The rate at which water is supplied to a factory is given by V(t) = 3t^2 - 16t + 10 m^3/hour where t = 0 at 6:00 AM. How much water is supplied between 9 AM and 1 PM.

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aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

The rate at which water is supplied to the factory given by

`(dV)/(dt)=3t^2-16t+10 m^3/(hr.)`

Thus volume of the water supplied from 6 AM to 1.00 PM.It is not given to us when t=0 what V= ? ,so we can not solve this as initial value problems.

`V=int_6^13(3t^2-16t+10)dt`

`=int_6^9(3t^2-16t+10)dt+int_9^13(3t^2-16t+10)dt`

`Thus `

`int_9^13(3t^2-16t+10)dt=int_6^13(3t^2-16t+10)dt-int_6^9(3t^2-16t+10)dt`

`=(t^3-8t^2+10t)_6^13-(t^3-8t^2+10t)_6^9`

`=(975-(-12))-(171-(-12))`

`=987-183`

`=804`  `m^3`

Thus volume of water supplied to factory from 9.00 AM to 01.00 PM

is  804 cubic m.

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The rate at which water is supplied to a factory is given by `V(t) = 3t^2 - 16t + 10` m^3/hour where t = 0 at 6:00 AM.

Assuming t is the no. of hours starting from 6:00 AM,

t=3 at 9 AM and t=7 at 1 PM.

Hence, the volume of water supplied between 9 AM and 1 PM is given by:

`int_3^7 3t^2 - 16t + 10 dt`

`=[3*t^3/3-16*t^2/2+10t]_3^7`

`=[t^3-8t^2+10t]_3^7`

`=(7^3-8*7^2+10*7)-(3^3-8*3^2+10*3)`

`=36`

Therefore, the volume of water supplied in the given interval of time is 36 m^3.

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