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can someone help me with (1.2)? The reference [1] provides a model of the rate o

ID: 3142829 • Letter: C

Question

can someone help me with (1.2)?

The reference [1] provides a model of the rate of production of methane (CH_4) by the archaebacteria Methanosarcina in the presence of nickel (Ni). Table 0.1 identifies the quantities involved in the model. Table 0.1: Quantities involved in the production of methane by Methanosarcina. The model relates such quantities by the following equations, numbered here as in reference [1, p. 138] K(t) = k_0 + k.1 middot A(t), dA/dt = beta middot K beta(t) = beta_0 + beta_1 middot A (t). This problem reveals that there is a time before which the model A accumulates all the available carbon on the planet. (1.1) Find a formula for A(t) in terms of t, k_0, k_1, beta_0, beta_1, with A(0) = 0, for t in an interval about 0. (1.2) Prove that there exists a time t_t > 0 such that for each amount Z > 0 there exists a time t_z

Explanation / Answer

dA / (B0 + B1*A) (k0 + k1*A) = dt

dA / (B0 k0 + (B1 k0 + k1 B0)*A + B1*k1*A2) = dt

dA / (A-m)2 - p2) = B1*K1 dt

p2 = (B0 k0) - [(B1 k0 + k1 B0)/ (2*B1*K1)]2

m = (B1 k0 + k1 B0)/ (2*B1*K1)

ln [(A-m-p)/ (A-m+p)] = 2p* B1*K1 t

let n = 2p* B1*K1

A-m-p / A-m+p = exp (nt)

A ( 1 - exp (nt)) = m (1 - exp (nt) + p (1 + exp (nt))

A = m + p (1 + exp (nt)) / (1 - exp (nt))

dA/dt = p (n exp (nt) (1 - exp (nt)) + n exp (nt) (1 + exp (nt))) / (1 + exp (nt)) * (1 - exp (nt))  

dA/dt = p ( 2n exp (nt) ) / (1 + exp (nt)) * (1 - exp (nt))

dA/dt > 0 for all values of t

so A is a monotonous function, which means there will always be a tt such that tt > tz when A > 0

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