Calc1231W7T2 - Brine and related topics

Wednesday, 5 September 2018

12:37 PM

Untitled picture.png Machine generated alternative text:
A tank holds 1500 litres of water in which 6000 grams of salt has been dissolved. Brine which contains 7 grams of salt per litre is run in at a rate of 15 litres per minute. (See
below for a picture of the tank).
The solution is thoroughly mixed and is extracted from the tank at the same rate. Let a: be the number of grams of salt in the tank after t minutes.
Based on the information given
• Initially there are 6000
O grams of salt in the tank.
• salt is being added to the tank at a rate of 105
• the number of litres in the tank at time t is 1500
O grams per minute
The salt is evenly distributed in the tank by mixing. Hence if there are a: grams of salt in the tank, then one litre taken from this tank contains x/1500
grams Of salt, and 15 litres contains —
grams of salt. The change in salt with respect to time is described by the differential equation
100
— inflow — outflow 105 — —.
dt
100

Untitled picture.png Machine generated alternative text:
Inflow
O
O
O
Outflow
O
O

Untitled picture.png Machine generated alternative text:
The previous question showed how to setup an initial value problem, such as
= 100 -
with the initial condition that 5000 when t O . Now we can solve this equation for x.
This differential equation is (select all that are true)
u exact
separable
linear
I-j unsolvable.
Hence, we compute that
100 '

Untitled picture.png Machine generated alternative text:
The previous question showed how to setup an initial value problem, such as
= 100 -
with the initial condition that 5000 when t O . Now we can solve this equation for x.
This differential equation is (select all that are true)
u exact
separable
linear
I-j unsolvable.
Hence, we compute that
100 '

Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings

Machine generated alternative text:
One method to solve
dc
= 15
is to write this as the linear differential equation
9
dt
dc
50
9
Multiplying by the integrating factor
and the right hand side as
exp(9V50)
+ —x = 15
dt
50
, we can write the left hand side as
9t
d
xe 50
dt
Then we integrate both sides of this expression with respect to t. The absence of initial conditions means a constant of integration remains in our answer. How do you prefer to write this?
(both answers are correct)
250
-9t/50-C
for some constant C
3
250
+ Ae¯9t/50 for some constant A.
3
The limiting value of as t DO is 250/3

Machine generated alternative text:
An unidentified virus infects a small town. Each month 50 new cases are reported; and through recovery or death 4% ofthe infected population no longer have the disease. Let V be the
number of people at time t who have the virus. Then
= 50
100
Initially, nobody has the disease. The solution to this initial value problem is:
Hence we can predict (to the nearest integer) the number of infected people after
. 4 months. V (4)
. 10 months: V(IO)
412
• 100 months: V(IOO)
1227
As time goes on, we expect that the number of infected people tends to
1250

Machine generated alternative text:
The effectiveness of a police force may be measured by its clearance rate: the number of charges laid in a month divided by the total number of unsolved crimes.
In Arachnid Boy's home town, new crimes are reported roughly 16 times per month, and while Arachnid boy is in town; the police clearance rate is 48%
Arachnid Boy comes back from his holiday and finds there are 150 unsolved crimes.
Let x be the number of unsolved crimes at the start of month t , with t = 0 representing the first month that Arachnid Boy is back from his holiday.
We can find an equation for x by solving the initial value problem
dc
16 -
dt
• initially, x
O t] t] crimes per month; and
The solution to this inital value problem is
161.48 +
As time drags on, the number of unsolved crimes approaches, to the nearest integer
lim x
33
Arachnid Boy by Dotdottronic on DeviantArt

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