Volume of a leaky tank
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
asked Feb 26 '16 at 11:57
user11128user11128
1335
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
asked Feb 26 '16 at 11:57
user11128user11128
1335
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
asked Feb 26 '16 at 11:57
user11128user11128
1335
A tank of water is initially leaking at a rate of $1$ litre per minute. The rate it continues to leak at is proportional to the volume in the tank. After $7$ minutes, the tank is half full. What is the capacity of the tank?
I set $dV/dt=-kV$ where $V$ is volume and $k$ is constant.
Plugging in the initial conditions of $dV/dt(t=0)=-1$ tells me $k=V_{t=0}^{-1}$ and so I have $dV/dt=-V_{t=0}^{-1} V$. Now we integrate to get $ln{V}=-V_{t=0}^{-1} t$ or $V=-e^{V^{-1}_{t=0}t}$.
Using $V(t=7)=frac{1}{2} V_{t=0}^{-1}$, we find $frac{1}{2} V_{t=0}^{-1} = -e^{7V_{t=0}^{-1}}$. Does this have any real solutions?
differential-equations
differential-equations
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
asked Feb 26 '16 at 11:57
user11128user11128
1335
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
asked Feb 26 '16 at 11:57
user11128user11128
1335
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
edited Mar 14 '18 at 14:59
Rodrigo de Azevedo
12.8k41855
12.8k41855
asked Feb 26 '16 at 11:57
user11128user11128
1335
asked Feb 26 '16 at 11:57
user11128user11128
1335
asked Feb 26 '16 at 11:57
user11128user11128
1335
1335
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
add a comment |
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
– user11128
Feb 26 '16 at 14:23
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
– GoodDeeds
Feb 26 '16 at 15:20
Yes I think we are to assume this. Don't know what the problem was then.....alas
– user11128
Feb 26 '16 at 15:51
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1673101%2fvolume-of-a-leaky-tank%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
add a comment |
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
add a comment |
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
You forgot the integration constant when solving the DE. The general solution to $frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$
where the constant $e^c=V_0$. You then have the equation $$frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
answered Feb 26 '16 at 12:06
Bobson DugnuttBobson Dugnutt
8,53331939
8,53331939
add a comment |
add a comment |
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
– user11128
Feb 26 '16 at 14:23
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
– GoodDeeds
Feb 26 '16 at 15:20
Yes I think we are to assume this. Don't know what the problem was then.....alas
– user11128
Feb 26 '16 at 15:51
add a comment |
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
– user11128
Feb 26 '16 at 14:23
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
– GoodDeeds
Feb 26 '16 at 15:20
Yes I think we are to assume this. Don't know what the problem was then.....alas
– user11128
Feb 26 '16 at 15:51
add a comment |
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
There should be a constant of integration when you integrate. Moreover, when you say $ln V=-ct$, you get $V=e^{-ct}$, not $V=-e^{ct}$.
Let the capacity be $V_0$.
$$frac{dV}{dt}=-kV$$
$$[ln V]_{V_0}^{V}=-k[t]_{0}^{t}$$
$$lnleft(frac{V}{V_0}right)=-kt$$
Moreover, at $t=0$, $frac{dV}{dt}=-1$ and $V=V_0$.
Thus,
$$-1=-kV_0$$
$$k=frac{1}{V_0}$$
which gives,
$$lnleft(frac{V}{V_0}right)=-frac{1}{V_0}t$$
At $t=0$, $V=frac{V_0}{2}$
Thus,
$$lnleft(frac{frac{V_0}2}{V_0}right)=-frac{1}{V_0}7$$
$$lnleft(frac{1}{2}right)=-frac{1}{V_0}7$$
$$V_0=frac{7}{ln 2}$$
with appropriate units.
Regarding the other part of the question, it can be seen by drawing a graph that $frac{x}{2}=-e^{7x}$ has a solution in the fourth quadrant. Clearly, if $x$ denoted the volume of the tank, then there would be no solution.
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
answered Feb 26 '16 at 12:14
GoodDeedsGoodDeeds
10.3k31335
10.3k31335
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
– user11128
Feb 26 '16 at 14:23
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
– GoodDeeds
Feb 26 '16 at 15:20
Yes I think we are to assume this. Don't know what the problem was then.....alas
– user11128
Feb 26 '16 at 15:51
add a comment |
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
– user11128
Feb 26 '16 at 14:23
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
– GoodDeeds
Feb 26 '16 at 15:20
Yes I think we are to assume this. Don't know what the problem was then.....alas
– user11128
Feb 26 '16 at 15:51
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
– user11128
Feb 26 '16 at 14:23
Hi thanks for this. Actually this was a mistake on my part when I re-did it today. I did it originally in a test a few days ago and got the same answer you have above and it was marked incorrect which got me puzzled. Oh well...
– user11128
Feb 26 '16 at 14:23
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
– GoodDeeds
Feb 26 '16 at 15:20
@user11128 There is one missing point - I assumed that the tank is initially full, but the question does not mention it anywhere.
– GoodDeeds
Feb 26 '16 at 15:20
Yes I think we are to assume this. Don't know what the problem was then.....alas
– user11128
Feb 26 '16 at 15:51
Yes I think we are to assume this. Don't know what the problem was then.....alas
– user11128
Feb 26 '16 at 15:51
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1673101%2fvolume-of-a-leaky-tank%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
