Water is flowing into and out of a small reservoir. The total amount of water can be modeled by R, where R(t)=20-15sint^2/25 cubic feet, t is measured in hours and 0

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  a) about -1.03 cubic feet per hour

  b) R'(t) = -6/5t·cos(t^2/25); the rate of flow into the reservoir

  c) 0.859 and 5.972 hours

  d) 9.838 hours

Step-by-step explanation:

The rest of the problem statement is in the first attachment. See the second attachment for the graphing calculator input/output.

a) The average rate of change on the interval [0, 8] is ...

  (R(8) -R(0))/(8 -0) ≈ -1.03004 . . . cubic feet per hour

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b) The derivative can be found using the chain rule.

  R'(t) = -15(2t/25)cos(t^2/25)

  R'(t) = -6/5t·cos(t^2/25)

The derivative of the volume is the rate of change of volume, that is, the flow rate into the reservoir.

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c) To find the time values when R' = 'average rate of change', we defined the constant a1 to be that average rate of change. The graphing calculator shows the curve R'(t)-a1, which has zeros at the times of interest. The times for which the instantaneous rate of change equals the average rate of change are t = 0.859 and t = 5.972 hours.

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d) The time of interest is beyond the domain of the function. We cannot use the given function to determine the time of interest.

However, if we extend the domain to include the time of interest, we find it is t = 9.838 hours.

yes, what he said

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