How to build a differential equation of quantity by time of a drug that is in the digestive system and...
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I'm giving that a child has swallowed 11 pills of 100 milligrams of a drug and was rushed into the emergency room after 2 hours by that time the drugs have passed from his stomach to his intestines.
The drug is absorbed in the blood steam at a proportional rate to the quantity in the digestive system,and the drug is removed from the blood at a proportional rate to the quantity that is in the blood.
(**)
It is known there is half-life for 5-hour blood absorption and half-life for removal from the blood (The time when the amount of the drug decreases by half in the blood,assuming there is no any drug absorbtion) is 6 Hours.
I'm trying to write a differential equation for the quantity of the drug in the digestive system and an equation of the quantity of the drug in the blood both equations are by time and with a starting condition.
I'm having trouble understanding (**) to write my first equation ,i'm trying to understand by how much time the quantity of the drug gets absorbed by the digestive system and delivered to the bloodstream to build my first equation.
From the given i can say that
G(0)=1,100 (the quantity of the drug in the digestive system at the initial time is 1,100 milligrams)
calculus differential-equations
asked Dec 1 at 13:25
user3133165
1758
add a comment |
up vote
2
down vote
favorite
I'm giving that a child has swallowed 11 pills of 100 milligrams of a drug and was rushed into the emergency room after 2 hours by that time the drugs have passed from his stomach to his intestines.
The drug is absorbed in the blood steam at a proportional rate to the quantity in the digestive system,and the drug is removed from the blood at a proportional rate to the quantity that is in the blood.
(**)
It is known there is half-life for 5-hour blood absorption and half-life for removal from the blood (The time when the amount of the drug decreases by half in the blood,assuming there is no any drug absorbtion) is 6 Hours.
I'm trying to write a differential equation for the quantity of the drug in the digestive system and an equation of the quantity of the drug in the blood both equations are by time and with a starting condition.
I'm having trouble understanding (**) to write my first equation ,i'm trying to understand by how much time the quantity of the drug gets absorbed by the digestive system and delivered to the bloodstream to build my first equation.
From the given i can say that
G(0)=1,100 (the quantity of the drug in the digestive system at the initial time is 1,100 milligrams)
calculus differential-equations
asked Dec 1 at 13:25
user3133165
1758
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm giving that a child has swallowed 11 pills of 100 milligrams of a drug and was rushed into the emergency room after 2 hours by that time the drugs have passed from his stomach to his intestines.
The drug is absorbed in the blood steam at a proportional rate to the quantity in the digestive system,and the drug is removed from the blood at a proportional rate to the quantity that is in the blood.
(**)
It is known there is half-life for 5-hour blood absorption and half-life for removal from the blood (The time when the amount of the drug decreases by half in the blood,assuming there is no any drug absorbtion) is 6 Hours.
I'm trying to write a differential equation for the quantity of the drug in the digestive system and an equation of the quantity of the drug in the blood both equations are by time and with a starting condition.
I'm having trouble understanding (**) to write my first equation ,i'm trying to understand by how much time the quantity of the drug gets absorbed by the digestive system and delivered to the bloodstream to build my first equation.
From the given i can say that
G(0)=1,100 (the quantity of the drug in the digestive system at the initial time is 1,100 milligrams)
calculus differential-equations
asked Dec 1 at 13:25
user3133165
1758
I'm giving that a child has swallowed 11 pills of 100 milligrams of a drug and was rushed into the emergency room after 2 hours by that time the drugs have passed from his stomach to his intestines.
The drug is absorbed in the blood steam at a proportional rate to the quantity in the digestive system,and the drug is removed from the blood at a proportional rate to the quantity that is in the blood.
(**)
It is known there is half-life for 5-hour blood absorption and half-life for removal from the blood (The time when the amount of the drug decreases by half in the blood,assuming there is no any drug absorbtion) is 6 Hours.
I'm trying to write a differential equation for the quantity of the drug in the digestive system and an equation of the quantity of the drug in the blood both equations are by time and with a starting condition.
I'm having trouble understanding (**) to write my first equation ,i'm trying to understand by how much time the quantity of the drug gets absorbed by the digestive system and delivered to the bloodstream to build my first equation.
