EXERCISE This table and plot display the population P of yea

EXERCISE This table and plot display the population P of yeast cells observed in a sample of grape juice being fermented into wine, measured over a time t The data looks like the relationship between P and is not a linear one. Explain why we would conclude that P does not follow the function P -mt (Draw on the graph if that helps) (minutes)P t (minutes) 12 37 112 341 1035 3143 9548 4 5 6 Population of yeast cells in grape juice 10000 r 8000 4000 2000 t (minutes)

Solution

1. P = mt is the equation of a straight line passing through the origin.Further, if the function P were a straight line, all the plotted observations would be on the straight line and the graph of the function, therefore would lokk like a straight line. However, the graph of P does not look like a straight line. Therefore, P is not a linear function.

Alternatively, on substituting P = 12 and t=1 in the equation P = mt, we get m = 12. Then, when t = 2, P (2) should be 12*2 24 and when t = 5, P(5) should be 12*5 = 60. However, we find that P(2) = 37 and P(5) = 1035. Similarly, none of the other observations fit in with the equation P = mt. Therefore, P is not a linear function.

2. From the graph plotted, it appears that there is a linear relationsip betweeb ln(P) and t. Let ln (P) = at + c. When t = 1, ln(P) = 2.50. Therefore a + c = 2.50...(1). Also, when t = 2, ln(P) = 2a + c = 3.61...(2). Then, on subtracting the 1st equation from the second equation, we get 2a + c - a - c = 3.61 - 2.50 or, a = 1.11. Then, from the 1st equation , we get 1.11 + c = 2.50 so that c = 2.50 - 1.11 = 1.39. Then ln(P) = 1.11t + 1.39 ...(3) Then, when t =0, we have ln(P) = 1.39. As per the graph, the intervept is about 1.38. The two obsevations are very close to each other. This is the source of the intercept.

3. When e^ln(P) = e^mt+b , we have on taking natural log of both the sides, ln(P)ln(e ) = (mt + b) ln( e) ( as log a^b = b loga) Further, since ln (e) = 1, we have ln(P) = mt + b . Therefore P = e^{mt + b}. Also, then P = e^b. e^mt as a^{x + y} = a^x . a^y . We have determined in the previous exercise that ln (P) = 1.11t + 1.39 ( 3rs equation). Therefore, P = e^{1.11t+ 1.39} = e^1.39. e^1.11t Now, we know that e^1.39 = 4.01485 = 4 (approximately). Therefore, P = 4e^1.11t Now, if P = P₀ et/ , where P₀ = 4, then = 1/1.11.= 0.09.From the 3rd equation above, we have ln(P) = 1.11t + 1.39 =( 1/ )t + 1.39. The slope of this straight line is 1/ . Thus the time constant iun the exponential function P = P₀ et/, is the reciprocal of the slope of the straight line  ln(P) = 1.11t + 1.39.