At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 15 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.)
At a time of t hours after noon
Put the ships on a grid, A along the x-axis and B along the y-axis
In my diagram I have a right-angled triangle with AB the hypotenuse, AO is 15t + 10 and BO = 15t
AB^2 = (15t)^2 + (15t+10)^2 = 450t^2 + 300t + 100
2 AB d(AB)/dt = 900t + 300
when t = 3 (3:00 pm)
AB^2 = 450(9) + 300(3) + 100 = 5050
AB=√5050
d(AB)/dt = (900t+300)/(2AB) = (450t + 150)?AB
= (450(3) + 150)/√5050 = appr 21.11 knots
after x hours, the distance y between the ships is
y^2 = (10+15x)^2 + (15x)^2
at 3 pm, x=3, so
y^2 = 55^2 + 45^2
y = √5050 = 71.06
2y dy/dt = 2(10+15x)(15) + 2(15x)(15) = 300(3x+1)
when x=3,
2(71.06) dy/dt = 3000
dy/dt = 3000/142.12 = 21.11 knots
To find the speed at which the distance between the ships is changing, we can use the concept of rates of change.
Let's break down the problem step-by-step:
Step 1: Find the position of each ship at 3 PM.
Since ship A is sailing west at 15 knots for 3 hours from noon to 3 PM, it would have traveled a distance of 15 knots/hour * 3 hours = 45 nautical miles west.
So, the final position of ship A at 3 PM would be 10 nautical miles (noon position) - 45 nautical miles (distance traveled west) = -35 nautical miles west.
Since ship B is sailing north at 15 knots for 3 hours, it would have traveled a distance of 15 knots/hour * 3 hours = 45 nautical miles north.
So, the final position of ship B at 3 PM would be 45 nautical miles (distance traveled north).
Step 2: Find the distance between the ships at 3 PM.
The distance between the ships can be calculated using the Pythagorean theorem, as they form a right-angled triangle.
Using the positions of ship A and ship B at 3 PM, we have:
Distance^2 = (-35 nautical miles)^2 + (45 nautical miles)^2
Distance^2 = 1225 nautical miles^2 + 2025 nautical miles^2
Distance^2 = 3250 nautical miles^2
Distance ≈ 57.0088 nautical miles
Step 3: Find the rate at which the distance between the ships is changing at 3 PM.
To find this rate, we can differentiate the distance equation with respect to time (t).
Using the chain rule, we have:
2 * Distance * d(Distance)/dt = 2 * (-35 nautical miles) * (-35 knots) + 2 * (45 nautical miles) * (15 knots)
d(Distance)/dt = (1225 nautical miles/1 hour) + (1350 nautical miles/1 hour)
d(Distance)/dt = 2575 nautical miles/1 hour
Therefore, the speed at which the distance between the ships is changing at 3 PM is 2575 knots/hour.
To find the speed at which the distance between the two ships is changing at 3 PM, we need to find the rate of change of the distance between them with respect to time.
Let's break down the problem:
Given:
- Ship A is located 10 nautical miles due west of ship B at noon.
- Ship A is sailing west at a speed of 15 knots, which means it covers a distance of 15 nautical miles every hour.
- Ship B is sailing north at a speed of 15 knots, which means it covers a distance of 15 nautical miles every hour.
We want to find the rate of change of the distance between the two ships at 3 PM.
To solve this problem, we can use the concept of relative velocity. At any given time, the velocity between two objects can be calculated by taking the difference in their velocities along each direction.
Let's consider a snapshot at 3 PM. By that time, 3 hours will have passed since noon.
- Ship A will have traveled 3 hours * 15 knots = 45 nautical miles west from its original position.
- Ship B will have traveled 3 hours * 15 knots = 45 nautical miles north from its original position.
Now, we have a right-angled triangle formed by the two ships and the distance between them. The distance (D) between the ships is the hypotenuse of this triangle.
By using the Pythagorean theorem, we can express D in terms of the distance traveled by each ship:
D^2 = (45)^2 + (45)^2
D^2 = 2025 + 2025
D^2 = 4050
D ≈ 63.64 nautical miles
Now, to find how fast the distance (D) is changing, we need to differentiate the equation D^2 = 4050 with respect to time (t) using implicit differentiation.
2D * dD/dt = 0
Now, substitute the known values:
2(63.64) * dD/dt = 0
Simplify the equation:
127.28 * dD/dt = 0
Now, solve for dD/dt (the rate at which the distance is changing):
dD/dt = 0 / 127.28
dD/dt = 0 knots
Therefore, at 3 PM, the speed at which the distance between the ships is changing is 0 knots.