Two ships pursue straight-line courses at steady speeds. At each time t (in hours, where 0 less-than-or-equal t less-than-or-equal 14), the positions of the two ships are given by:

Question

Two ships pursue straight-line courses at steady speeds. At each time t (in hours, where 0 less-than-or-equal t less-than-or-equal 14), the positions of the two ships are given by:

Ship A (10t+35, 20t+25)
Ship B (11t+15, 22t+15)
These coordinates are with reference to a particular coordinate system, with distance measured in kilometres.

What is the closest distance to which the ships approach each other and when does this closest distance occur? Give you answers exactly.

Answer 1

Clearly the distance z between the ships can be found using

z^2 = (Ax-Bx)^2 + (Ay-By)^2
= (-t+20)^2 + (-2t+10)^2
= 5t^2 - 80t + 500

2z dz/dt = 10t-80
dz/dt = 0 when t=8.

So, just evaluate z at t=8.

Answer 2

To find the closest distance to which the ships approach each other, we need to compute the distance between the two ships at different times and find the minimum distance.

The distance between two points (x1, y1) and (x2, y2) in a coordinate system is given by the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Let's calculate the distance between the two ships at a general time 't':
d(t) = sqrt(((11t + 15) - (10t + 35))^2 + ((22t + 15) - (20t + 25))^2)
= sqrt((11t - 10t - 20)^2 + (22t - 20t - 10)^2)
= sqrt(t^2 + (-2t - 10)^2)
= sqrt(t^2 + 4t^2 + 40t + 100)
= sqrt(5t^2 + 40t + 100)

To find the exact minimum distance, we can take the derivative of d(t) with respect to t and set it equal to 0.

d'(t) = (1/2) * (5t^2 + 40t + 100)^(-1/2) * (10t + 40)
= (10t + 40) / (2 * sqrt(5t^2 + 40t + 100))

Setting d'(t) = 0 yields:
10t + 40 = 0
10t = -40
t = -4

Since the time 't' must be between 0 and 14, we discard the negative solution.

So, the closest distance occurs when t = -4. However, since distance cannot be negative, we can ignore this result.

Now, we need to consider the boundary points, t = 0 and t = 14, to determine if they have a smaller distance.

For t = 0:
d(0) = sqrt((15 - 35)^2 + (15 - 25)^2)
= sqrt((-20)^2 + (-10)^2)
= sqrt(400 + 100)
= sqrt(500)
= 10 * sqrt(5) km (approximately)

For t = 14:
d(14) = sqrt(((11 * 14) + 15 - (10 * 14) - 35)^2 + ((22 * 14) + 15 - (20 * 14) - 25)^2)
= sqrt((155 - 25)^2 + (308 - 250)^2)
= sqrt(130^2 + 58^2)
= sqrt(16900 + 3364)
= sqrt(20264)
= 142.42 km (approximately)

Therefore, the closest distance the ships approach each other is 10 * sqrt(5) km (approximately) and it occurs at t = 0.