# Distance physics question

1. Aug 6, 2006

### 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

2. Aug 6, 2006

### 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.

3. Aug 6, 2006

### 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

4. Aug 6, 2006

### 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
$$\int_a^b\sqrt{1+ \frac{dy}{dx}}dx[/itex] In particular, for a "parabolic trajectory", we can take y= ax2 so that $\frac{dy}{dx}= 2ax$ and the integral can be written as [tex]\int_a^b\sqrt{1+ 4a^2x^2}dx[/itex] That can be done by a trigonometric substitution: let $2ax= tan\theta$. Then $2adx= sec^2\theta d\theta$ and [tex]\sqrt{1+ 4a^2x^2}= \sqrt{1+ tan^2\theta}= \sqrt{sec^2\theta}= sec\theta$$

The integral becomes
$$\int_{arctan a}^{arctan b} sec^3\theta d\theta= \int_{arctan a}^{arctan b} \frac{d\theta}{cos^3\theta}$$
Since cosine is to an odd power, multiply both numerator and denominator by $cos\theta$
$$\int_{arctan a}^{arctan b}\frac{cos\theta d\theta}{cos^4\theta}=\int_{arctan a}^{arctan b}\frac{cos\theta d\theta}{(1- sin^2\theta)^2}$$

Now make the substitution $u= sin\theta$ so that $du= cos\theta d\theta$. The integral becomes
$$\int_{sin(arctan a)}^{sin(arctan b)}\frac{du}{(1- u^2)^2}$$
which can be done by partial fractions.
(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 $sin(arctan b)$.)

5. Aug 7, 2006

### 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 ?

Thanks!

6. Aug 7, 2006

### d_leet

Yes it should be.

7. Aug 7, 2006

### Zeit

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

8. Aug 7, 2006

### d_leet

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.

9. Aug 7, 2006