What? Drones?

Calculus Level 5

A cube is centered at the origin. Eight drones are positioned at its eight vertices. Specifically, their positions are given by, $r_1 = (1, -1, -1) ,r_2 = (1, 1, -1), r_3 = (-1, 1, -1), r_4 = (-1, -1, -1)$ $r_5 = (1, -1, 1), r_6 = (1, 1, 1), r_7 = (-1, 1, 1), r_8 = (-1, -1,1)$

At $t = 0$ drone $k$ starts flying towards drone $1 + (k \mod 8)$ , so that drone $1$ flies towards drone $2$ , and drone $2$ flies towards drone $3$ , …,etc., and finally, drone $8$ flies towards drone $1$ .

To make the problem interesting, we'll add one more feature, as follows, for each $k = 1, 2, …,8$ the velocity of the $k$ -th drone is specified as,

$\dot{r}_k = (0.1 k) (r_j - r_k) \hspace{15pt} \text{ where } j = 1 + (k \mod 8)$

The drones trajectories converge towards a single point $(x, y, z)$ . Find $\lfloor 1000(x - y - z) \rfloor$

The answer is 1065.

2 solutions

Mark Hennings
Dec 19, 2018

The family of vector-valued differential equations $\dot{\mathbf{r}}_k \; = \; \tfrac{1}{10}k\big(\mathbf{r}_{k+1} - \mathbf{r}_k\big) \hspace{2cm} 1 \le k \le 8$ imply that $\frac{d}{dt}\left(\sum_{k=1}^8 k^{-1}\mathbf{r}_k\right) \; = \; \tfrac{1}{10}\sum_{k=1}^8 \big(\mathbf{r}_{k+1} - \mathbf{r}_k\big) \; = \; \mathbf{0}$ so that the vector $\sum_{k=1}^8 k^{-1}\mathbf{r}_k$ remains constant throughout the motion. Given that the drones all converge to the point $\mathbf{x}$ , we deduce that $\left(\sum_{k=1}^8 k^{-1}\right) \mathbf{x} \; = \; \sum_{k=1}^8 k^{-1}\mathbf{r}_k(0) \; = \; \frac{1}{840}\left(\begin{array}{c}853 \\ -121 \\ -1217 \end{array}\right)$ so that $\mathbf{x} \; = \; \frac{1}{2243}\left(\begin{array}{c}853 \\ -121 \\ -1217 \end{array}\right)$ which makes the answer $\left\lfloor 1000 \times \frac{853 + 121 + 1217}{2243}\right\rfloor \; = \; \left\lfloor \frac{811000}{761} \right\rfloor \; = \; \boxed{1065}$ The $8 \times 8$ matrix that determines these vector-valued differential equations, namely $\left(\begin{array}{cccccccc}-1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -2 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -3 & 3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -4 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -5 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -6 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -7 & 7 \\ 8 & 0 & 0 & 0 & 0 & 0 & 0 & -8 \end{array} \right)$ has one eigenvalue of $0$ , and its other seven eigenvalues have negative real part. This implies that the drones do indeed converge to a single point (corresponding to the eigenvector associated with the eigenvalue of $0$ ). I shall omit the details.

Excellent solution. Thank you for sharing it.

Hosam Hajjir - 2 years, 5 months ago

Hosam Hajjir
Dec 19, 2018

The system is described by $8 \times 3$ first-order linear differential equations with constant coefficients. While it is possible to obtain a closed-form solution, it is much easier to solve numerically using any differential equation solver. I used the Runge-Kutta 4th-Order method for integrating a vector first-order differential equation. The result is that the trajectories converge to the point $(x, y, z)=(0.3736,-0.1590, -0.5331)$ , and this makes the answer $\lfloor 1000( 0.3736+0.1590 + 0.5331) \rfloor = \boxed{1065}$ .

I have another solution which I wrote earlier, but I deleted thet.I would like to write it again.(dR/dt)=0.1k(Rj–Rk).These implies that (1/k)(dR/dt)=0.1(Rj–Rk).If we add all this equalities for k=1,2....,8 we get summation of (dR/dt) for k=1 to 8 is equals to 0.Now integrating both sides we get summation of R/k from k=1 to 8=c,where c is a constant which is constant irrespective of R.Thus if we put initial values of there coordinates then we get c(x,y,z) {as R and c represents tuple of coordinates}.Now when all the drones come at a same point then R (x,y,z) is same for all drones .Thus R(1+1/2+1/3+1/4+1/5+1/6+1/7+1/8)=c. From the known value of c which we determined earlier we get this R(x,y,z). And thus 1000{x―y―z}=1065

Alapan Das - 1 year, 10 months ago

What? Drones?

The answer is 1065.

2 solutions

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