Question

The weight of a population of yeast is given by a differentiable function y, where y(t) is measured in grams and t is measured in days. The weight of the yeast population increases according to the equation dy/dt= ky, where k is a constant. At time t=0, the weight of the yeast population is 120 grams and is increasing at the rate of 24 grams per day. Which of the following is an expression for y(t)?
a. 120e^24t
b. 120e^t/5
c. e^t/5+119
d. 24t+ 120

Answer 1

Answer:

b) $y = 120e^{t/5}$

Step-by-step explanation:

given that the weight of a population of yeast is given by a differentiable function y, where y(t) is measured in grams and t is measured in days. The weight of the yeast population increases according to the equation

$\frac{dy}{dt} =ky$ for a constant k.

Separate the variables and integrate

$ln y=k t +C\\y = Ae^{kt}$

Use the fact when t=0 y = 120

$120 = A$

$y = 120e^{kt}$

When t =1, rate of grwoth is 24 gms.

i.e. $144 = 120e^{k*1}\\k = ln 144/120 = 0.1823~0.2$

Hence option b i.e.

$y = 120e^{t/5}$