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

Respuesta :

Answer:

b) [tex]y = 120e^{t/5}[/tex]

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

[tex]\frac{dy}{dt} =ky[/tex] for a constant k.

Separate the variables and integrate

[tex]ln y=k t +C\\y = Ae^{kt}[/tex]

Use the fact when t=0 y = 120

[tex]120 = A[/tex]

[tex]y = 120e^{kt}[/tex]

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

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

Hence option b i.e.

[tex]y = 120e^{t/5}[/tex]

The weight of a population of yeast is given by a differentiable function y, where [tex]\rm y (t)= 120e^{{t}/{5}}[/tex] and this can be determined by integrating the given data.

Given :

  • 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.

According to the given data:

[tex]\rm \dfrac{dy}{dt}= Ky[/tex]

[tex]\rm \dfrac{1}{y}dy=Kdt[/tex]

Now, integrate the above equation. The above expression becomes:

ln(y) = Kt + C

[tex]\rm y = Ae^{Kt}[/tex]   --- (1)

Now, at t=0, y = 120. Substitute these values in the above equation.

[tex]\rm 120 = Ae^{0}[/tex]

A = 120

At (t = 1) the rate of growth is 24 grams per day.

[tex]\rm 144=120e^{K}[/tex]

[tex]\rm K=ln\dfrac{144}{120}[/tex]

K = 0.2 (Approx)

So, substituting the known terms in equation (1).

[tex]\rm y (t)= 120e^{{t}/{5}}[/tex]

So, the correct option is b).

For more information, refer to the link given below:

https://brainly.com/question/22008756

ACCESS MORE