To give you an example, say you're paragliding and in this scenario, your velocity and height starts as:

$t_0 : V_{y-initial} = 0\,m/s,\quad V_{x-initial} = 15\,m/s, \quad h_{initial} = 180\,m$

From this state, you release a rock that you were holding. Now assume no air resistance for the rock, and as such, $V_x = 15\,m/s$ for its entire fall. $|V_y| = |gt|$ where we assume $g = 10\,m/s^2$.

When the rock has hit the ground, how far has it travelled?

  • Q1: Which functions am I seeking? Velocity or distance? (To calculate total distance travelled)

To find the time $t_f$ when the rock has hit the ground, we put:

$$S_y(t) = 180 \iff \int V_y(t) = 5t^2 = 180 \implies t = 6\,s$$

Length of curve, i.e. total distance travelled is then calculated as:

$$L = \int^6_0\sqrt{V_x(t)^2+V_y(t)^2} = \int^6_0\sqrt{15^2+(10t)^2} \approx 209\,m$$

  • Q2: Is this answer even reasonable? Here's how I think: It must fall at least $180$ meters, which it fulfills. However, during the $6$ seconds, the minimum distance travelled in $x$-axis is $15\cdot6 = 90\,m$. This means the distance travelled must be greater than $180+90\,m$. It does not fulfill this. (I'm not sure where I'm wrong here).

I can solve for $V_{tot}$ as it's a net velocity between $V_x$ and $V_y$. Pythagoras' gives us,

$$V_{tot} = \sqrt{V_x^2+V_y^2} = \sqrt{15^2+(10t)^2}$$

And as such, it's a function we're looking to find the length of:

$$L = \int^6_0\sqrt{1+\sqrt{15^2+(10t)^2}}$$

  • Q3: Why is this wrong?

1 Answer 1


A1: I do not really understand you question. You did it correctly in the next paragraph.

A2: Your argument for why the length is wrong is wrong. $209\text{ m}$ is correct. When you add the lengths you assume they are in the same direction, but they are perpendicular. This means the answer should be atleast and just a bit more than $\sqrt{180^2+90^2}\approx 201$ by using pythagoras.

A3: That formula is for non-parametric curves (I do not know the proper name). I you want to use that you have to convert it from parametric to non-parametric. \begin{align*} S_y&=180-5t^2\\ S_x&=15t \end{align*} We want to express $S_y$ as a function of $S_x$. This gives us \begin{align*} S_y&=180-5\left(\frac{S_x}{15}^2\right)\\ \frac{dS_y}{dS_x}&=-\frac 2{45}S_x \end{align*} Now we use the formula \begin{align*} L&=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)}\,dx\\ &=\int_0^{90}\sqrt{1+\left(-\frac 2{45}x\right)}\\ &\approx 209 \end{align*}

  • $\begingroup$ Nailed it, good work. $\endgroup$ Dec 8, 2014 at 11:10
  • $\begingroup$ Sorry for late response. You made everything clear. Thanks a lot :) $\endgroup$
    – B. Lee
    Dec 8, 2014 at 17:30

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