Mixing
How to calculate the length of a parabolic arc if two end points & angle at both ends are specified?
$begingroup$
First I confess I my mathematics knowledge is not very good. So first can an expert please first confirm if this question is actually solvable? Because elsewhere on net I have read that you actually need to specify 3 points to find the length of a parabolic arc.
Imagine this is a plane which takes off at an angle of 30 degrees & gradually makes an angle of 60 degrees by the time it is at a distance of 3 km from the point where it left the ground.
geometry analytic-geometry conic-sections
asked Dec 14 '16 at 20:33
AtulAtul
111
$endgroup$
add a comment |
$begingroup$
First I confess I my mathematics knowledge is not very good. So first can an expert please first confirm if this question is actually solvable? Because elsewhere on net I have read that you actually need to specify 3 points to find the length of a parabolic arc.
Imagine this is a plane which takes off at an angle of 30 degrees & gradually makes an angle of 60 degrees by the time it is at a distance of 3 km from the point where it left the ground.
geometry analytic-geometry conic-sections
asked Dec 14 '16 at 20:33
AtulAtul
111
$endgroup$
$begingroup$
Do you really need a parabolic arc? A circular arc is easier.
$endgroup$
– Aretino
Dec 14 '16 at 20:53
add a comment |
$begingroup$
First I confess I my mathematics knowledge is not very good. So first can an expert please first confirm if this question is actually solvable? Because elsewhere on net I have read that you actually need to specify 3 points to find the length of a parabolic arc.
Imagine this is a plane which takes off at an angle of 30 degrees & gradually makes an angle of 60 degrees by the time it is at a distance of 3 km from the point where it left the ground.
geometry analytic-geometry conic-sections
asked Dec 14 '16 at 20:33
AtulAtul
111
$endgroup$
First I confess I my mathematics knowledge is not very good. So first can an expert please first confirm if this question is actually solvable? Because elsewhere on net I have read that you actually need to specify 3 points to find the length of a parabolic arc.
Imagine this is a plane which takes off at an angle of 30 degrees & gradually makes an angle of 60 degrees by the time it is at a distance of 3 km from the point where it left the ground.
geometry analytic-geometry conic-sections
geometry analytic-geometry conic-sections
asked Dec 14 '16 at 20:33
AtulAtul
111
asked Dec 14 '16 at 20:33
AtulAtul
111
asked Dec 14 '16 at 20:33
AtulAtul
111
asked Dec 14 '16 at 20:33
AtulAtul
111
asked Dec 14 '16 at 20:33
AtulAtul
111
111
$begingroup$
Do you really need a parabolic arc? A circular arc is easier.
$endgroup$
– Aretino
Dec 14 '16 at 20:53
add a comment |
$begingroup$
Do you really need a parabolic arc? A circular arc is easier.
$endgroup$
– Aretino
Dec 14 '16 at 20:53
$begingroup$
Do you really need a parabolic arc? A circular arc is easier.
$endgroup$
– Aretino
Dec 14 '16 at 20:53
$begingroup$
Do you really need a parabolic arc? A circular arc is easier.
$endgroup$
– Aretino
Dec 14 '16 at 20:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Without loss of generality we can suppose that the coordinate are chosen in such a way that the parabola has equation $ y=ax^2+bx$. So we have $y'=2ax+b$.
From the condition that the starting angle is $30°$ we have $y'(0)=tan 30°=frac{sqrt{3}}{3}$ and this gives $b=frac{sqrt{3}}{3}$.
Now the problem is to find a point $P=(p,y(p))$ such that the tangent at this point has a slope of $y'(p)=sqrt{3}=tan 60°$ and a distance from the origin $PO=3$.
This gives the system
$$
begin{cases}
sqrt{p^2+left(ap^2+frac{sqrt{3}}{3}pright)^2}=3\
2ap+frac{sqrt{3}}{3}=sqrt{3}
end{cases}
$$
Find $a=frac{sqrt{3}}{3p}$ from the second equation and substitute in the first equation. With a bit of algebra you can find $p$ and solve the problem.
