Respuesta :
Answer:
(a)2000 of cells were present initially.
(b)The differential equation of the given exponential model is
[tex]\frac{dP}{dt}=0.3P[/tex]
(c)The population will be double after 2.31 hours.
(d)The number of cell will be 8,000 after 4.62 hours.
Step-by-step explanation:
Given that,
[tex]P(t)= 2000e^{0.3 t}[/tex]
where P(t) is number of cell and t is time in hours.
(a)
Initial time means t=0.
Putting t=0 i the given expression
[tex]P(0)= 2000e^{0.3 \times 0}[/tex]
[tex]\Rightarrow P(0)= 2000[/tex]
2000 of cells were present initially.
(b)
The differential equation of population growth is
[tex]\frac{dP}{dt}=kP[/tex]
The solution of the above equation is
[tex]P(t)=Ce^{kt}[/tex]
Comparing the given exponential model to the above solution
So, k= 0.3 and C=2000
The differential equation of the given exponential model is
[tex]\frac{dP}{dt}=0.3P[/tex]
(c)
If the population double then P(t) = 2× P(0) =2×2000
[tex]P(t)= 2000e^{0.3 t}[/tex]
[tex]\Rightarrow 2\times 2000=2000e^{0.3t}[/tex]
[tex]\Rightarrow e^{0.3t}=2[/tex]
Taking ln both sides
[tex]\Rightarrow ln (e^{0.3t})=ln2[/tex]
[tex]\Rightarrow {0.3t=ln2[/tex]
[tex]\Rightarrow t=\frac{ln2}{0.3}[/tex]
[tex]\Rightarrow t=2.31[/tex] h
The population will be double after 2.31 hours.
(d)
Now P(t)=8,000
[tex]P(t)= 2000e^{0.3 t}[/tex]
[tex]\Rightarrow 8,000= 2000e^{0.3 t}[/tex]
[tex]\Rightarrow e^{0.3 t}=\frac{8,000}{2,000}[/tex]
[tex]\Rightarrow e^{0.3 t}=4[/tex]
Taking ln function both sides
[tex]\Rightarrow ln(e^{0.3 t})=ln 4[/tex]
[tex]\Rightarrow t=\frac{ln 4}{0.3}[/tex]
[tex]\Rightarrow t=4.62[/tex] h
The number of cell will be 8,000 after 4.62 hours.
