How to calculate the length of a parabolic arc if two end points & angle at both ends are specified? 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.
 A: 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
$$
\begin{cases}
\sqrt{p^2+\left(ap^2+\frac{\sqrt{3}}{3}p\right)^2}=3\\
2ap+\frac{\sqrt{3}}{3}=\sqrt{3}
\end{cases}
$$
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{3\sqrt{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:
$$
y'=\frac{2\sqrt{7}}{9}x + \frac{\sqrt{3}}{3}x
$$
The ''final'' point  have coordinates:
$$
P=(p,y(p))= \left(\frac{3\sqrt{3}}{\sqrt{7}},\frac{6}{\sqrt{7}} \right)
$$ 
So the arc length from $O$ to $P$ is given by the integral:
$$
\int_0^p \sqrt{1+[y'(x)]^2}dx= \int_0^p\sqrt{1+(2ax+b)^2}dx
$$
This integral can be evaluated using first the substitution $u=2ax+b$ that gives:
$$
\int \sqrt{1+(2ax+b)^2}dx =\frac{1}{2a}\int\sqrt{1+u^2} du
$$
than (with a bit of work) using the trigonometric substitution $u=\tan v$.
A: 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
$$\int_0^x\sqrt{1+(2ax+b)^2}\,dx=\int\cosh t\frac{\cosh t}{2a}dt=\frac{t+\cosh t\sinh t}{4a}
\\=\left.\frac{\text{arsinh}(2ax+b)+\sqrt{1+(2ax+b)^2}(2ax+b)}{4a}\right|_0^x$$
by setting $2ax+b=\sinh t$.
