Differential equations and optics
There is a problem in Mathematical Methods by Mary L. Boas book (Prob.8.4.20) that I couldn't understand what it asking me to do.
Find the shape of a mirror which has the property that rays from a point $O$ on the axis are reflected into a parallel beam.
Hint: Take the point $O$ at the origin. Show from the figure that $tan 2theta = y/x$. Use the formula for $tan 2theta$ to express this in terms of $tan theta = dx/dy$ and solve the resulting differential equation.
(Hint: See Problem 16.)
I think it is asking me to use $tan 2theta = y/x$ to solve the differential equation $tan theta=dy/dx$. Is this correct?
If not, is there any idea that may help me to understand how to write a differential equation that will represents the case?
differential-equations
edited yesterday
The Pointer
2,60021334
asked yesterday
Nour
6
New contributor
Nour 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 |
There is a problem in Mathematical Methods by Mary L. Boas book (Prob.8.4.20) that I couldn't understand what it asking me to do.
Find the shape of a mirror which has the property that rays from a point $O$ on the axis are reflected into a parallel beam.
Hint: Take the point $O$ at the origin. Show from the figure that $tan 2theta = y/x$. Use the formula for $tan 2theta$ to express this in terms of $tan theta = dx/dy$ and solve the resulting differential equation.
(Hint: See Problem 16.)
I think it is asking me to use $tan 2theta = y/x$ to solve the differential equation $tan theta=dy/dx$. Is this correct?
If not, is there any idea that may help me to understand how to write a differential equation that will represents the case?
differential-equations
edited yesterday
The Pointer
2,60021334
asked yesterday
Nour
6
New contributor
Nour 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 |
There is a problem in Mathematical Methods by Mary L. Boas book (Prob.8.4.20) that I couldn't understand what it asking me to do.
Find the shape of a mirror which has the property that rays from a point $O$ on the axis are reflected into a parallel beam.
Hint: Take the point $O$ at the origin. Show from the figure that $tan 2theta = y/x$. Use the formula for $tan 2theta$ to express this in terms of $tan theta = dx/dy$ and solve the resulting differential equation.
(Hint: See Problem 16.)
I think it is asking me to use $tan 2theta = y/x$ to solve the differential equation $tan theta=dy/dx$. Is this correct?
If not, is there any idea that may help me to understand how to write a differential equation that will represents the case?
differential-equations
edited yesterday
The Pointer
2,60021334
asked yesterday
Nour
6
New contributor
Nour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There is a problem in Mathematical Methods by Mary L. Boas book (Prob.8.4.20) that I couldn't understand what it asking me to do.
Find the shape of a mirror which has the property that rays from a point $O$ on the axis are reflected into a parallel beam.
Hint: Take the point $O$ at the origin. Show from the figure that $tan 2theta = y/x$. Use the formula for $tan 2theta$ to express this in terms of $tan theta = dx/dy$ and solve the resulting differential equation.
(Hint: See Problem 16.)
I think it is asking me to use $tan 2theta = y/x$ to solve the differential equation $tan theta=dy/dx$. Is this correct?
If not, is there any idea that may help me to understand how to write a differential equation that will represents the case?
differential-equations
differential-equations
edited yesterday
The Pointer
2,60021334
asked yesterday
Nour
6
New contributor
Nour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
The Pointer
2,60021334
asked yesterday
Nour
6
New contributor
Nour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
The Pointer
2,60021334
edited yesterday
The Pointer
2,60021334
edited yesterday
The Pointer
2,60021334
2,60021334
asked yesterday
Nour
6
New contributor
Nour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Nour
6
asked yesterday
Nour
6
6
New contributor
Nour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nour is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nour 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 |
add a comment |
2 Answers
2
active
oldest
votes
A picture of ray trace while under reflection is helpful.
$$ tan 2theta = frac{PX}{XO}=frac{y}{x}$$
$$ phi= pi/2-theta,quad tan phi= cot theta = frac{1}{tan theta}= frac{dy}{dx}$$
In order to set up DE of reflector now find
$$ frac{dy}{dx}=tan phi = cot theta =f(tan 2 theta)= f(x,y) $$ using double angle formula.
answered yesterday
Narasimham
20.6k52158
add a comment |
The relation $tan2θ=y/x$ defines what $θ$ is. This can be used in the differential equation $tanθ=dy/dx$ to eliminate $θ$.