Substituting $a=frac{sqrt{3}}{3p}$, the first equation gives $p=frac{3sqrt{3}}{sqrt{7}} $, so we have $a=frac{sqrt{7}}{9}$ and the equation of the parabola is:
$$
y=frac{sqrt{7}}{9}x^2 + frac{sqrt{3}}{3}x
$$
with derivative:
$$
y'=frac{2sqrt{7}}{9}x + frac{sqrt{3}}{3}x
$$
The ''final'' point have coordinates:
$$
P=(p,y(p))= left(frac{3sqrt{3}}{sqrt{7}},frac{6}{sqrt{7}} right)
$$
So the arc length from $O$ to $P$ is given by the integral:
$$
int_0^p sqrt{1+[y'(x)]^2}dx= int_0^psqrt{1+(2ax+b)^2}dx
$$
This integral can be evaluated using first the substitution $u=2ax+b$ that gives:
$$
int sqrt{1+(2ax+b)^2}dx =frac{1}{2a}intsqrt{1+u^2} du
$$
than (with a bit of work) using the trigonometric substitution $u=tan v$.
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
How does finding the focus length help me find the arc length? mathworld.wolfram.com/ParabolicSegment.html doesn't mention what role 'p' plays in the formula they give for the parabolic segment?
$endgroup$
– Atul
Dec 14 '16 at 21:35
$begingroup$
I should have mentioned that I have no prior knowledge of Parabolas. I thought there would be some simple piece of knowledge that I'm missing which will help me solve the question. It's clear now that I need to study Parabolas thoroughly before I can solve this problem.
$endgroup$
– Atul
Dec 14 '16 at 21:39
$begingroup$
In your figure $3$ is the distance from the starting point (the origin), not the arc length.
$endgroup$
– Emilio Novati
Dec 14 '16 at 21:42
$begingroup$
I know that. I am confused. Is the 'p' that I found out my required answer?
$endgroup$
– Atul
Dec 14 '16 at 21:53
$begingroup$
No, $p$ is the abscissa of the final point, from which you can find the arc length. I've added to my answer.
$endgroup$
– Emilio Novati
Dec 15 '16 at 11:03
|
show 2 more comments
$begingroup$
The equation of the parabola is $$y=ax^2+bx$$ and the given conditions allow you to write
$$begin{cases}y'(0)=b=tan30°=dfrac1{sqrt3},\y'(x)=2ax+b=tan60°=sqrt3,\x^2+(ax^2+bx)^2=3^2.end{cases}$$
Then noting that from the second equation $ax=dfrac1{sqrt3}$ we have
$$x^2+frac{4x^2}3=9$$ and $$x=sqrt{frac{27}7}.$$
Now the length is given by the integral
$$int_0^xsqrt{1+(2ax+b)^2},dx=intcosh tfrac{cosh t}{2a}dt=frac{t+cosh tsinh t}{4a}
\=left.frac{text{arsinh}(2ax+b)+sqrt{1+(2ax+b)^2}(2ax+b)}{4a}right|_0^x$$
by setting $2ax+b=sinh t$.
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Without loss of generality we can suppose that the coordinate are chosen in such a way that the parabola has equation $ y=ax^2+bx$. So we have $y'=2ax+b$.
From the condition that the starting angle is $30°$ we have $y'(0)=tan 30°=frac{sqrt{3}}{3}$ and this gives $b=frac{sqrt{3}}{3}$.
Now the problem is to find a point $P=(p,y(p))$ such that the tangent at this point has a slope of $y'(p)=sqrt{3}=tan 60°$ and a distance from the origin $PO=3$.
This gives the system
$$
begin{cases}
sqrt{p^2+left(ap^2+frac{sqrt{3}}{3}pright)^2}=3\
2ap+frac{sqrt{3}}{3}=sqrt{3}
end{cases}
$$
Find $a=frac{sqrt{3}}{3p}$ from the second equation and substitute in the first equation. With a bit of algebra you can find $p$ and solve the problem.
