At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 21 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
D^2 = x^2 + y^2
2 D dD/dt = 2 x dx/dt + 2 y dy/dt
To find the rate at which the distance between the ships is changing at 4 PM, we can use the concept of related rates.
Let's start by visualizing the situation. We have two ships, A and B. Ship A is located 30 nautical miles due west of ship B. Ship A is sailing west at 21 knots (21 nautical miles per hour), and ship B is sailing north at 15 knots (15 nautical miles per hour).
First, let's represent the position of ship A at time t as x(t) and the position of ship B as y(t).
At noon, ship A is 30 nautical miles due west of ship B, so we have:
x(0) = 30
y(0) = 0
Now, we need to find the relationship between x(t) and y(t) that describes the changing distance between the ships. We can use the Pythagorean theorem to relate x(t), y(t), and the distance d(t) between the ships:
d(t)^2 = x(t)^2 + y(t)^2
To find the rate at which the distance is changing with respect to time, we differentiate both sides of the equation with respect to t:
2d(t)d'(t) = 2x(t)x'(t) + 2y(t)y'(t)
We know that ship A is sailing west at 21 knots, so its velocity is -21 (negative sign indicates westward direction). Ship B is sailing north at 15 knots, so its velocity is +15 (positive sign indicates northward direction). Therefore, we have:
x'(t) = -21
y'(t) = 15
To find the rate at which the distance between the ships is changing at 4 PM (t = 4), we substitute the given values into the equation:
2d(4)d'(4) = 2x(4)x'(4) + 2y(4)y'(4)
We know that at noon (t = 0), x(0) = 30 and y(0) = 0. To find x(4) and y(4), we can use the velocities and the time elapsed since noon:
x(4) = x(0) + x'(t) * t = 30 + (-21) * 4 = -54
y(4) = y(0) + y'(t) * t = 0 + 15 * 4 = 60
Plugging in these values, we get:
2d(4)d'(4) = 2(-54)(-21) + 2(60)(15)
2d(4)d'(4) = 2268 + 1800
2d(4)d'(4) = 4068
Finally, divide by 2d(4) to solve for d'(4):
d'(4) = 4068 / (2d(4))
d'(4) = 4068 / (2√((-54)^2 + (60)^2))
By calculating the square root in the denominator and further simplifications, we can determine the rate at which the distance between the ships is changing at 4 PM.