A bucket filled with 10 gallons of water starts to leak at t=0. The volume of th

Question

A bucket filled with 10 gallons of water starts to leak at t=0. The volume of the water is given as V(t) =10 (1-t/100)^2 until the bucket is empty at t= 100. What is the rate of leaking water from the bucket at exactly 60 seconds? When is the instantaneous rate of change of V equal to the average rate of change of V from t=0 to t = 100. I factored the volume as v(t) =10 -t/5 + (t^2/1000) and set up the first question as v'(60)= 1/5 + 1/100 *60 but I dont think that is correct. Any help would be great!

Explanation / Answer

V(t) =10 (1-t/100)^2 ,,,> V'(t) = 2* 10 *(1-t/100)*(-1/100) so at t= 60 V't = 0.08 and for the second part........ average leaking rate = (10-0)/100 = 0.1 now this rate should be equal to the instantaneous rate so ...... 0.1 = 2* 10 *(1-t/100)*(-1/100 ) this gives......... t = = 50

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