Substituting $a=frac{sqrt{3}}{3p}$, the first equation gives $p=frac{3sqrt{3}}{sqrt{7}} $, so we have $a=frac{sqrt{7}}{9}$ and the equation of the parabola is:
$$
y=frac{sqrt{7}}{9}x^2 + frac{sqrt{3}}{3}x
$$
with derivative:
$$
y'=frac{2sqrt{7}}{9}x + frac{sqrt{3}}{3}x
$$
The ''final'' point have coordinates:
$$
P=(p,y(p))= left(frac{3sqrt{3}}{sqrt{7}},frac{6}{sqrt{7}} right)
$$
So the arc length from $O$ to $P$ is given by the integral:
$$
int_0^p sqrt{1+[y'(x)]^2}dx= int_0^psqrt{1+(2ax+b)^2}dx
$$
This integral can be evaluated using first the substitution $u=2ax+b$ that gives:
$$
int sqrt{1+(2ax+b)^2}dx =frac{1}{2a}intsqrt{1+u^2} du
$$
than (with a bit of work) using the trigonometric substitution $u=tan v$.
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
How does finding the focus length help me find the arc length? mathworld.wolfram.com/ParabolicSegment.html doesn't mention what role 'p' plays in the formula they give for the parabolic segment?
$endgroup$
– Atul
Dec 14 '16 at 21:35
$begingroup$
I should have mentioned that I have no prior knowledge of Parabolas. I thought there would be some simple piece of knowledge that I'm missing which will help me solve the question. It's clear now that I need to study Parabolas thoroughly before I can solve this problem.
$endgroup$
– Atul
Dec 14 '16 at 21:39
$begingroup$
In your figure $3$ is the distance from the starting point (the origin), not the arc length.
$endgroup$
– Emilio Novati
Dec 14 '16 at 21:42
$begingroup$
I know that. I am confused. Is the 'p' that I found out my required answer?
$endgroup$
– Atul
Dec 14 '16 at 21:53
$begingroup$
No, $p$ is the abscissa of the final point, from which you can find the arc length. I've added to my answer.
$endgroup$
– Emilio Novati
Dec 15 '16 at 11:03
|
show 2 more comments
$begingroup$
Without loss of generality we can suppose that the coordinate are chosen in such a way that the parabola has equation $ y=ax^2+bx$. So we have $y'=2ax+b$.
From the condition that the starting angle is $30°$ we have $y'(0)=tan 30°=frac{sqrt{3}}{3}$ and this gives $b=frac{sqrt{3}}{3}$.
Now the problem is to find a point $P=(p,y(p))$ such that the tangent at this point has a slope of $y'(p)=sqrt{3}=tan 60°$ and a distance from the origin $PO=3$.
This gives the system
$$
begin{cases}
sqrt{p^2+left(ap^2+frac{sqrt{3}}{3}pright)^2}=3\
2ap+frac{sqrt{3}}{3}=sqrt{3}
end{cases}
$$
Find $a=frac{sqrt{3}}{3p}$ from the second equation and substitute in the first equation. With a bit of algebra you can find $p$ and solve the problem.
Substituting $a=frac{sqrt{3}}{3p}$, the first equation gives $p=frac{3sqrt{3}}{sqrt{7}} $, so we have $a=frac{sqrt{7}}{9}$ and the equation of the parabola is:
$$
y=frac{sqrt{7}}{9}x^2 + frac{sqrt{3}}{3}x
$$
with derivative:
$$
y'=frac{2sqrt{7}}{9}x + frac{sqrt{3}}{3}x
$$
The ''final'' point have coordinates:
$$
P=(p,y(p))= left(frac{3sqrt{3}}{sqrt{7}},frac{6}{sqrt{7}} right)
$$
So the arc length from $O$ to $P$ is given by the integral:
$$
int_0^p sqrt{1+[y'(x)]^2}dx= int_0^psqrt{1+(2ax+b)^2}dx
$$
This integral can be evaluated using first the substitution $u=2ax+b$ that gives:
$$
int sqrt{1+(2ax+b)^2}dx =frac{1}{2a}intsqrt{1+u^2} du
$$
than (with a bit of work) using the trigonometric substitution $u=tan v$.
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
How does finding the focus length help me find the arc length? mathworld.wolfram.com/ParabolicSegment.html doesn't mention what role 'p' plays in the formula they give for the parabolic segment?
$endgroup$
– Atul
Dec 14 '16 at 21:35
$begingroup$
I should have mentioned that I have no prior knowledge of Parabolas. I thought there would be some simple piece of knowledge that I'm missing which will help me solve the question. It's clear now that I need to study Parabolas thoroughly before I can solve this problem.
$endgroup$
– Atul
Dec 14 '16 at 21:39
$begingroup$
In your figure $3$ is the distance from the starting point (the origin), not the arc length.