From the given i can say that
G(0)=1,100 (the quantity of the drug in the digestive system at the initial time is 1,100 milligrams)
calculus differential-equations
calculus differential-equations
asked Dec 1 at 13:25
user3133165
1758
asked Dec 1 at 13:25
user3133165
1758
asked Dec 1 at 13:25
user3133165
1758
asked Dec 1 at 13:25
user3133165
1758
asked Dec 1 at 13:25
user3133165
1758
1758
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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accepted
You get as equations
begin{align}
dot G &= -k_1G\
dot L &= k_1G-k_2L
end{align}
and from the half-life times $e^{-5k_1}=0.5$ and $e^{-6k_2}=0.5$, where time is measured in hours.
answered Dec 1 at 13:32
LutzL
54.3k41953
So the first equation is quantity of the drug in the digestive system, and the second one is the quantity that is in the bloodstream?
– user3133165
Dec 1 at 13:37
Could you please let me understand the $e^{-5k_{1}}=0.5 $ and $e^{-6k_{2}}=0.5$ this is the part that is giving me a hard time understanding
– user3133165
Dec 1 at 13:39
Yes. I interpreted the naming convention to take the second consonant of "digestive" and "blood", you might want to take more conventional variable names, if such a convention exists.
– LutzL
Dec 1 at 13:39
1
The solution is $G(t)=e^{-k_1t}G(0)$. For $G(t)=0.5G(0)$ you need thus $e^{-k_1t}=0.5$. $t=5$ is given, which allows to compute $k_1$.
– LutzL
Dec 1 at 13:41
I'm trying to solve the first equation by separating and then integrating is this correct?
– user3133165
Dec 1 at 13:50
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You get as equations
begin{align}
dot G &= -k_1G\
dot L &= k_1G-k_2L
end{align}
and from the half-life times $e^{-5k_1}=0.5$ and $e^{-6k_2}=0.5$, where time is measured in hours.
answered Dec 1 at 13:32
LutzL
54.3k41953
So the first equation is quantity of the drug in the digestive system, and the second one is the quantity that is in the bloodstream?
– user3133165
Dec 1 at 13:37
Could you please let me understand the $e^{-5k_{1}}=0.5 $ and $e^{-6k_{2}}=0.5$ this is the part that is giving me a hard time understanding
– user3133165
Dec 1 at 13:39
Yes. I interpreted the naming convention to take the second consonant of "digestive" and "blood", you might want to take more conventional variable names, if such a convention exists.
– LutzL
Dec 1 at 13:39
1
The solution is $G(t)=e^{-k_1t}G(0)$. For $G(t)=0.5G(0)$ you need thus $e^{-k_1t}=0.5$. $t=5$ is given, which allows to compute $k_1$.
– LutzL
Dec 1 at 13:41
I'm trying to solve the first equation by separating and then integrating is this correct?
– user3133165
Dec 1 at 13:50
|
show 2 more comments
up vote
1
down vote
accepted
You get as equations
begin{align}
dot G &= -k_1G\
dot L &= k_1G-k_2L
end{align}
and from the half-life times $e^{-5k_1}=0.5$ and $e^{-6k_2}=0.5$, where time is measured in hours.
answered Dec 1 at 13:32
LutzL
54.3k41953
So the first equation is quantity of the drug in the digestive system, and the second one is the quantity that is in the bloodstream?
– user3133165
Dec 1 at 13:37
Could you please let me understand the $e^{-5k_{1}}=0.5 $ and $e^{-6k_{2}}=0.5$ this is the part that is giving me a hard time understanding
– user3133165
Dec 1 at 13:39
Yes. I interpreted the naming convention to take the second consonant of "digestive" and "blood", you might want to take more conventional variable names, if such a convention exists.
– LutzL
Dec 1 at 13:39
1
The solution is $G(t)=e^{-k_1t}G(0)$. For $G(t)=0.5G(0)$ you need thus $e^{-k_1t}=0.5$. $t=5$ is given, which allows to compute $k_1$.
– LutzL
Dec 1 at 13:41
I'm trying to solve the first equation by separating and then integrating is this correct?
