.

Step3: Differentiate $P^\prime(t)$ to get $P^{\prime\prime}(t)$

Differentiate $P^\prime(t)=5 + 30t^{2}$ with respect to $t$. Using the power rule again, $P^{\prime\prime}(t)=\frac{d}{dt}(5)+\frac{d}{dt}(30t^{2})=0+60t = 60t$.

Answer:

$60t$