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