$endgroup$
– Emilio Novati
Dec 14 '16 at 21:42
$begingroup$
I know that. I am confused. Is the 'p' that I found out my required answer?
$endgroup$
– Atul
Dec 14 '16 at 21:53
$begingroup$
No, $p$ is the abscissa of the final point, from which you can find the arc length. I've added to my answer.
$endgroup$
– Emilio Novati
Dec 15 '16 at 11:03
|
show 2 more comments
$begingroup$
Without loss of generality we can suppose that the coordinate are chosen in such a way that the parabola has equation $ y=ax^2+bx$. So we have $y'=2ax+b$.
From the condition that the starting angle is $30°$ we have $y'(0)=tan 30°=frac{sqrt{3}}{3}$ and this gives $b=frac{sqrt{3}}{3}$.
Now the problem is to find a point $P=(p,y(p))$ such that the tangent at this point has a slope of $y'(p)=sqrt{3}=tan 60°$ and a distance from the origin $PO=3$.
This gives the system
$$
begin{cases}
sqrt{p^2+left(ap^2+frac{sqrt{3}}{3}pright)^2}=3\
2ap+frac{sqrt{3}}{3}=sqrt{3}
end{cases}
$$
Find $a=frac{sqrt{3}}{3p}$ from the second equation and substitute in the first equation. With a bit of algebra you can find $p$ and solve the problem.
Substituting $a=frac{sqrt{3}}{3p}$, the first equation gives $p=frac{3sqrt{3}}{sqrt{7}} $, so we have $a=frac{sqrt{7}}{9}$ and the equation of the parabola is:
$$
y=frac{sqrt{7}}{9}x^2 + frac{sqrt{3}}{3}x
$$
with derivative:
$$
y'=frac{2sqrt{7}}{9}x + frac{sqrt{3}}{3}x
$$
The ''final'' point have coordinates:
$$
P=(p,y(p))= left(frac{3sqrt{3}}{sqrt{7}},frac{6}{sqrt{7}} right)
$$
So the arc length from $O$ to $P$ is given by the integral:
$$
int_0^p sqrt{1+[y'(x)]^2}dx= int_0^psqrt{1+(2ax+b)^2}dx
$$
This integral can be evaluated using first the substitution $u=2ax+b$ that gives:
$$
int sqrt{1+(2ax+b)^2}dx =frac{1}{2a}intsqrt{1+u^2} du
$$
than (with a bit of work) using the trigonometric substitution $u=tan v$.
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
Without loss of generality we can suppose that the coordinate are chosen in such a way that the parabola has equation $ y=ax^2+bx$. So we have $y'=2ax+b$.
From the condition that the starting angle is $30°$ we have $y'(0)=tan 30°=frac{sqrt{3}}{3}$ and this gives $b=frac{sqrt{3}}{3}$.
Now the problem is to find a point $P=(p,y(p))$ such that the tangent at this point has a slope of $y'(p)=sqrt{3}=tan 60°$ and a distance from the origin $PO=3$.
This gives the system
$$
begin{cases}
sqrt{p^2+left(ap^2+frac{sqrt{3}}{3}pright)^2}=3\
2ap+frac{sqrt{3}}{3}=sqrt{3}
end{cases}
$$
Find $a=frac{sqrt{3}}{3p}$ from the second equation and substitute in the first equation. With a bit of algebra you can find $p$ and solve the problem.
Substituting $a=frac{sqrt{3}}{3p}$, the first equation gives $p=frac{3sqrt{3}}{sqrt{7}} $, so we have $a=frac{sqrt{7}}{9}$ and the equation of the parabola is:
$$
y=frac{sqrt{7}}{9}x^2 + frac{sqrt{3}}{3}x
$$
with derivative:
$$
y'=frac{2sqrt{7}}{9}x + frac{sqrt{3}}{3}x
$$
The ''final'' point have coordinates:
$$
P=(p,y(p))= left(frac{3sqrt{3}}{sqrt{7}},frac{6}{sqrt{7}} right)
$$
So the arc length from $O$ to $P$ is given by the integral:
$$
int_0^p sqrt{1+[y'(x)]^2}dx= int_0^psqrt{1+(2ax+b)^2}dx
$$
This integral can be evaluated using first the substitution $u=2ax+b$ that gives:
$$
int sqrt{1+(2ax+b)^2}dx =frac{1}{2a}intsqrt{1+u^2} du
$$
than (with a bit of work) using the trigonometric substitution $u=tan v$.
