Class notes
Written Assignment 2
- Course
- MATH 100
- Institution
- University Of British Columbia (UBC )
In Math 100 of UBC, there is 5 written assignment in total which weigh 20% of your final course.
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ASSIGNMENT 2Trung Duong Nguyen - 14624761
Syifa Nadhira Nurul Izzah - 63842660
Hussein Al Aaref - 19722024
1. Consider the differences between the types of questions on the previous written assignment, Assignment
1, and the types of questions on your high school math assignments. In one or two paragraphs, describe
one major difference, and explain how you might approach assignments in this course differently. Be
specific and provide details.
(All assignments starting from this one will include a reflection question designed to encourage you to
think deeply about how mathematics is done.)
We think that the main difference between the previous assignment and the assignments from high
school is the amount of instruction that we gain from the professors. In high school, whenever there is
an assignment, either minor or significant, the teachers will always talk about it, at least once, but for
the written assignments, we did not have that much instruction other than “The written assignment is
due this weekend”. This is obviously not an easy difference, but we think that it is a necessary change,
to practice being relied less on the teacher/professor.
To get used to the change, we simply change the way we approach the assignment, from waiting for
instructions, to actually taking a read of the assignment and looking for confusion. We decided for each
of the members to do their own version and before the deadline, we will create a final version, based on
each people’s opinion. We agreed that this is a good plan since everyone will have a chance to take a look
at the assignment and try to understand it.
Sea lice are a parasite that feed on salmon. They occur naturally in low densities on wild salmon, but
crowded conditions on salmon farms are ideal for sea louse outbreaks. Sea lice from salmon farms can spread
onto wild migrating salmon. To help protect wild salmon, the Canadian federal government requires salmon
farms to treat their fish with a parasiticide (or else to harvest the fish) when sea lice counts on a salmon
farm reach a threshold of three motile lice per fish.
A model for sea louse population on a salmon farm before and after treatment is given by
(
P eat if t < T
p(t) = −bt
,
Qe if t ≥ T
where p(t) is the number of motile lice per fish at time t, T is the time of treatment, and P, Q, a, b > 0 are
constant parameters.
2. Find the value of Q such that p(t) is continuous for all t.
From the function, we notice that lim − p(t) = P eaT and lim + p(t) = Qe−bT , since there is no restriction
t−>T t−>T
for p(t) that stop us to sub in the limit value for t (it is also one of the familiar function, which is the
, exponential function). For p(t) to be continuous for all t, lim p(t) for all a must exist, which also means
t−>a
that the two calculated limit must be equal:
P eaT = Qe−bT
ln(P eaT ) = ln(Qe−bT )
ln(P ) + ln(eaT ) = ln(Q) + ln(e−bT )
ln(P ) + aT ∗ ln(e) = ln(Q) − bT ∗ ln(e)
ln(P ) + aT + bT = ln(Q)
ln(P ) + T (a + b) = ln(Q)
ln(P ) + ln(eT (a+b) ) = ln(Q)
ln(P ∗ eT (a+b) ) = ln(Q)
P eT (a+b) = Q
− > the desired value calculated
In the winter of 2005-2006, the sea louse populations for two salmon farms in the Broughton Archipelago
were fitted to the model above. The population graphs are shown in Figure 1 at the bottom of this document.
3. (a) Use the limit definition of derivative to show that p(t) satisfies the differential equation
(
ap(t) if t < T
p′ (t) = .
−bp(t) if t > T
ch − 1
(You may use without proof the fact that lim = log(c) for all c > 0.)
h→0 h
From the original function and the differential equation, we can see clearly that p(t) has 2 ”parts”,
which we so worked on individually. We want to firstly calculate lim− p(t) to evaluate the first
x→T
branch of the function:
p(t + h) − p(t)
lim p(t) = lim
x→T − h→0 h
P ea(t+h) − P eat
lim− p(t) = lim
x→T h→0 h
P eat+ah − P eat
lim− p(t) = lim
x→T h→0 h
P eat ∗ eah − P eat
lim p(t) = lim
x→T − h→0 h
P eat (eah − 1)
lim− p(t) = lim
x→T h→0 h
a ∗ P eat (eah − 1)
lim p(t) = lim
x→T − h→0 a∗h
eah − 1
lim− p(t) = lim aP eat ∗ lim
x→T h→0 ah→0 ah
lim p(t) = ap(t) ∗ log(e)
x→T −
lim p(t) = ap(t)
x→T −
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