You can use the double angle relation
$$
tan2θ=frac{2tanθ}{1-tan^2θ}.
$$
Problem 16 is about an DE (perhaps even this one) that is of the Clairaut type $u=xu'+f(u')$ after substitution.
answered yesterday
LutzL
55.8k42054
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
– Nour
9 hours ago
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
– LutzL
3 hours ago
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
});
}
});
Nour 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%2f3050388%2fdifferential-equations-and-optics%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
A picture of ray trace while under reflection is helpful.
$$ tan 2theta = frac{PX}{XO}=frac{y}{x}$$
$$ phi= pi/2-theta,quad tan phi= cot theta = frac{1}{tan theta}= frac{dy}{dx}$$
In order to set up DE of reflector now find
$$ frac{dy}{dx}=tan phi = cot theta =f(tan 2 theta)= f(x,y) $$ using double angle formula.
answered yesterday
Narasimham
20.6k52158
add a comment |
A picture of ray trace while under reflection is helpful.
$$ tan 2theta = frac{PX}{XO}=frac{y}{x}$$
$$ phi= pi/2-theta,quad tan phi= cot theta = frac{1}{tan theta}= frac{dy}{dx}$$
In order to set up DE of reflector now find
$$ frac{dy}{dx}=tan phi = cot theta =f(tan 2 theta)= f(x,y) $$ using double angle formula.
answered yesterday
Narasimham
20.6k52158
add a comment |
A picture of ray trace while under reflection is helpful.
$$ tan 2theta = frac{PX}{XO}=frac{y}{x}$$
$$ phi= pi/2-theta,quad tan phi= cot theta = frac{1}{tan theta}= frac{dy}{dx}$$
In order to set up DE of reflector now find
$$ frac{dy}{dx}=tan phi = cot theta =f(tan 2 theta)= f(x,y) $$ using double angle formula.
answered yesterday
Narasimham
20.6k52158
A picture of ray trace while under reflection is helpful.
$$ tan 2theta = frac{PX}{XO}=frac{y}{x}$$
$$ phi= pi/2-theta,quad tan phi= cot theta = frac{1}{tan theta}= frac{dy}{dx}$$
In order to set up DE of reflector now find
$$ frac{dy}{dx}=tan phi = cot theta =f(tan 2 theta)= f(x,y) $$ using double angle formula.
answered yesterday
Narasimham
20.6k52158
answered yesterday
Narasimham
20.6k52158
answered yesterday
Narasimham
20.6k52158
answered yesterday
Narasimham
20.6k52158
20.6k52158
add a comment |
add a comment |
The relation $tan2θ=y/x$ defines what $θ$ is. This can be used in the differential equation $tanθ=dy/dx$ to eliminate $θ$.
You can use the double angle relation
$$
tan2θ=frac{2tanθ}{1-tan^2θ}.
$$
Problem 16 is about an DE (perhaps even this one) that is of the Clairaut type $u=xu'+f(u')$ after substitution.
answered yesterday
LutzL
55.8k42054
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
– Nour
9 hours ago
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
– LutzL
3 hours ago
add a comment |
The relation $tan2θ=y/x$ defines what $θ$ is. This can be used in the differential equation $tanθ=dy/dx$ to eliminate $θ$.
You can use the double angle relation
$$
tan2θ=frac{2tanθ}{1-tan^2θ}.
$$
Problem 16 is about an DE (perhaps even this one) that is of the Clairaut type $u=xu'+f(u')$ after substitution.
answered yesterday
LutzL
55.8k42054
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
– Nour
9 hours ago
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
– LutzL
3 hours ago
add a comment |
The relation $tan2θ=y/x$ defines what $θ$ is. This can be used in the differential equation $tanθ=dy/dx$ to eliminate $θ$.
You can use the double angle relation
$$
tan2θ=frac{2tanθ}{1-tan^2θ}.
$$
Problem 16 is about an DE (perhaps even this one) that is of the Clairaut type $u=xu'+f(u')$ after substitution.
answered yesterday
LutzL
55.8k42054
The relation $tan2θ=y/x$ defines what $θ$ is. This can be used in the differential equation $tanθ=dy/dx$ to eliminate $θ$.
You can use the double angle relation
$$
tan2θ=frac{2tanθ}{1-tan^2θ}.
$$
Problem 16 is about an DE (perhaps even this one) that is of the Clairaut type $u=xu'+f(u')$ after substitution.
answered yesterday
LutzL
55.8k42054
edited 3 hours ago
answered yesterday
LutzL
55.8k42054
answered yesterday
LutzL
55.8k42054
answered yesterday
LutzL
55.8k42054
55.8k42054
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
– Nour
9 hours ago
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
– LutzL
3 hours ago
add a comment |
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
– Nour
9 hours ago
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
– LutzL
3 hours ago
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
– Nour
9 hours ago
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
– Nour
9 hours ago
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
– LutzL
3 hours ago
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
– LutzL
3 hours ago
add a comment |
Nour is a new contributor. Be nice, and check out our Code of Conduct.
Nour is a new contributor. Be nice, and check out our Code of Conduct.
Nour is a new contributor. Be nice, and check out our Code of Conduct.
Nour 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%2f3050388%2fdifferential-equations-and-optics%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
