Differential equations and optics
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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?
ordinary-differential-equations
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
asked Dec 23 '18 at 14:34
NourNour
6
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add a comment |
$begingroup$
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?
ordinary-differential-equations
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
asked Dec 23 '18 at 14:34
NourNour
6
$endgroup$
add a comment |
$begingroup$
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?
ordinary-differential-equations
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
asked Dec 23 '18 at 14:34
NourNour
6
$endgroup$
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?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
asked Dec 23 '18 at 14:34
NourNour
6
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
asked Dec 23 '18 at 14:34
NourNour
6
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
edited Dec 23 '18 at 16:02
The Pointer
2,61821438
2,61821438
asked Dec 23 '18 at 14:34
NourNour
6
asked Dec 23 '18 at 14:34
NourNour
6
asked Dec 23 '18 at 14:34
NourNour
6
6
add a comment |
add a comment |
2 Answers
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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 Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
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 Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
$endgroup$
$begingroup$
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
$endgroup$
– Nour
Dec 25 '18 at 1:43
$begingroup$
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
$endgroup$
– LutzL
Dec 25 '18 at 8:19
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
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 Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
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 Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
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 Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
$endgroup$
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 Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
answered Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
answered Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
answered Dec 23 '18 at 17:40
NarasimhamNarasimham
20.8k52158
20.8k52158
add a comment |
add a comment |
$begingroup$
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 Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
$endgroup$
$begingroup$
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
$endgroup$
– Nour
Dec 25 '18 at 1:43
$begingroup$
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
$endgroup$
– LutzL
Dec 25 '18 at 8:19
add a comment |
$begingroup$
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 Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
$endgroup$
$begingroup$
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
$endgroup$
– Nour
Dec 25 '18 at 1:43
$begingroup$
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
$endgroup$
– LutzL
Dec 25 '18 at 8:19
add a comment |
$begingroup$
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 Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
$endgroup$
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 Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
edited Dec 25 '18 at 8:18
answered Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
answered Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
answered Dec 23 '18 at 14:47
LutzLLutzL
57.9k42054
57.9k42054
$begingroup$
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
$endgroup$
– Nour
Dec 25 '18 at 1:43
$begingroup$
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
$endgroup$
– LutzL
Dec 25 '18 at 8:19
add a comment |
$begingroup$
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
$endgroup$
– Nour
Dec 25 '18 at 1:43
$begingroup$
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
$endgroup$
– LutzL
Dec 25 '18 at 8:19
$begingroup$
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
$endgroup$
– Nour
Dec 25 '18 at 1:43
$begingroup$
I think the differential equation would be for $y(x)$, because $tan theta = dy/dx neq dxdy$.
$endgroup$
– Nour
Dec 25 '18 at 1:43
$begingroup$
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
$endgroup$
– LutzL
Dec 25 '18 at 8:19
$begingroup$
Yes, I was misled by the typo in the quoted problem where it is $dx/dy$. It is correct in the original.
$endgroup$
– LutzL
Dec 25 '18 at 8:19
add a comment |
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