– user3133165
Dec 1 at 13:50
|
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You get as equations
begin{align}
dot G &= -k_1G\
dot L &= k_1G-k_2L
end{align}
and from the half-life times $e^{-5k_1}=0.5$ and $e^{-6k_2}=0.5$, where time is measured in hours.
answered Dec 1 at 13:32
LutzL
54.3k41953
You get as equations
begin{align}
dot G &= -k_1G\
dot L &= k_1G-k_2L
end{align}
and from the half-life times $e^{-5k_1}=0.5$ and $e^{-6k_2}=0.5$, where time is measured in hours.
answered Dec 1 at 13:32
LutzL
54.3k41953
answered Dec 1 at 13:32
LutzL
54.3k41953
answered Dec 1 at 13:32
LutzL
54.3k41953
answered Dec 1 at 13:32
LutzL
54.3k41953
54.3k41953
So the first equation is quantity of the drug in the digestive system, and the second one is the quantity that is in the bloodstream?
– user3133165
Dec 1 at 13:37
Could you please let me understand the $e^{-5k_{1}}=0.5 $ and $e^{-6k_{2}}=0.5$ this is the part that is giving me a hard time understanding
– user3133165
Dec 1 at 13:39
Yes. I interpreted the naming convention to take the second consonant of "digestive" and "blood", you might want to take more conventional variable names, if such a convention exists.
– LutzL
Dec 1 at 13:39
1
The solution is $G(t)=e^{-k_1t}G(0)$. For $G(t)=0.5G(0)$ you need thus $e^{-k_1t}=0.5$. $t=5$ is given, which allows to compute $k_1$.
– LutzL
Dec 1 at 13:41
I'm trying to solve the first equation by separating and then integrating is this correct?
– user3133165
Dec 1 at 13:50
|
show 2 more comments
So the first equation is quantity of the drug in the digestive system, and the second one is the quantity that is in the bloodstream?
– user3133165
Dec 1 at 13:37
Could you please let me understand the $e^{-5k_{1}}=0.5 $ and $e^{-6k_{2}}=0.5$ this is the part that is giving me a hard time understanding
– user3133165
Dec 1 at 13:39
Yes. I interpreted the naming convention to take the second consonant of "digestive" and "blood", you might want to take more conventional variable names, if such a convention exists.
– LutzL
Dec 1 at 13:39
1
The solution is $G(t)=e^{-k_1t}G(0)$. For $G(t)=0.5G(0)$ you need thus $e^{-k_1t}=0.5$. $t=5$ is given, which allows to compute $k_1$.
– LutzL
Dec 1 at 13:41
I'm trying to solve the first equation by separating and then integrating is this correct?
– user3133165
Dec 1 at 13:50
So the first equation is quantity of the drug in the digestive system, and the second one is the quantity that is in the bloodstream?
– user3133165
Dec 1 at 13:37
So the first equation is quantity of the drug in the digestive system, and the second one is the quantity that is in the bloodstream?
– user3133165
Dec 1 at 13:37
Could you please let me understand the $e^{-5k_{1}}=0.5 $ and $e^{-6k_{2}}=0.5$ this is the part that is giving me a hard time understanding
– user3133165
Dec 1 at 13:39
Could you please let me understand the $e^{-5k_{1}}=0.5 $ and $e^{-6k_{2}}=0.5$ this is the part that is giving me a hard time understanding
– user3133165
Dec 1 at 13:39
Yes. I interpreted the naming convention to take the second consonant of "digestive" and "blood", you might want to take more conventional variable names, if such a convention exists.
– LutzL
Dec 1 at 13:39
Yes. I interpreted the naming convention to take the second consonant of "digestive" and "blood", you might want to take more conventional variable names, if such a convention exists.
– LutzL
Dec 1 at 13:39
1
1
The solution is $G(t)=e^{-k_1t}G(0)$. For $G(t)=0.5G(0)$ you need thus $e^{-k_1t}=0.5$. $t=5$ is given, which allows to compute $k_1$.
– LutzL
Dec 1 at 13:41
The solution is $G(t)=e^{-k_1t}G(0)$. For $G(t)=0.5G(0)$ you need thus $e^{-k_1t}=0.5$. $t=5$ is given, which allows to compute $k_1$.
– LutzL
Dec 1 at 13:41
I'm trying to solve the first equation by separating and then integrating is this correct?
– user3133165
Dec 1 at 13:50
I'm trying to solve the first equation by separating and then integrating is this correct?
– user3133165
Dec 1 at 13:50
|
show 2 more comments
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