How far can a car travel along a parabolic trajectory between two points?

[Zeit]

Hi,

I don't know if I'm in the the good section to ask my question, I don't really know which type of mathematics my question is about.

If I'm in a car, according to a parabolic trajectory, which would be the distance traversed by the car between locations A and B?

Thanks a lot

Zeit

NB : Sorry for the mistakes

 

[neutrino]

Your question is not very clear. A car is following a parabolic trajectory with you in it but what(or where) are points A and B? If this question is from some textbook please provide it as it is in the book.

 

[Zeit]

Excuse me, I have some (many) problems to express myself in English.

No, it is not a question from a textbook, but from my mind. In fact, this example of a car is just another way to say : what is the equation to know the distance between any points of a parabola? Not the equation d² = (x'-x)² + (y'-y)², which is for a line, but the "curved distance".

Thanks

 

[HallsofIvy]

You posted this in the "general math" section but the answer requires calculus so this may not help you. This is a definitely "non-trivial" problem!

In general, if y= f(x), the "curved distance" (arclength) on the graph of y= f(x), from (a, f(a)) to (b, f(b)) is given by
\int_a^b\sqrt{1+ \frac{dy}{dx}}dx[/itex]<br /> <br /> In particular, for a &quot;parabolic trajectory&quot;, we can take y= ax² so that \frac{dy}{dx}= 2ax and the integral can be written as <br /> \int_a^b\sqrt{1+ 4a^2x^2}dx[/itex]&lt;br /&gt; &lt;br /&gt; That can be done by a trigonometric substitution: let 2ax= tan\theta. Then 2adx= sec^2\theta d\theta and &lt;br /&gt; \sqrt{1+ 4a^2x^2}= \sqrt{1+ tan^2\theta}= \sqrt{sec^2\theta}= sec\theta&lt;br /&gt; &lt;br /&gt; The integral becomes&lt;br /&gt; \int_{arctan a}^{arctan b} sec^3\theta d\theta= \int_{arctan a}^{arctan b} \frac{d\theta}{cos^3\theta}&lt;br /&gt; Since cosine is to an odd power, multiply both numerator and denominator by cos\theta&lt;br /&gt; \int_{arctan a}^{arctan b}\frac{cos\theta d\theta}{cos^4\theta}=\int_{arctan a}^{arctan b}\frac{cos\theta d\theta}{(1- sin^2\theta)^2}&lt;br /&gt; &lt;br /&gt; Now make the substitution u= sin\theta so that du= cos\theta d\theta. The integral becomes&lt;br /&gt; \int_{sin(arctan a)}^{sin(arctan b)}\frac{du}{(1- u^2)^2}&lt;br /&gt; which can be done by partial fractions.&lt;br /&gt; (To find sin(arctan a), draw a right triangle with opposite side of length a, near side of length b, so that the angle is [/itex]arctan a[/itex]. By the Pythagorean theorem, the hypotenus has length \sqrt{a^1+ 1}. sin(arctan a)= \frac{a}{\sqrt{a^2+1|}}. Do the same for sin(arctan b).)

 

[Zeit]

Thanks HallsofIvy for the explanation.

But, in the first equation, you wrote dy/dx. Then, you wrote that because y = ax², dy/dx = 2ax and you substitute dy/dx in the first equation for 4a²x² in the second. My question is : in the first equation, isn't (dy/dx)² ?

[...] the "curved distance" (arclength) [...]

Thanks!

 

[d_leet]

Thanks HallsofIvy for the explanation.

But, in the first equation, you wrote dy/dx. Then, you wrote that because y = ax², dy/dx = 2ax and you substitute dy/dx in the first equation for 4a²x² in the second. My question is : in the first equation, isn't (dy/dx)² ?



Thanks!

Yes it should be.

 

[Zeit]

Hi

In particular, for a "parabolic trajectory", we can take y= ax² so that dy/dx = 2ax

If y has more parameters, like y = a(x-h)²+k, what would be dy/dx? Or if y = ax²+bx+c, would dy/dx equal 2ax+b ?

Thanks

 

[d_leet]

Hi



If y has more parameters, like y = a(x-h)²+k, what would be dy/dx? Or if y = ax²+bx+c, would dy/dx equal 2ax+b ?

Thanks

That would be correct. And in the first case where you have y = a(x-h)²+k then dy/dx would be 2a(x-h) which shouldn't be hard to see.

 

[Zeit]

Thanks for the answer.