we know dA(t) = Rin - ROLE dt in where R = concentration of salt in inflow × input rate = (1gm/ltr)×(4 ltr/min) = 4gm/min = Roi concentration of salt in out flow output rate = A(t) 200 gm/ltrx (4 ltr/min) = A(t) 50 gm/min dA(t) A(t) =4- dt 50 dA(t) 200-A(t) dt 50 dA(t) 1 -dt = 200-A(t) 50 Above equation is in variable separable form .. on integrating dA(t) = (1/1 dt + C C is constant of integration. 200-A(t) 50 1 In(200 A(t)) = · -t+C 50 1 1 In = -t+C 200-A(t) 1 200-A(t) 1 200-A(t) =e50° 1 200 A(t)= Pe 50* P = ec 50 by taking exponential power 介 A(t) 200-Pe 50 P is a constant.
we know dA(t) = Rin - ROLE dt in where R = concentration of salt in inflow × input rate = (1gm/ltr)×(4 ltr/min) = 4gm/min = Roi concentration of salt in out flow output rate = A(t) 200 gm/ltrx (4 ltr/min) = A(t) 50 gm/min dA(t) A(t) =4- dt 50 dA(t) 200-A(t) dt 50 dA(t) 1 -dt = 200-A(t) 50 Above equation is in variable separable form .. on integrating dA(t) = (1/1 dt + C C is constant of integration. 200-A(t) 50 1 In(200 A(t)) = · -t+C 50 1 1 In = -t+C 200-A(t) 1 200-A(t) 1 200-A(t) =e50° 1 200 A(t)= Pe 50* P = ec 50 by taking exponential power 介 A(t) 200-Pe 50 P is a constant.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
A tank contains 200 liters of fluid in which 30 grams of
salt is dissolved. Brine containing 1 gram of salt per liter
is then pumped into the tank at a rate of 4 L/min; the
well-mixed solution is pumped out at the same rate. Find
the number A(t) of grams of salt in the tank at time t.
I've attached a photo of the solution to this problem. My question is why it's Rin-Rout instead of Rout - Rin. I don't understand the difference between using Rin-Rout and Rout-Rin. Please provide a clear explanation for this part.
Transcribed Image Text:we know
dA(t)
= Rin - ROLE
dt
in
where R = concentration of salt in inflow × input rate = (1gm/ltr)×(4 ltr/min)
= 4gm/min
=
Roi concentration of salt in out flow output rate =
A(t)
200
gm/ltrx (4 ltr/min)
=
A(t)
50
gm/min
dA(t) A(t)
=4-
dt
50
dA(t) 200-A(t)
dt
50
dA(t) 1
-dt
=
200-A(t) 50
Above equation is in variable separable form
.. on integrating
dA(t) = (1/1 dt + C C is constant of integration.
200-A(t) 50
1
In(200 A(t)) = ·
-t+C
50
1
1
In
=
-t+C
200-A(t)
1
200-A(t)
1
200-A(t)
=e50°
1
200 A(t)= Pe 50*
P = ec
50
by taking exponential power
介
A(t) 200-Pe 50
P is a constant.
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