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Horizontal Asymptote of a Function with no X values
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I am working on some simple differential equation work and it seems I've completely forgotten the rules of limits.
I have a differential equation that looks like so:
$fracdydx = fracy6(4 -y)$
Plotting the slope field it's clear to see if you start at the point (0, 6) the solution function will approach $y = 4$ as $x$ tends to infinity. This makes sense, because the derivative goes to 0 at that point.
However I am having difficulty justifying this with limits:
$lim_limitsx to inf fracy6(4 -y) = ...$
It doesn't seem there are any substitution rules I can do here to prove this is a horizontal asymptote even though it clearly is from the graph.
What am I missing here in order to prove this to myself?
differential-equations limits
asked Aug 3 at 18:20
user581515
132
add a comment |Â
up vote
2
down vote
favorite
I am working on some simple differential equation work and it seems I've completely forgotten the rules of limits.
I have a differential equation that looks like so:
$fracdydx = fracy6(4 -y)$
Plotting the slope field it's clear to see if you start at the point (0, 6) the solution function will approach $y = 4$ as $x$ tends to infinity. This makes sense, because the derivative goes to 0 at that point.
However I am having difficulty justifying this with limits:
$lim_limitsx to inf fracy6(4 -y) = ...$
It doesn't seem there are any substitution rules I can do here to prove this is a horizontal asymptote even though it clearly is from the graph.
What am I missing here in order to prove this to myself?
differential-equations limits
asked Aug 3 at 18:20
user581515
132
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am working on some simple differential equation work and it seems I've completely forgotten the rules of limits.
I have a differential equation that looks like so:
$fracdydx = fracy6(4 -y)$
Plotting the slope field it's clear to see if you start at the point (0, 6) the solution function will approach $y = 4$ as $x$ tends to infinity. This makes sense, because the derivative goes to 0 at that point.
However I am having difficulty justifying this with limits:
$lim_limitsx to inf fracy6(4 -y) = ...$
It doesn't seem there are any substitution rules I can do here to prove this is a horizontal asymptote even though it clearly is from the graph.
What am I missing here in order to prove this to myself?
differential-equations limits
asked Aug 3 at 18:20
user581515
132
I am working on some simple differential equation work and it seems I've completely forgotten the rules of limits.
I have a differential equation that looks like so:
$fracdydx = fracy6(4 -y)$
Plotting the slope field it's clear to see if you start at the point (0, 6) the solution function will approach $y = 4$ as $x$ tends to infinity. This makes sense, because the derivative goes to 0 at that point.
However I am having difficulty justifying this with limits:
$lim_limitsx to inf fracy6(4 -y) = ...$
It doesn't seem there are any substitution rules I can do here to prove this is a horizontal asymptote even though it clearly is from the graph.
What am I missing here in order to prove this to myself?
differential-equations limits
asked Aug 3 at 18:20
user581515
132
asked Aug 3 at 18:20
user581515
132
asked Aug 3 at 18:20
user581515
132
asked Aug 3 at 18:20
user581515
132
132
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
While you might not see $x$ explicitly, $y$ is a function of $x$,
$$fracdydx= fracy6(4-y)$$
$$int frac6y(4-y), dy=int,dx$$
$$ ln(4-y) - ln (y)=-frac23x+C$$
As $x$ goves to $infty$, RHS goes to $-infty$, hence LHS has to go to $-infty$ as well. Hence $y to 4$.
We can also represent it as
$$frac4-yy=Aexpleft( -frac23 xright)$$
$$frac4y=Aexpleft( -frac23 xright)+1$$
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
Thank you for your edit. Where did A come from? Other than that I think I'm following so far. As x goes to infinity $Aexp(-frac23x)$ goes to zero. Which after some algebra shows $y = 4$. So we first need to find the differentials via integration, and then use those to prove the limit?
– user581515
Aug 3 at 18:36
I don't think it is a must but it is just a possible way. I take exponential on both sides $A$ comes from exponentiation of $C$.
