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 production of phosphate rock (in thou of metric tons) is given approx by f(t)...

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good2beme | Student, Undergraduate | Honors

Posted June 12, 2012 at 2:13 AM via web

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 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 change of production between 2000 and 2012.

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

   

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 12, 2012 at 5:01 AM (Answer #1)

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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.`

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