How to calculate the length of a parabolic arc if two end points & angle at both ends are specified?

Question

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.

Answer 1

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.

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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 $u=\tan v$.