# Thread: Arc length for parabolic projectile.

1. ## Arc length for parabolic projectile.

I'm stumped by this one and my engineer flatmates weren't any help.

Someone is attempting to jump a 100m canyon on his motorbike. the ramp is inclined at 30degrees and the top of the ramp is level with the ground on the other side of the canyon.

By assuming that the man and his bike is a point mass and there's no wind resistance, it can be shown that the flight path taken by the man's bike is a parabola

$y=\frac{-2gx^2}{3v^2}+\frac{x}{\sqrt{3}}$

where g=10ms^-2 is the gravitation constant, and v his launch speed (in ms^-1.

Use the arc length formula to write down a definite integral which gives the lengh of the flight path in terms of v. then evaluate the integral to get an expression for the arc length in terms of v.

I played with it for a while with not much luck. I think I have to parametrize the equation into something like x=vcos(30)t, y=vsin(30)t -(1/2)gt^2 then use the parametric arc length formula. Then there's the trig substitution to worry about, i guess it'll be something substituted for tan then integrating sec. Any help would be wonderful.

2. Originally Posted by Rombie
I'm stumped by this one and my engineer flatmates weren't any help.

Someone is attempting to jump a 100m canyon on his motorbike. the ramp is inclined at 30degrees and the top of the ramp is level with the ground on the other side of the canyon.

By assuming that the man and his bike is a point mass and there's no wind resistance, it can be shown that the flight path taken by the man's bike is a parabola

$y=\frac{-2gx^2}{3v^2}+\frac{x}{\sqrt{3}}$

where g=10ms^-2 is the gravitation constant, and v his launch speed (in ms^-1.

Use the arc length formula to write down a definite integral which gives the lengh of the flight path in terms of v. then evaluate the integral to get an expression for the arc length in terms of v.

I played with it for a while with not much luck. I think I have to parametrize the equation into something like x=vcos(30)t, y=vsin(30)t -(1/2)gt^2 then use the parametric arc length formula. Then there's the trig substitution to worry about, i guess it'll be something substituted for tan then integrating sec. Any help would be wonderful.
See here, the integral that you need is the first one in the Modern Methods section.

You will see that to get the integrand you will need to compute $f'(x)=\frac{dy}{dx}$ in this case.

RonL

3. Thanks for that but I know the integral I want, i'm just confused by the extra starting variable. dy/dx has a nasty form straight out and I wonder if i can parametrize the equation to make it simpler. Also I don't know what range to integrate over. from 0 to a? a being where y = 0 at the impact.

4. For the derivative i get
$\frac{dy}{dx}=\frac{-4gx}{3v^2}+\frac{1}{\sqrt{3}}$

and the arc length integral.
$\int{\sqrt{1+(\frac{-4gx}{3v^2}+\frac{1}{\sqrt{3}})^2}}$

What range should I be integrating over? from x=0 to where f(x)=0(again but at the impact)? The integral looks pretty nasty. Is it even in terms of v? I think doing it parametrically might help. If that is the simplest integral i can generate how should i go about evaluating it? substitute tan(theta) for the derivative? I'm just concerned about the "in terms of v" bit. Does that mean the answer for the arc length should be some polynomial of v?

Then my parametric attempts are just as hamfisted. My textbook says I can use these equations. When I differentiate them there is still a leftover term of t that I don't want.
$x=vcos(30)t$
$y=vsin(30)t + (1/2)gt^2$

Any help appreciated. Thanks.

5. Originally Posted by Rombie
For the derivative i get
$\frac{dy}{dx}=\frac{-4gx}{3v^2}+\frac{1}{\sqrt{3}}$

and the arc length integral.
$\int{\sqrt{1+(\frac{-4gx}{3v^2}+\frac{1}{\sqrt{3}})^2}}$

What range should I be integrating over? from x=0 to where f(x)=0(again but at the impact)? The integral looks pretty nasty. Is it even in terms of v? I think doing it parametrically might help. If that is the simplest integral i can generate how should i go about evaluating it? substitute tan(theta) for the derivative? I'm just concerned about the "in terms of v" bit. Does that mean the answer for the arc length should be some polynomial of v?

