# Boat A's position is given by x_a(t)=3-t, y_a(t)=2t-4 and boat B's position is given by x_b(t)=4-3t, y_b(t)=3-2t. The distance units are kilometres and the time units are hours. At what time are the boats closest to each other?

Jun 13, 2017

The time is $= \frac{3}{2} h$

#### Explanation:

I tried it this way :

The position of $A$ is ${r}_{A} = \left(3 - t , 2 t - 4\right)$

The position of $B$ is ${r}_{B} = \left(4 - 3 t , 3 - 2 t\right)$

The difference of their positions is

$= {r}_{A} - {r}_{B} = \left(3 - t - 4 + 3 t , 2 t - 4 - 3 + 2 t\right)$

$= \left(- 1 + 2 t , 4 t - 7\right)$

The distance is

$p = \sqrt{{\left(- 1 + 2 t\right)}^{2} + {\left(4 t - 7\right)}^{2}}$

$= \sqrt{1 - 4 t + 4 {t}^{2} + 49 - 56 t + 16 {t}^{2}}$

$= \sqrt{20 {t}^{2} - 60 t + 50}$

To calculate the closest distance, we calculate the first derivative

$\frac{\mathrm{dp}}{\mathrm{dt}} = \frac{1}{2 \sqrt{20 {t}^{2} - 60 t + 50}} \cdot \left(40 t - 60\right)$

When

$\frac{\mathrm{dp}}{\mathrm{dt}} = 0$, $\implies$, $40 t - 60 = 0$

$t = \frac{60}{40} = \frac{3}{2} h$