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 ' 




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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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