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
edited Dec 15 '16 at 11:02
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
answered Dec 14 '16 at 20:58
Emilio NovatiEmilio Novati
52.2k43474
52.2k43474
$begingroup$
How does finding the focus length help me find the arc length? mathworld.wolfram.com/ParabolicSegment.html doesn't mention what role 'p' plays in the formula they give for the parabolic segment?
$endgroup$
– Atul
Dec 14 '16 at 21:35
$begingroup$
I should have mentioned that I have no prior knowledge of Parabolas. I thought there would be some simple piece of knowledge that I'm missing which will help me solve the question. It's clear now that I need to study Parabolas thoroughly before I can solve this problem.
$endgroup$
– Atul
Dec 14 '16 at 21:39
$begingroup$
In your figure $3$ is the distance from the starting point (the origin), not the arc length.
$endgroup$
– Emilio Novati
Dec 14 '16 at 21:42
$begingroup$
I know that. I am confused. Is the 'p' that I found out my required answer?
$endgroup$
– Atul
Dec 14 '16 at 21:53
$begingroup$
No, $p$ is the abscissa of the final point, from which you can find the arc length. I've added to my answer.
$endgroup$
– Emilio Novati
Dec 15 '16 at 11:03
|
show 2 more comments
$begingroup$
How does finding the focus length help me find the arc length? mathworld.wolfram.com/ParabolicSegment.html doesn't mention what role 'p' plays in the formula they give for the parabolic segment?
$endgroup$
– Atul
Dec 14 '16 at 21:35
$begingroup$
I should have mentioned that I have no prior knowledge of Parabolas. I thought there would be some simple piece of knowledge that I'm missing which will help me solve the question. It's clear now that I need to study Parabolas thoroughly before I can solve this problem.
$endgroup$
– Atul
Dec 14 '16 at 21:39
$begingroup$
In your figure $3$ is the distance from the starting point (the origin), not the arc length.
$endgroup$
– Emilio Novati
Dec 14 '16 at 21:42
$begingroup$
I know that. I am confused. Is the 'p' that I found out my required answer?
$endgroup$
– Atul
Dec 14 '16 at 21:53
$begingroup$
No, $p$ is the abscissa of the final point, from which you can find the arc length. I've added to my answer.
$endgroup$
– Emilio Novati
Dec 15 '16 at 11:03
$begingroup$
How does finding the focus length help me find the arc length? mathworld.wolfram.com/ParabolicSegment.html doesn't mention what role 'p' plays in the formula they give for the parabolic segment?
$endgroup$
– Atul
Dec 14 '16 at 21:35
$begingroup$
How does finding the focus length help me find the arc length? mathworld.wolfram.com/ParabolicSegment.html doesn't mention what role 'p' plays in the formula they give for the parabolic segment?
$endgroup$
– Atul
Dec 14 '16 at 21:35
$begingroup$
I should have mentioned that I have no prior knowledge of Parabolas. I thought there would be some simple piece of knowledge that I'm missing which will help me solve the question. It's clear now that I need to study Parabolas thoroughly before I can solve this problem.
$endgroup$
– Atul
Dec 14 '16 at 21:39
$begingroup$
I should have mentioned that I have no prior knowledge of Parabolas. I thought there would be some simple piece of knowledge that I'm missing which will help me solve the question. It's clear now that I need to study Parabolas thoroughly before I can solve this problem.
$endgroup$
– Atul
Dec 14 '16 at 21:39
$begingroup$
In your figure $3$ is the distance from the starting point (the origin), not the arc length.
$endgroup$
– Emilio Novati
Dec 14 '16 at 21:42
$begingroup$
In your figure $3$ is the distance from the starting point (the origin), not the arc length.
$endgroup$
– Emilio Novati
Dec 14 '16 at 21:42
$begingroup$
I know that. I am confused. Is the 'p' that I found out my required answer?
$endgroup$
– Atul
Dec 14 '16 at 21:53
$begingroup$
I know that. I am confused. Is the 'p' that I found out my required answer?