– Siong Thye Goh
Aug 3 at 18:38
add a comment |Â
up vote
0
down vote
You are not missing anything.
The differential equation that you have $$ fracdydx = fracy6(4 -y)$$
has two equilibrium points, namely $y=0$ and $y=4.$ both of which are represented by horizontal line.
The solution $y=0$ is unstable and the solution $y=4$ is stable.
The solution $y=4$ is actually an attractor which means nearby points will approach $y=4$ as you have notices with the point $(0,6).$
The solution $y=0$ is a repeller which means nearby points will scape from it.
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
While you might not see $x$ explicitly, $y$ is a function of $x$,
$$fracdydx= fracy6(4-y)$$
$$int frac6y(4-y), dy=int,dx$$
$$ ln(4-y) - ln (y)=-frac23x+C$$
As $x$ goves to $infty$, RHS goes to $-infty$, hence LHS has to go to $-infty$ as well. Hence $y to 4$.
We can also represent it as
$$frac4-yy=Aexpleft( -frac23 xright)$$
$$frac4y=Aexpleft( -frac23 xright)+1$$
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
Thank you for your edit. Where did A come from? Other than that I think I'm following so far. As x goes to infinity $Aexp(-frac23x)$ goes to zero. Which after some algebra shows $y = 4$. So we first need to find the differentials via integration, and then use those to prove the limit?
– user581515
Aug 3 at 18:36
I don't think it is a must but it is just a possible way. I take exponential on both sides $A$ comes from exponentiation of $C$.
– Siong Thye Goh
Aug 3 at 18:38
add a comment |Â
up vote
1
down vote
accepted
While you might not see $x$ explicitly, $y$ is a function of $x$,
$$fracdydx= fracy6(4-y)$$
$$int frac6y(4-y), dy=int,dx$$
$$ ln(4-y) - ln (y)=-frac23x+C$$
As $x$ goves to $infty$, RHS goes to $-infty$, hence LHS has to go to $-infty$ as well. Hence $y to 4$.
We can also represent it as
$$frac4-yy=Aexpleft( -frac23 xright)$$
$$frac4y=Aexpleft( -frac23 xright)+1$$
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
Thank you for your edit. Where did A come from? Other than that I think I'm following so far. As x goes to infinity $Aexp(-frac23x)$ goes to zero. Which after some algebra shows $y = 4$. So we first need to find the differentials via integration, and then use those to prove the limit?
– user581515
Aug 3 at 18:36
I don't think it is a must but it is just a possible way. I take exponential on both sides $A$ comes from exponentiation of $C$.
– Siong Thye Goh
Aug 3 at 18:38
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
While you might not see $x$ explicitly, $y$ is a function of $x$,
$$fracdydx= fracy6(4-y)$$
$$int frac6y(4-y), dy=int,dx$$
$$ ln(4-y) - ln (y)=-frac23x+C$$
As $x$ goves to $infty$, RHS goes to $-infty$, hence LHS has to go to $-infty$ as well. Hence $y to 4$.
We can also represent it as
$$frac4-yy=Aexpleft( -frac23 xright)$$
$$frac4y=Aexpleft( -frac23 xright)+1$$
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
While you might not see $x$ explicitly, $y$ is a function of $x$,
$$fracdydx= fracy6(4-y)$$
$$int frac6y(4-y), dy=int,dx$$
$$ ln(4-y) - ln (y)=-frac23x+C$$
As $x$ goves to $infty$, RHS goes to $-infty$, hence LHS has to go to $-infty$ as well. Hence $y to 4$.
We can also represent it as
$$frac4-yy=Aexpleft( -frac23 xright)$$
$$frac4y=Aexpleft( -frac23 xright)+1$$
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
edited Aug 3 at 18:29
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
answered Aug 3 at 18:24
Siong Thye Goh
76.5k134794
76.5k134794
Thank you for your edit. Where did A come from? Other than that I think I'm following so far. As x goes to infinity $Aexp(-frac23x)$ goes to zero. Which after some algebra shows $y = 4$. So we first need to find the differentials via integration, and then use those to prove the limit?
