Dead leaves accumulate on the ground in a forest at a rate of 3 grams per square

Question

Dead leaves accumulate on the ground in a forest at a rate of 3 grams per square centimeter per year. At the same time, these leaves decompose at a continuous rate of 55 percent per year. A. Write a differential equation for the total quantity Q of dead leaves (per square centimeter) at time t: dQdt= B. Sketch a solution to your differential equation showing that the quantity of dead leaves tends toward an equilibrium level. Assume that initially (t=0) there are no leaves on the ground. What is the initial quantity of leaves? Q(0)= What is the equilibrium level? Qeq= Does the equilibrium value attained depend on the initial condition? A. yes B. no

Explanation / Answer

The rate at which the total amount of leaves changes as a function of time is simply given by:

dQ/dt = rate leaves accumulate - rate leaves decompose.

Think of it this way: what would be your net progress in the forward direction if you too x steps per minute forward, and y steps per minute backward. It'd be (x-y) steps/minutes, right? This is the same idea. Leave pile up, but leave disappear (by decomposition). The net rate of pileup is the difference between those two rates.

In this case, the leaves accumulate at a constant rate of 4 gm/(yr*cm^2).

The leaves decompose at a constant precentage rate. That is, 0.55 of the leaves present in a given year decompose in that year. If Q is the amount of leaves present, then the rate they disappear is given by (0.55/yr)*Q. So:

dQ/dt = 4 gm/(yr*cm^2) - (0.55/yr)*Q

Let's write this as:

dQ/dt = a - b*Q (where a = 4 gm/(yr*cm^2) and b = (0.55/yr))

dQ/dt = -b*(Q - a/b)

This is a separable differential equation:

dQ/(Q - a/b) = -b dt

ln(Q - a/b) = -b*t + c

where c is the constant of integration. Let's define this constant in terms of the amount of leaves present on the ground (per cm^2, at time t = 0. Call this quantity Qo). Then:

ln(Qo -a/b) = c

So

ln(Q - a/b) = ln(Qo -a/b) - b*t

ln((Q - a/b)/(Qo - a/b)) = -b*t

Q - a/b = (Qo - a/b)*exp(-b*t)

Q(t) = a/b + (Qo - a/b)*exp(-b*t)

In this case

Q(t) = (4cm/(yr*cm^2))/(0.55/yr) + (Qo - (4cm/(yr*cm^2))/(0.55/yr))*exp(-0.55*t/yr)

Q(t) = (7.27 + (Qo - 7.27)*exp(-0.55*t/yr))*(cm/cm^2)

you are missing something from this question. a percentage should be given, and the answer would be dQ/dt = 55 - (percentage/100)Q

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.