At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 20 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.)
At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 21 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.)
To find how fast the distance between the ships is changing at 4 PM, we can use the concept of rates of change and apply the Pythagorean theorem.
Let's break down the information given:
- Ship A is 10 nautical miles due west of ship B at noon.
- Ship A is sailing west at a speed of 17 knots.
- Ship B is sailing north at a speed of 20 knots.
To calculate the position of the ships at 4 PM, we need to find how far each ship has traveled in their respective directions. Since we know the speed and time, we can use the formula:
Distance = Speed x Time
For Ship A:
The distance traveled by Ship A = Speed of Ship A x Time
= 17 knots x 4 hours
= 68 nautical miles
Since Ship A is traveling west, it has moved 68 nautical miles due west of its original position.
For Ship B:
The distance traveled by Ship B = Speed of Ship B x Time
= 20 knots x 4 hours
= 80 nautical miles
Since Ship B is traveling north, it has moved 80 nautical miles due north.
Now, we can use the Pythagorean theorem to calculate the distance between the two ships at 4 PM:
Distance between the ships = √((distance traveled by Ship A)^2 + (distance traveled by Ship B)^2)
= √((68)^2 + (80)^2)
= √((4624) + (6400))
= √(11024)
= 104.9 nautical miles (rounded to one decimal place)
To find how fast the distance is changing at 4 PM, we need to calculate the rate of change, which is the derivative of the distance with respect to time.
Rate of change = d(Distance between the ships)/d(Time)
To differentiate the Pythagorean theorem, we use the chain rule:
d(Distance between the ships)/d(Time) = (2*(68)*(d(68)/d(Time))) + (2*(80)*(d(80)/d(Time)))
= 2*(68)*(d(68)/d(Time)) + 2*(80)*(d(80)/d(Time))
Now let's calculate the derivatives:
d(68)/d(Time) = 0
d(80)/d(Time) = 0
Therefore, the rate of change = 2*(68)*(0) + 2*(80)*(0)
= 0
Hence, at 4 PM, the distance between the ships is not changing; it remains constant at 104.9 nautical miles.