– user581515
Aug 3 at 18:36
I don't think it is a must but it is just a possible way. I take exponential on both sides $A$ comes from exponentiation of $C$.
– Siong Thye Goh
Aug 3 at 18:38
add a comment |Â
Thank you for your edit. Where did A come from? Other than that I think I'm following so far. As x goes to infinity $Aexp(-frac23x)$ goes to zero. Which after some algebra shows $y = 4$. So we first need to find the differentials via integration, and then use those to prove the limit?
– user581515
Aug 3 at 18:36
I don't think it is a must but it is just a possible way. I take exponential on both sides $A$ comes from exponentiation of $C$.
– Siong Thye Goh
Aug 3 at 18:38
Thank you for your edit. Where did A come from? Other than that I think I'm following so far. As x goes to infinity $Aexp(-frac23x)$ goes to zero. Which after some algebra shows $y = 4$. So we first need to find the differentials via integration, and then use those to prove the limit?
– user581515
Aug 3 at 18:36
Thank you for your edit. Where did A come from? Other than that I think I'm following so far. As x goes to infinity $Aexp(-frac23x)$ goes to zero. Which after some algebra shows $y = 4$. So we first need to find the differentials via integration, and then use those to prove the limit?
– user581515
Aug 3 at 18:36
I don't think it is a must but it is just a possible way. I take exponential on both sides $A$ comes from exponentiation of $C$.
– Siong Thye Goh
Aug 3 at 18:38
I don't think it is a must but it is just a possible way. I take exponential on both sides $A$ comes from exponentiation of $C$.
– Siong Thye Goh
Aug 3 at 18:38
add a comment |Â
up vote
0
down vote
You are not missing anything.
The differential equation that you have $$ fracdydx = fracy6(4 -y)$$
has two equilibrium points, namely $y=0$ and $y=4.$ both of which are represented by horizontal line.
The solution $y=0$ is unstable and the solution $y=4$ is stable.
The solution $y=4$ is actually an attractor which means nearby points will approach $y=4$ as you have notices with the point $(0,6).$
The solution $y=0$ is a repeller which means nearby points will scape from it.
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
add a comment |Â
up vote
0
down vote
You are not missing anything.
The differential equation that you have $$ fracdydx = fracy6(4 -y)$$
has two equilibrium points, namely $y=0$ and $y=4.$ both of which are represented by horizontal line.
The solution $y=0$ is unstable and the solution $y=4$ is stable.
The solution $y=4$ is actually an attractor which means nearby points will approach $y=4$ as you have notices with the point $(0,6).$
The solution $y=0$ is a repeller which means nearby points will scape from it.
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are not missing anything.
The differential equation that you have $$ fracdydx = fracy6(4 -y)$$
has two equilibrium points, namely $y=0$ and $y=4.$ both of which are represented by horizontal line.
The solution $y=0$ is unstable and the solution $y=4$ is stable.
The solution $y=4$ is actually an attractor which means nearby points will approach $y=4$ as you have notices with the point $(0,6).$
The solution $y=0$ is a repeller which means nearby points will scape from it.
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
You are not missing anything.
The differential equation that you have $$ fracdydx = fracy6(4 -y)$$
has two equilibrium points, namely $y=0$ and $y=4.$ both of which are represented by horizontal line.
The solution $y=0$ is unstable and the solution $y=4$ is stable.
The solution $y=4$ is actually an attractor which means nearby points will approach $y=4$ as you have notices with the point $(0,6).$
The solution $y=0$ is a repeller which means nearby points will scape from it.
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 18:32
Mohammad Riazi-Kermani
27k41850
27k41850
add a comment |Â
add a comment |Â
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