The rate of flow of water into a dam is given by R=500+20t L h^-1. If there is 15 000L of water initially in the dam, how much water will there be in the dam after 10 hours?

Question

Answer: 21 000L

Answer 1

If R' = 500 +20 t

then:

Rate of flow of water into the dam:

dR / dt = 500 + 20 t

dR = 500 dt + 20 t ∙ dt

Integrating both sides:

R = 500 t + 20 t² / 2 + C

R = 10 t² + 500 t + C

C = integration constant

Initial condition:

t = 0 , R = 15 000

R = 10 t² + 500 t + C

15 000 = 10 ∙ 0² + 500 ∙ 0 + C

15 000 = C

C = 15 000

R = 10 t² + 500 t + C

R = 10 t² + 500 t + 15 000

After 10 hours:

R = 10 ∙ 10² + 500 ∙ 10 + 15 000

R = 10 ∙ 100 + 5 000 + 15 000

R = 1 000 + 5 000 + 15 000

R = 21 000 L