The rate of flow of water into a dam is given by R'= 500+20t Lh^-1. If there is 15,000L of water initially in the dam, how much water will there be in the dam after 10 hours?

answer is 21,000L

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Rate of flow of water into the dam, R'= 500+20t = dR/dt

or dR = (500 + 20t) dt

Integrating both sides:

R = 500t + 20 * (t^2/2) + c , c being integration constant

Initial condition: t = 0, R = 15,000

So 15,000 = 500*0 + 20*(0^2/2) + c

or c = 15,000

Plugging in the value of c = 15,000

the amount of water going into the dam at time t is given by:

R = 500t + 20 * (t^2/2) + 15,000

So after 10 hours,

R = 500*10 + 20 * (10^2/2) + 15,000

= 5,000 + 1,000 + 15,000

= **21,000 L**

We have given

`(dR)/(dt)=500+20t`

`dR=(500+20t)dt`

on integration

`R=500t+20t^2/2+c`

`R=500t+10t^2+c`

where c is integrating constant,to be determined with initiatial condition.

t=0 , R=15000

15000=500 x 0+ 10 x 0+c

c=15000 L

Water in dam at any time t is

R=500t+10t^2+15000

Water in dam after 10 hrs.

R=500 x 10+ 10 x100 +15000

R=5000+1000+15000

**R=21000 L**

**Ans. There will be 21000 L water in the dam after 10 houurs .**

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