# production of phosphate rock (in thou of metric tons) is given approx by f(t) =–890t2 + 3100t+ 43000;where t is time in years and t = 0 corresponds to 1995. (a) Find the average rate of...

production of phosphate rock (in thou of metric tons) is given approx by *f*(*t*) =–890*t*2 + 3100*t*+ 43000;where *t* is time in years and *t* = 0 corresponds to 1995.

(a) Find the average rate of change of production between 2000 and 2012.

(b) Find the instantaneous rate of change of production in 2012.

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a) You need to evaluate the average rate of change of production such that:

`A(t) = (f(t_2) - f(t_1))/(t_2 - t_1)`

The problem provides the informations `f(t_2) = 2012, f(t_1) = 2000` and f(t) = 1995 for t = 0.

Since you know that f(t) = 1995 for t = 0, you may evaluate `t_2` and `t_1` such that:

`2000 = t + t_1=gt 2000 = 1995 + t_1 =gt t_1 = 5`

`2012 = t + t_2=gt 2000 = 1995 + t_2 =gt t_2 = 7`

You need to substitute the given values in formula A(t) such that:

`A = (2012 - 2000)/(7-2)`

`A = 2/2 =gt A = 1`

**Hence, evaluating the average rate of change of production between 2000 and 2012 yields A = 1**.

b) You need to evaluate the instantaneous rate of change of production at the moment t= 2012, hence you need to evaluate the derivative of function at t = 2012 such that:

`f'(t) = -890*2*t + 3100`

`f'(t) = -1780*t + 3100`

You need to substitute 2012 for t in equation `f'(t) = -1780*t + 3100` such that:

`f'(2012) = -1780*2012 + 3100`

`f'(2012) = 3578260`

**Hence, evaluating the instantaneous rate of change of production at the moment t= 2012 yields `f'(2012) = 3578260.` **