Two ships set sail at the same time - Leaving Cert Mathematics - Question 9 - 2020

To establish this function, we find the positions of the two ships at time t (in hours):

1. The position of Ship A is given by:

   ext{Distance}_A = 15t ext{ km east of Port A}.

   Therefore, the total distance from Port A to Ship A becomes: 90 - 15t.

2. The position of Ship B is given by:

   ext{Distance}_B = 30t ext{ km south of Port B}.

Now, using the Pythagorean theorem for the distance s between the ships, we have:

$s^2 = (90 - 15t)^2 + (30t)^2$

Expanding the equation:

$s^2 = (90 - 15t)(90 - 15t) + (30t)(30t)$
$= 8100 - 2700t + 225t^2 + 900t^2$
$= 8100 - 2700t + 1125t^2$

Thus, we arrive at:

$s(t) = (1125t^2 - 2700t + 8100)^{1/2}$.

To find the minimum distance, we differentiate s(t):

1. Start with: $s(t) = (1125t^2 - 2700t + 8100)^{1/2}$

2. Differentiate s(t):
   Apply the chain rule: $\frac{ds}{dt} = \frac{1}{2}(1125t^2 - 2700t + 8100)^{-1/2} \times (2250t - 2700)$

3. Set ds/dt to zero to find critical points: $2250t - 2700 = 0$
   $2250t = 2700$
   $t = 1.2 ext{ hours}$

4. To find the distance at this value of t, substitute back into s(t):
   $s(1.2) = (1125(1.2)^2 - 2700(1.2) + 8100)^{1/2}$ $= (1125(1.44) - 3240 + 8100)^{1/2}$ $= (1620 - 3240 + 8100)^{1/2}$
   $= (7200)^{1/2} = 84.85 ext{ km}$

Thus, the distance between the ships at this time is approximately 84.9 km.