At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 24 knots and ship B is sailing north at 16 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
P is the starting point for ship B
after 3 hrs (3 PM)
B is north of P by ... 3 * 16 NM
A is west of P by ... [20 + (3 * 24)] NM
APB is a right triangle , hypot AB
BP^2 + AP^2 = AB^2
differentiating
... 2 BP dBP/dt + 2 AP dAP/dt = 2 AB dAB/DT
cancel the 2's , plug in known quantities
To find how fast the distance between the ships is changing at 3 PM, we need to use the concept of relative velocity.
First, let's find the position of each ship at 3 PM.
Since ship A is sailing west at a constant speed of 24 knots for 3 hours, it will have traveled a distance of 24 knots/hour * 3 hours = 72 nautical miles to the west. Therefore, ship A will be 20 nautical miles + 72 nautical miles = 92 nautical miles due west of its original position at 3 PM.
Similarly, ship B is sailing north at a constant speed of 16 knots for 3 hours, which means it will have traveled a distance of 16 knots/hour * 3 hours = 48 nautical miles to the north. Therefore, ship B will be 48 nautical miles north of its original position at 3 PM.
Now, to find the distance between the ships at 3 PM, we can use the Pythagorean theorem. The distance between two points in a plane is given by the formula:
distance = √((change in x)^2 + (change in y)^2)
In this case, the change in x (horizontal distance) is 92 nautical miles, and the change in y (vertical distance) is 48 nautical miles. So, the distance between the ships at 3 PM is:
distance = √((92)^2 + (48)^2) ≈ √(8464 + 2304) ≈ √10768 ≈ 103.77 nautical miles
Now, to find how fast the distance between the ships is changing at 3 PM, we need to find the rate of change of the distance with respect to time.
Let's assume time t represents hours past noon. So at t = 3, the rate of change of the distance between the ships can be found using the chain rule of differentiation:
d(distance)/dt = (∂distance/∂x) * (dx/dt) + (∂distance/∂y) * (dy/dt),
where
- (∂distance/∂x) is the partial derivative of the distance with respect to x (horizontal distance),
- (∂distance/∂y) is the partial derivative of the distance with respect to y (vertical distance),
- (dx/dt) is the rate of change of the x-coordinate of one ship with respect to time (ship A's velocity),
- (dy/dt) is the rate of change of the y-coordinate of the other ship with respect to time (ship B's velocity).
In this case, (∂distance/∂x) = change in x / distance = 92 / 103.77 ≈ 0.886,
(∂distance/∂y) = change in y / distance = 48 / 103.77 ≈ 0.462,
(dx/dt) = rate of change of x-coordinate = -24 knots (since ship A is traveling west, the x-coordinate decreases),
(dy/dt) = rate of change of y-coordinate = 16 knots (since ship B is traveling north, the y-coordinate increases).
Using these values, we can calculate the rate of change of the distance between the ships at 3 PM:
d(distance)/dt = (0.886) * (-24) + (0.462) * 16
≈ -21.26 + 7.39
≈ -13.87 knots
Therefore, the distance between the ships is changing at a rate of approximately 13.87 knots per hour at 3 PM.