The amount of a certain bacteria y in a petri dish grows according to the equation dy / dt = ky, where k is a constant and t is measured in hours.

Question

If the amount of bacteria triples in 10 hours, k is approximately what value? Chioces : (a) -1.204, (b) -0.110 , (c) 0.110 (d) 1.204 (e) 0.3

Answer 1

Integrate dy/y=k: ln(y)=k; y=ae^(kt) where a is a constant. When t=0, y=a; when t=10, y=3a, 3a/a=ae^(10k)/a; 3=e^(10k) and k is approximately ln(3)/10=0.10986 or 0.110 (answer (c)).

CHECK: y=ae^0=a; y=ae^1.1=3a (approx).