Rotational volume and differential equation
A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.
$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.
$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$
I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$
But I don't really get anywhere from here. Am I going at this from the wrong direction?
The answer is supposed to be $f(x)=Kx^4$, where K is a constant
calculus integration volume differential
asked 2 days ago
Micaele
32
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.
$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.
$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$
I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$
But I don't really get anywhere from here. Am I going at this from the wrong direction?
The answer is supposed to be $f(x)=Kx^4$, where K is a constant
calculus integration volume differential
asked 2 days ago
Micaele
32
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
– B. Goddard
2 days ago
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
– Micaele
2 days ago
add a comment |
A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.
$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.
$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$
I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$
But I don't really get anywhere from here. Am I going at this from the wrong direction?
The answer is supposed to be $f(x)=Kx^4$, where K is a constant
calculus integration volume differential
asked 2 days ago
Micaele
32
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A container with the shape of some function y=f(x) is rotated around the y-axis. It's filled with a fluid and has a hole in the bottom where the fluid leaks out. The rate of liquid flowing out should be proportional to the square root of the height of liquid in the container.
Determine the function f(x) such that the height of liquid in the container has a constant negative slope.
$v(t)$ is the volume of water in the container at time t
$v'(t)$ is the rate of change in volume at time t
$h(t)$ is the height of the water level in the container.
$h'(t)$ is the rate of change of the water level and its constant.
$v'(t)=-k*sqrt{h(t)}$
$v(t)=piint^{h(t)}_{0}f^2(x) dx$
I feel as if maybe the fundamental theorem of calculus would get involved here.
$v'(t)=pi f^2(h(t))$
But I don't really get anywhere from here. Am I going at this from the wrong direction?
The answer is supposed to be $f(x)=Kx^4$, where K is a constant
calculus integration volume differential
calculus integration volume differential
asked 2 days ago
Micaele
32
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Micaele
32
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Micaele
32
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Micaele
32
asked 2 days ago
Micaele
32
32
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Micaele is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
– B. Goddard
2 days ago
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
– Micaele
2 days ago
add a comment |
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
– B. Goddard
2 days ago
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
– Micaele
2 days ago
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
– B. Goddard
2 days ago
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
– B. Goddard
2 days ago
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
– Micaele
2 days ago
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
– Micaele
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by
$$ V(y) = pi int_0^y g(s)^2 ds$$
By the fundamental theorem of calculus:
$$ frac{dV}{dy} = pi g(y)^2 $$
By the chain rule
$$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$
where $frac{dy}{dt}=-c$ is some negative constant
It is also given that
$$ frac{dV}{dt} = -ksqrt{y} $$
Therefore
$$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$
Inverting gives
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered 2 days ago
Dylan
12.4k31026
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
});
}
});
Micaele is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3060965%2frotational-volume-and-differential-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by
$$ V(y) = pi int_0^y g(s)^2 ds$$
By the fundamental theorem of calculus:
$$ frac{dV}{dy} = pi g(y)^2 $$
By the chain rule
$$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$
where $frac{dy}{dt}=-c$ is some negative constant
It is also given that
$$ frac{dV}{dt} = -ksqrt{y} $$
Therefore
$$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$
Inverting gives
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered 2 days ago
Dylan
12.4k31026
add a comment |
Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by
$$ V(y) = pi int_0^y g(s)^2 ds$$
By the fundamental theorem of calculus:
$$ frac{dV}{dy} = pi g(y)^2 $$
By the chain rule
$$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$
where $frac{dy}{dt}=-c$ is some negative constant
It is also given that
$$ frac{dV}{dt} = -ksqrt{y} $$
Therefore
$$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$
Inverting gives
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered 2 days ago
Dylan
12.4k31026
add a comment |
Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by
$$ V(y) = pi int_0^y g(s)^2 ds$$
By the fundamental theorem of calculus:
$$ frac{dV}{dy} = pi g(y)^2 $$
By the chain rule
$$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$
where $frac{dy}{dt}=-c$ is some negative constant
It is also given that
$$ frac{dV}{dt} = -ksqrt{y} $$
Therefore
$$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$
Inverting gives
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered 2 days ago
Dylan
12.4k31026
Let $x = g(y)$ be the inverse function of $y=f(x)$. The volume of the liquid, w.r.t the height $y$ is given by
$$ V(y) = pi int_0^y g(s)^2 ds$$
By the fundamental theorem of calculus:
$$ frac{dV}{dy} = pi g(y)^2 $$
By the chain rule
$$ frac{dV}{dt} = frac{dV}{dy}frac{dy}{dt} = -cpi g(y)^2 $$
where $frac{dy}{dt}=-c$ is some negative constant
It is also given that
$$ frac{dV}{dt} = -ksqrt{y} $$
Therefore
$$ -cpi g(y)^2 = -ksqrt{y} implies x = g(y) = sqrt{frac{k}{cpi}}y^{1/4} $$
Inverting gives
$$ y = f(x) = left(frac{cpi}{k}right)^2x^4 = Kx^4 $$
answered 2 days ago
Dylan
12.4k31026
answered 2 days ago
Dylan
12.4k31026
answered 2 days ago
Dylan
12.4k31026
answered 2 days ago
Dylan
12.4k31026
12.4k31026
add a comment |
add a comment |
Micaele is a new contributor. Be nice, and check out our Code of Conduct.
Micaele is a new contributor. Be nice, and check out our Code of Conduct.
Micaele is a new contributor. Be nice, and check out our Code of Conduct.
Micaele is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3060965%2frotational-volume-and-differential-equation%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
You've set this up as if the rotation is about the $x$-axis. I think your second equation should be integral of $2pi f.$
– B. Goddard
2 days ago
That's right, that's something that needs to change. But if I don't have an equation, how can I express x in forms of y? All I know is that there is a function y=f(x).
– Micaele
2 days ago