$endgroup$
– Atul
Dec 14 '16 at 21:53
$begingroup$
No, $p$ is the abscissa of the final point, from which you can find the arc length. I've added to my answer.
$endgroup$
– Emilio Novati
Dec 15 '16 at 11:03
$begingroup$
No, $p$ is the abscissa of the final point, from which you can find the arc length. I've added to my answer.
$endgroup$
– Emilio Novati
Dec 15 '16 at 11:03
|
show 2 more comments
$begingroup$
The equation of the parabola is $$y=ax^2+bx$$ and the given conditions allow you to write
$$begin{cases}y'(0)=b=tan30°=dfrac1{sqrt3},\y'(x)=2ax+b=tan60°=sqrt3,\x^2+(ax^2+bx)^2=3^2.end{cases}$$
Then noting that from the second equation $ax=dfrac1{sqrt3}$ we have
$$x^2+frac{4x^2}3=9$$ and $$x=sqrt{frac{27}7}.$$
Now the length is given by the integral
$$int_0^xsqrt{1+(2ax+b)^2},dx=intcosh tfrac{cosh t}{2a}dt=frac{t+cosh tsinh t}{4a}
\=left.frac{text{arsinh}(2ax+b)+sqrt{1+(2ax+b)^2}(2ax+b)}{4a}right|_0^x$$
by setting $2ax+b=sinh t$.
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
The equation of the parabola is $$y=ax^2+bx$$ and the given conditions allow you to write
$$begin{cases}y'(0)=b=tan30°=dfrac1{sqrt3},\y'(x)=2ax+b=tan60°=sqrt3,\x^2+(ax^2+bx)^2=3^2.end{cases}$$
Then noting that from the second equation $ax=dfrac1{sqrt3}$ we have
$$x^2+frac{4x^2}3=9$$ and $$x=sqrt{frac{27}7}.$$
Now the length is given by the integral
$$int_0^xsqrt{1+(2ax+b)^2},dx=intcosh tfrac{cosh t}{2a}dt=frac{t+cosh tsinh t}{4a}
\=left.frac{text{arsinh}(2ax+b)+sqrt{1+(2ax+b)^2}(2ax+b)}{4a}right|_0^x$$
by setting $2ax+b=sinh t$.
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
The equation of the parabola is $$y=ax^2+bx$$ and the given conditions allow you to write
$$begin{cases}y'(0)=b=tan30°=dfrac1{sqrt3},\y'(x)=2ax+b=tan60°=sqrt3,\x^2+(ax^2+bx)^2=3^2.end{cases}$$
Then noting that from the second equation $ax=dfrac1{sqrt3}$ we have
$$x^2+frac{4x^2}3=9$$ and $$x=sqrt{frac{27}7}.$$
Now the length is given by the integral
$$int_0^xsqrt{1+(2ax+b)^2},dx=intcosh tfrac{cosh t}{2a}dt=frac{t+cosh tsinh t}{4a}
\=left.frac{text{arsinh}(2ax+b)+sqrt{1+(2ax+b)^2}(2ax+b)}{4a}right|_0^x$$
by setting $2ax+b=sinh t$.
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
$endgroup$
The equation of the parabola is $$y=ax^2+bx$$ and the given conditions allow you to write
$$begin{cases}y'(0)=b=tan30°=dfrac1{sqrt3},\y'(x)=2ax+b=tan60°=sqrt3,\x^2+(ax^2+bx)^2=3^2.end{cases}$$
Then noting that from the second equation $ax=dfrac1{sqrt3}$ we have
$$x^2+frac{4x^2}3=9$$ and $$x=sqrt{frac{27}7}.$$
Now the length is given by the integral
$$int_0^xsqrt{1+(2ax+b)^2},dx=intcosh tfrac{cosh t}{2a}dt=frac{t+cosh tsinh t}{4a}
\=left.frac{text{arsinh}(2ax+b)+sqrt{1+(2ax+b)^2}(2ax+b)}{4a}right|_0^x$$
by setting $2ax+b=sinh t$.
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
edited Jan 29 at 9:27
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 9:20
Yves DaoustYves Daoust
131k676229
131k676229
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$begingroup$
Do you really need a parabolic arc? A circular arc is easier.
$endgroup$
– Aretino
Dec 14 '16 at 20:53