1. [13 marks] A river flows at the rate of 60,000 m3 /day into a full reservoir of volume 400,000 m3...

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1. [13 marks] A river flows at the rate of 60,000 m3 /day into a full reservoir of volume 400,000 m3 , and the water in the reservoir is pumped out at the same rate for a town’s water supply. The river water contains certain bacteria at concentration 0.02 g/m3 , and the initial concentration of the bacteria in the reservoir is 0.005 g/m3 . (a) Let y(t) be the amount of bacteria (in grams) in the reservoir at time t (days). Write down and solve an appropriate differential equation for y(t) along with the appropriate initial condition. (b) Using the solution in part (a), determine what happens to the concentration of bacteria in the reservoir as t ? 8. (c) The water from the reservoir is not safe to drink if the bacteria concentration reaches 0.01 g/m3 . How many days does it take for the bacteria concentration to reach this level? Express the answer in exact form and to 3 decimal places. 2. [10 marks] Find the derivatives of each of the following, showing the working and simplifying. (a) f(x) = (4 + 3 sin2 x) 5/3 (b) g(t) = tan(3t + 4e -2t ) (c) y = y(x), where x 2 sin y + cos(x + 3y) = 1 3. [11 marks] Find the following integrals exactly. Use calculus and show all working. (a) ? (4 cos 2x - sin px) dx (b) ? v p 0 x cos(x 2 - p 4 ) dx (c) ? 2x 3 (1 + x 8 ) -1 dx 4. [11 marks] The top of a buoy moves vertically (due to waves) such that its height (in metres) above sea level is h(t) = 0.8 sin(p(t - 1)) + 2 where t = 0 is time in seconds. (a) Using the formula for h(t), find its period. Sketch the graph of h(t) for 0 = t = 4. (b) Find the time at which the top of the buoy reaches its maximum height for the 100th time, showing the reasoning. (c) Find h ' (t) and evaluate h ' (4/3) exactly. Is the top of the buoy moving up or down at t = 4/3 and why? (d) Find the smallest value of t = 0 for which h ' (t) is a maximum, showing the working.

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MAS182: Applied Mathematics Semester 1, 2013 Assignment 5 Due by: 4:00pm Monday, 20 May 2013 3 1. [13 marks] A river ?ows at the rate of 60,000 m /day into a full reservoir of volume 3 400,000 m , and the water in the reservoir is pumped out at the same rate for a town’s 3 water supply. The river water contains certain bacteria at concentration 0.02 g/m , and 3 the initial concentration of the bacteria in the reservoir is 0.005 g/m . (a) Let y(t) be the amount of bacteria (in grams) in the reservoir at time t (days). Write down and solve an appropriate di?erential equation for y(t) along with the appropriate initial condition. (b) Using the solution in part (a), determine what happens to the concentration of bacteria in the reservoir as t!1. (c) Thewaterfromthereservoirisnotsafetodrinkifthebacteriaconcentrationreaches 3 0.01 g/m . How many days does it take for the bacteria concentration to reach this level? Express the answer in exact form and to 3 decimal places. 2. [10 marks] Find the derivatives of each of the following, showing the working and simpli- fying. 2 5=3 (a) f(x) = (4+3sin x) -2t (b) g(t) = tan(3t+4e ) 2 (c) y = y(x), where x siny+cos(x+3y) = 1 3. [11 marks] Find the following integrals exactly. Use calculus and show all working. ? (a) (4cos2xsinx)dx v ? 2 (b) xcos(x )dx 4 0 ? 3 8 -1 (c) 2x (1+x ) dx 4. [11 marks] The top of a buoy moves vertically (due to waves) such that its height (in metres) above sea level is h(t) = 0:8sin((t1))+2 where t 0 is time in seconds. (a) Using the formula for h(t), ?nd its period. Sketch the graph of h(t) for 0 t 4. (b) Find the time at which the top of the buoy reaches its maximum height for the 100th time, showing the reasoning. ' ' (c) Find h(t) and evaluate h(4=3) exactly. Is the top of the buoy moving up or down at t = 4=3 and why? ' (d) Find the smallest value of t 0 for which h(t) is a maximum, showing the working. Notes: 10% of the marks for this assignment are reserved for presentation. ...

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Robert · Accepted Answer

Solution 
Rate of river Flow =  ????
????
  = 60,000 m3/day 
Total Volume = V = 400,000 m3 
Rate of change of concentration is  
dC/dt ∝ -C 
dC/C = -kdt  
where k is constant which is equivalent to dv/vdt 
on integrating both sides we get, 
logC = kt + A 
at t=0 C = 0.05 
hence  
log0.05 = A  
so,  
logC = kt + log0.05 
logC/0.05 = kt 
C/0.05 = ekt 
C = 0.05ekt 
If y(t) be the amount of bacteria then we can write  
y(t) / V = 0.05ekt 
y(t) = 0.05Vekt
Solution 
At t tends to infinity, reservoir will be completely full of water that means  
C = 0.05e∞ 
Since e∞ is infinity then bacterias concentration will be infinite.
Solution 
C = 0.05ekt 
Given that  
C = 0.01g/m3 
0.01 = 0.05 ekt 
ekt = 1/5 
kt = log0.2 
t = 1.609/dv/V 
t = 1.609 * 400000/60000 
t = 10.72 days
Solution 
On diffrentiating both sides with respect to x we get 
f’(x) = ??(4+3 sin
2 ??)5/3
????
  
We will need to diffrentiate it by chain rule 
= ??(4+3 sin
2 ??)5/3
????
 ∗ ??(4+3 sin
2 ??)
????
  
=5
3
(4 + 3 sin2 ??)2/3* (0+6sinx* ??????????
????
) 
f’(x)  = ??
??
(?? + ?? ???????? ??)??/??*6sinx*cosx
Solution 
On diffrentiating both sides with respect to t we get 
g'(t) = ???????? (3??+4??
−2??)
????
 
We will need to diffrentiate it by chain rule 
= ???????? (3??+4??
−2??)
????
∗ ??(3??+4??
−2??)
????
 
= 1
1+ (3??+4??−2??)2
*(3+4*-2??−2??) 
g'(t) = (?? − ????
−????)
??+ (????+????−????)??

Solution 
On diffrentiating both sides with respect to x we get 
y'(x) = ??  (??
2????????+cos (??+3??))
??

1. [13 marks] A river flows at the rate of 60,000 m3 /day into a full reservoir of volume 400,000 m3 , and the water in the reservoir is pumped out at the same rate for a town’s water supply. The...

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