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kati45 [8]
3 years ago
6

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

the amount of litter present, in grams per square meter, as a function of time t in years. If the litter falls at a constant rate of L grams per square meter per year, and if it decays at a constant proportional rate of k per year, then the limiting value of A is R = L/k. For this exercise and the next, we suppose that at time t = 0, the forest floor is clear of litter.
Required:
If D is the difference between the limiting value and A, so that D = R - A, then D is an exponential function of time. Find the initial value of D in terms of R.
Mathematics
1 answer:
Travka [436]3 years ago
4 0

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L  grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C'      (C' = kC)

taking exponents of both sides, we have

L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt}      (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k}  - \frac{C"}{k} e^{kt}

When t = 0, A(0) = 0 (since the forest floor is initially clear)

A = \frac{L}{k}  - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{0}\\\frac{L}{k}  = \frac{C"}{k} \\C" = L

A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

when t = 0(at initial time), the initial value of D =

D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

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-3 x - 5 = 36. What is X?
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Answer:

x = -13,67

Step-by-step explanation:

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3 years ago
Consider the equation 3x + 1 = 2x + 5
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Find the distance between the points L(7,-1) and M(-2, 4).
zysi [14]

Answer:

The answer is

\sqrt{106}  \:  \:  \: or \:  \:  \: 10.3 \:  \:  \:  \: units

Step-by-step explanation:

The distance between two points can be found by using the formula

d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  } \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

L(7,-1) and M(-2, 4)

The distance between them is

|LM|  =   \sqrt{ ({7 + 2})^{2}  + ( { - 1 - 4})^{2} }  \\  =  \sqrt{ {9}^{2}  +  ({ - 5})^{2} }  \\  =  \sqrt{81 + 25}  \\  =  \sqrt{106}  \:  \:  \:  \:  \:  \:  \:  \\  = 10.29563014...

We have the final answer as

\sqrt{106}  \:  \:  \: or \:  \:  \: 10.3 \:  \:  \:  \: units

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