Question

A population grows from an initial size of 10 people to an amount P(t), given by P(t)=10(3+0.5t+t^3) where t is measured in years from 1996. Find the acceleration in the population t years from 1996.

a)60t people per year^2
b)30t people per year^2
c)600t people per year
d)36t people per year^2

Answer 1

Since the required is the acceleration in the population, the second derivative of the equation is needed.
The first derivative has to be determined first.
dP(t)/dt = 10(0.5+3t^2)
Next, the second derivative can be determined using the first derivative:
d2P(t) = 10(6t) = 60t
Since it's acceleration, the unit is per year^2

THE ANSWER IS a) 60t people per year^2 

Answer 2

The acceleration in the population t years from 1996 is 60t people per year^2.

What is a Function?

A function assigns the value of each element of one set to the other specific element of another set.

It is given that a population grows from an initial size of 10 people to an amount P(t), given by P(t)=10(3+0.5t+t³) where t is measured in years from 1996.

Now, in order to find the acceleration in the population t years from 1996, we need to find the second derivative of the population function, therefore,

$P(t)=10(3+0.5t+t³)\\\\P'(t)=10(0.5+3t^2)\\\\P''(t)=10(6t)\\\\P''(t)=60t$

Hence, the acceleration in the population t years from 1996 is 60t people per year^2.

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