# Thread: another application problem (population)

1. ## another application problem (population)

The US prison population followed the curve

N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88

(0 < t < 39)

in the years 1950 – 1989. Here t is the number of years since 1950, and N is the number of prisoners in thousands. When to the nearest year was the prison population decreasing most rapidly? When was it increasing most rapidly?

The took the derivative N(t) and set it equal to zero to find stationary points. I ended up getting answers the answers 16 and 9 but I know those are the wrong answers

2. Originally Posted by Kenneth
The US prison population followed the curve

N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88

(0 < t < 39)

in the years 1950 – 1989. Here t is the number of years since 1950, and N is the number of prisoners in thousands. When to the nearest year was the prison population decreasing most rapidly? When was it increasing most rapidly?

The took the derivative N(t) and set it equal to zero to find stationary points. I ended up getting answers the answers 16 and 9 but I know those are the wrong answers
N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88 -------(i)

That is a cubic function. If you graph that, you'd see that it is like a strectched slightly vertical S. Meaning, the graph goes up, goes down, then goes up again.
One of the characteristics of the cubic curve is it goes upward very fast, slows down to stop at the relative maximum point, goes downward, picking speed downward until it wants to go upward again. This is at the inflection point. This is where the graph attains its maximum downward speed. Then passing that inflection point, the graph will change curvature, preparing to go upward again. So it slows down on its downward plunge until it reaches the relative minimum point, after which it shoots up for the sky, picking up speeds as time goes by, because it won't go downward anymore.

So,
a)The graph goes downward fastest at the inflection point. We can get that through the second derivative of N(t). At the inflection point, N''(t) = 0.
b) maximum speed upward, in the range of the function, is either at the initial point, at t=0, or at terminal point, at t=39. We can get the rates through the first derivatve of N(t)

------------
N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88 -------(i)
N'(t) = 0.084702t^2 -2.1844t +13.029
N''(t) = 0.169404t -2.1844
Set that to zero,
t = 2.1844 /0.169404 = 12.89 or 13 years after 1950.
Therefore, the prison population was decreasing most rapidly in 1963. ----answer.

-------------
N(t) = 0.028234t^3 – 1.0922t^2 + 13.029t + 146.88 -------(i)
N'(t) = 0.084702t^2 -2.1844t +13.029

N'(0) = +13.029
N'(39) = 0.084702(39^2) -2.1844(39) +13.029 = +56.67
So, the slope of the curve is greater at t=0 than at t=39.
Therefore, the prison population was increasing most rapidly in 1950. ----answer.

3. Are you sure that you didn't mean to the say the slope of the curve is greater at t = 39 than t = 0? Because in the back of my book it says that the answer is 1989 was the year that the population was increasing mostly rapidly.

Thanks for explaining the first part of the answer by the way.

4. Originally Posted by Kenneth
Are you sure that you didn't mean to the say the slope of the curve is greater at t = 39 than t = 0? Because in the back of my book it says that the answer is 1989 was the year that the population was increasing mostly rapidly.

Thanks for explaining the first part of the answer by the way.
Yes, hey, was I wrong in part b). I thought the slope at t=0 was +130.29 when in fact it was only +13.029.
Since +56.69 gradient at t=39 is steeper than the 13.029 gradient at t=0, then the answer should be in 1989 when the prison population increased the fastest.

I need to clean my reading glasses sometime.