Answer:
i) ∴∫[tex]\frac{dA}{A}[/tex] = ∫ [tex]\frac{-1}{10} dt[/tex] ------ ( 1 )
ii) C = 4.303
Explanation:
Given data:
water in tank = 10 liters
concentration of salt flowing in = 0.5 liters/min
10 grams of salt is left in tank after 20 minutes
mixed water is drained at 1 liter per minute
i) Adequate mathematical model for the scenario
lets assume ; A( t ) amount of salt after t minute
A( t ) = 10grams where t = 20 minutes
differentiate A(t)
dA / dt = ( rate in ) - ( rate out )
= 0.5 * 0 - A/10
∴∫[tex]\frac{dA}{A}[/tex] = ∫ [tex]\frac{-1}{10} dt[/tex] ------ ( 1 )
ii) hence; In A = [tex]\frac{-1}{10} t + C[/tex] ----- ( 2 )
at t = 20 , A = 10grams
find C = In 10 + 2
= 2.303 + 2 = 4.303
back to ( 2 )
In A = [tex]\frac{-t}{10} + 4.303[/tex]
∴ A = e [tex]\frac{-t}{10} + 4.303[/tex]