Then my parametric attempts are just as hamfisted. My textbook says I can use these equations. When I differentiate them there is still a leftover term of t that I don't want.
$x=vcos(30)t$
$y=vsin(30)t + (1/2)gt^2$

Any help appreciated. Thanks.
The origin of the coordinate system is ground level on the take-off side of
the canyon. Therefore ground level corresponds to $y=0$, and so
the $x$'s corresponding to the two ends of the trajectory are the roots of:

$
\frac{-2gx^2}{3v^2}+\frac{x}{\sqrt{3}}=0
$

which are:

$x=0 \wedge x=\frac{\sqrt{3}v^2}{2~g}$

RonL

6. Originally Posted by Rombie
For the derivative i get
$\frac{dy}{dx}=\frac{-4gx}{3v^2}+\frac{1}{\sqrt{3}}$

and the arc length integral.
$\int{\sqrt{1+(\frac{-4gx}{3v^2}+\frac{1}{\sqrt{3}})^2}}$

What range should I be integrating over? from x=0 to where f(x)=0(again but at the impact)? The integral looks pretty nasty. Is it even in terms of v? I think doing it parametrically might help. If that is the simplest integral i can generate how should i go about evaluating it? substitute tan(theta) for the derivative? I'm just concerned about the "in terms of v" bit. Does that mean the answer for the arc length should be some polynomial of v?

Then my parametric attempts are just as hamfisted. My textbook says I can use these equations. When I differentiate them there is still a leftover term of t that I don't want.
$x=vcos(30)t$
$y=vsin(30)t + (1/2)gt^2$

Any help appreciated. Thanks.
To simplify this consider the arec length of the curve:

$
y=-ax^2+bx,\ \ \ a,b > 0
$

between $x=0$ and $x=b/a$, which gives you the integral:

$
S=\int_0^{b/a} \sqrt{1+(-2a~x^+b)^2} ~dx
$

Now make the change of variable: $u=2a x-b$, and then the integral becomes:

$
S=\frac{1}{2a}\int_{-b}^{b} \sqrt{1+u^2} ~du
$

Now evaluate this integral, and substitute appropriate values for $a$ and $b$ to give the arc length you require (which will be still be pretty horrible).

RonL

7. Thanks a lot, that was very helpful. Except that I found the other root was $x=\frac{\sqrt{3}v^2}{2~g}$ I substituted u for $tan(\theta)$, then used the identity $tan^2(\theta) + 1 = sec^2(\theta)$ to get $sec^3(\theta)$

So my integral was $S=\frac{1}{2a}\int_{tan(-b)}^{tan(2b)} sec^3(\theta) ~d\theta$

Which gave $\frac{1}{2a}(\frac{1}{2}(sec(\theta)tan(\theta))+\ frac{1}{2}(\ln|sec(\theta)+tan(\theta)|)$
from theta = tan(-b) to theta=tan(2b).

hmm now i'm not sure if I did the substitutions right. should the upper and lower be $tan^{-1}(2b)$ and $tan^{-1}(-b)$?

How did you get 2b as the upper? When i sub b/a into 2ax-b i get b.

Thanks a lot for the help I feel like I am slowly getting a hold on this problem.

8. Originally Posted by Rombie

How did you get 2b as the upper? When i sub b/a into 2ax-b i get b.
So do I now

RonL

9. Originally Posted by Rombie
hmm now i'm not sure if I did the substitutions right. should the upper and lower be $tan^{-1}(2b)$ and $tan^{-1}(-b)$?

How did you get 2b as the upper? When i sub b/a into 2ax-b i get b.

Thanks a lot for the help I feel like I am slowly getting a hold on this problem.
Yes the limits should be:

$\tan^{-1}(b)$ and $\tan^{-1}(-b)$?

I must say that I first just looked the integral up rather than do it by hand.

RonL

10. Originally Posted by Rombie
I'm stumped by this one and my engineer flatmates weren't any help.

Someone is attempting to jump a 100m canyon on his motorbike. the ramp is inclined at 30degrees and the top of the ramp is level with the ground on the other side of the canyon.

By assuming that the man and his bike is a point mass and there's no wind resistance, it can be shown that the flight path taken by the man's bike is a parabola

$y=\frac{-2gx^2}{3v^2}+\frac{x}{\sqrt{3}}$

where g=10ms^-2 is the gravitation constant, and v his launch speed (in ms^-1.

Use the arc length formula to write down a definite integral which gives the lengh of the flight path in terms of v. then evaluate the integral to get an expression for the arc length in terms of v.

I played with it for a while with not much luck. I think I have to parametrize the equation into something like x=vcos(30)t, y=vsin(30)t -(1/2)gt^2 then use the parametric arc length formula. Then there's the trig substitution to worry about, i guess it'll be something substituted for tan then integrating sec. Any help would be wonderful.
Now just to provide some reference I have calculated the initial speed to just clear the canyon, this is the solution to:

$\frac{\sqrt{3}v^2}{2~g}=100$,

which has solution $v=\sqrt{\frac{200g}{\sqrt{3}}}$.

Now I can numericaly integrate the arc length integral to get:

$
S \approx 105.307
$

RonL

11. Thanks, I get the same when I evaluate. You've been a big help.