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

To derive the distance function, we calculate the position of each ship at time t in hours:

• Ship A's eastward distance: $x(t) = 15t$
• Ship B's southward distance: $y(t) = 30t$

Now, the relative positions of the ships are:

• Ship A's distance from Port B: $90 - 15t$
• Ship B's distance from Port B: $30t$

Using the Pythagorean theorem, the distance (s) between the ships can be expressed as: $s^{2} = (90 - 15t)^{2} + (30t)^{2}$ Expanding this: s^{2} = (90^{2} - 2 \cdot 90 \cdot 15t + (15t)^{2}) + (30t)^{2 $s^{2} = 8100 - 2700t + 225t^{2} + 900t^{2}$ $s^{2} = 1125t^{2} - 2700t + 8100$

Thus, the function is:

s(t) = (1125t^{2} - 2700t + 8100)^{rac{1}{2}}

valid for $0 \leq t \leq 6$.

To find when the ships are closest to each other, we need to minimize the distance function s(t).

First, we differentiate s(t) with respect to t:

Using the chain rule: $\frac{ds}{dt} = \frac{1}{2}(1125t^{2} - 2700t + 8100)^{-\frac{1}{2}}(2250t - 2700)$ Setting the derivative equal to zero for minimization: $2250t - 2700 = 0$ Solving for t gives: $2250t = 2700 \Rightarrow t = \frac{2700}{2250} = 1.2 ext{ hours}$

Now, we substitute t = 1.2 back into the distance function: $s(1.2) = (1125(1.2)^{2} - 2700(1.2) + 8100)^{\frac{1}{2}}$ Calculating this: $= (1125(1.44) - 3240 + 8100)^{\frac{1}{2}}$ $= (1620 - 3240 + 8100)^{\frac{1}{2}}$ $= (6480)^{\frac{1}{2}} \approx 80.5 ext{ km}$

Therefore, the ships are closest to each other at approximately 80.5 km.