If fog is so bad that I can’t see for short distance what should I do
2 answers:
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Answer / Explanation:
To solve this , we start by recalling the conversion principle:
Thus, dc/dt = Rate of C in - Rate of C out,
Therefore, for the number of chemical C(t) , i gram, in the pond at time (t), the rate of flow of C is modeled as:
Rate of Cin = Cin F,
Where Cin is the concentration of the chemical going in ( in this case, Cin = 2 grams per gallon), and F is the rate of flow of water into the pond (in this case, F = 5 million gallons per year). We then have that
Rate of C in = 10 million grams per year
Therefore the rate of Cout = c(t)F, where c(t) = C(t) / V is the concentration of the chemical in the pond ( here, we are assuming instant mixing). The volume, (V) of the water in the pond is 10 million gallons (we are assuming that the rate of flow of water into the pond is the same as the rate of flow out, so that the volume of water in the pond remains constant). Thus,
Rate of C out = 1/2 C(t) million grams per year.
Thus, the differential equation describing the evolution of C =
C(t) in time is: dc/dt = 10 - C/2 in millions of grams per year.
Now to plot the solution using Maple and to describe in words the effect of the variation in the incoming chemical,
we have: C(t) = 20 +ce ∧ -t/2, where c is an arbitrary constant since there was no chemical in the pond initially.
Therefore, C(t) = 0 , then C = -20
So that Q(t) = 20 ( 1 - e ∧ -t/2)
Kindly find the expression of the above statement C = C(t) below:
Note: In the diagram, Q is represented by C
Therefore, C = Q
