Question

Question 18: A fungus is spreading on the surface of a wall. A biologist proposes a model for the rate at which the fungus is spreading on the wall. The total surface area of the wall is 9 m². The surface area that is affected at time t hours is x m². The biologist proposes that the rate of change of x is proportional to the product of the surface area that is affected and the surface area that is unaffected. Write 9 down a differential equation for this model. If x= 1 when t= 0, show that x=- * 1+8e-9ht a) Determine the value of k if it is found that, after 1 hour, 1.8 m² of surface area of the wall is affected by fungus. b) By using the value of k found in (a), find the time taken for 50% of the area of the wall to be affected.

Answer 1

affected at Solution: Let the surface area that is time t hours is a m?.. By the given data, de =kx(9-x) where & is proportional - constant ox, d = cat OY, [+ o) dx = kät entertaling we get, ☆ [ln(x) – en (9-x)] =kt te where cis arbitrary constant Now, given that x=1 when t=0 Then, t[ lnca) - n(8)] = C. or, e = -in(?). so, t[en(x) - (9-x) ) = kt - § 40(8) or, ten (e) =kt 9-x a 8x = eekt om, go-x - égkt & te9K+ = 1+82387 a, x= Beskt. (proved?

Solndfon: @ Griven that, 1.8 x= 1.8 when t=1 Then from @ we ret, 1.8 = 14 Bek or, 1+889K = = 5 er, 8ē9K = 4 on, ek = egy = 2 a, 9x = ln(2) OV, K = ence) en - 9 ". The valne of k is en & Now 50% of the area of tere wall = 3 / 2 = 4.5 m² from @ we aret, 1 - 148€SKE O, 1+86-9kt = 2 or, é okt = $ ar, e9xt = 8 : In (2) OV, 9kt = In (8) en (8) using (2) a, t = en (8) = . 2ncz? i. The required time is 3 hours.