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We describe the trajectory of a thrown object (neglecting friction and similiar effects) with the curve

$$k(t) = \left(v_0\cos(\beta)t,\,v_0\sin(\beta)t-\frac{g}{2}t^2\right)$$

with $t\geq 0$. $v_0>0$ is the initial velocity, $\beta\in[0,2\pi]$ the initial angle and $g$ the constant of gravitation. Compute

$$s(t) = \int_0^t \|k'(u)\|\,\mathrm du.$$

I do fail with simple transformations. First the derivative is very simple $$k'(t) = (v_0\cos(\beta),\,v_0\sin(\beta)-gt)$$ and then I wanted to compute the length so my first steps were $$\begin{align}\|k'(t)\| &= \sqrt{v_0^2\cos^2(\beta)+v_0^2\sin^2(\beta)-2gtv_0\sin(\beta)+g^2t^2}\\ &= \sqrt{v_0^2-2gtv_0\sin(\beta)+g^2t^2}\end{align}$$ where I factorized $v_0$ and got $\sin^2(\beta)+\cos^2(\beta)=1$. However this still seems very difficult to integrate, so I am looking for better simplifications or good techniques on how to integrate this properly.

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Try completing the square so that you get something of the form $\sqrt{(a+t)^2+b}$, and then use substitution. –  Eckhard Jan 14 '13 at 13:32
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2 Answers 2

up vote 1 down vote accepted

Hint: Use the method called "completing the square". In your case

$$\tag{1} v_0^2-2guv_0\sin\beta+g^2u^2= \left(gu-v_0 \sin \beta \right)^2+v_0^2 \cos ^2\beta . $$ This suggests the substitution $$y= g u-v_0 \sin \beta .$$

Spoiler below:

We need to calculate $$s(t) = \int_0^t \|k'(u)\|\, du.$$ Using (1) for the expression $ \|k'(u)\|$ and performing the above-mentioned substitution yields $$\begin{align}s(t) &= \int_0^t \sqrt{v_0^2-2guv_0\sin\beta+g^2u^2}\, du \\ &= \int_0^t \sqrt{\left(gu-v_0 \sin \beta \right)^2+v_0^2 \cos ^2\beta\, }du\\&=\int_{- v_0 \sin\beta}^{gt- v_0 \sin\beta} \sqrt{y^2 + v_0^2 \cos ^2\beta}.\end{align} $$ The last integral is usually solved via the substitution $y= v_0\cos\beta \sinh x$. We obtain $$\begin{align}s(t) &= v_0|\cos \beta|\int_{-\mathop{\rm asinh} \tan \beta}^{\mathop{\rm asinh} (g t/(v_0 \cos \beta)- \tan \beta)} \overbrace{\cosh^2 x}^{(1+\cosh 2x)/2}\,dx\\&= v_0|\cos \beta| \left(\frac{1}{2} x+ \frac{1}{4}\sinh(2x) \right)\Bigg|_{x=-\mathop{\rm asinh} \tan \beta}^{\mathop{\rm asinh} (g t/(v_0 \cos \beta)- \tan \beta)}\end{align}$$

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Actually I never saw this method, but I think that this yields $\int \|k'(u)\|\,\mathrm du=\frac{1}{2}g\left(u-\frac{v_0\sin(\beta)}{g}\right)^2+v_0\cos(\beta)u+\text{c‌​onstant}$. Did I understand this substitution right? –  Christian Ivicevic Jan 14 '13 at 13:40
    
@ChristianIvicevic: no, not exactly. See the edit of my answer... –  Fabian Jan 14 '13 at 14:48
    
I don't know whether this is correct, but we never did such complicated computations in class. I tried to solve this problem now by not factorizing so that I have $\sqrt{v_0^2\cos^2\beta+(v_0\sin\beta-gt)^2}$ and simplify that with other (much easier) approaches. Do you have some other suggestions? –  Christian Ivicevic Jan 14 '13 at 17:19
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You're trying to calculate the arclength of a parabola. This isn't easy, but it can be done. See http://en.wikipedia.org/wiki/Parabola#Length_of_an_arc_of_a_parabola

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