At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?
To find how fast the distance between the ships is changing at 4:00 PM, we can use the concept of rates of change.
First, let's break down the problem:
- At noon, Ship A is 130 km west of Ship B.
- Ship A is sailing east at a speed of 25 km/h.
- Ship B is sailing north at a speed of 15 km/h.
- We need to find the rate at which the distance between the ships is changing at 4:00 PM.
To solve this problem, we can use the Pythagorean theorem and find an expression for the distance between the two ships at any given time. Then we can differentiate this expression with respect to time to find how the distance is changing.
Let:
- x be the distance traveled by Ship A (eastward) in km from noon.
- y be the distance traveled by Ship B (northward) in km from noon.
From the given information, we have:
- x = 25t (since Ship A is traveling east at a constant speed of 25 km/h)
- y = 15t (since Ship B is traveling north at a constant speed of 15 km/h)
Using the Pythagorean theorem, the distance between the two ships is given by:
D = sqrt((130 + x)^2 + y^2)
To find how the distance is changing at 4:00 PM, we need to differentiate D with respect to time:
dD/dt = d(sqrt((130 + x)^2 + y^2))/dt
Now let's differentiate:
dD/dt = [(130 + x)^2 + y^2]^-1/2 * (2(130 + x)dx/dt + 2ydy/dt)
We have:
- dx/dt = 25 km/h (since Ship A is traveling east at a constant speed of 25 km/h)
- dy/dt = 15 km/h (since Ship B is traveling north at a constant speed of 15 km/h)
Substituting the values:
dD/dt = [(130 + 25t)^2 + (15t)^2]^-1/2 * (2(130 + 25t)(25) + 2(15t)(15))
To find the rate of change at 4:00 PM, substitute t = 4 into the expression for dD/dt:
dD/dt = [(130 + 25(4))^2 + (15(4))^2]^-1/2 * (2(130 + 25(4))(25) + 2(15(4))(15))
Now calculate the value of dD/dt to find how fast the distance between the ships is changing at 4:00 PM.
To find how fast the distance between the ships is changing at 4:00 PM, we need to use the concept of rates and the Pythagorean theorem.
Step 1: Determine the position of each ship at 4:00 PM.
To find the position of ship A at 4:00 PM, we need to calculate how far it has traveled from noon to 4:00 PM.
Ship A's speed is 25 km/h, and from noon to 4:00 PM is a total of 4 hours. Therefore, ship A will have traveled 4 hours x 25 km/h = 100 km east.
Since ship A started 130 km west of ship B, at 4:00 PM, it will be 130 km - 100 km = 30 km west of ship B.
Ship B's position remains the same since it is sailing north, and its speed does not affect its east-west position.
So at 4:00 PM, ship A is 30 km west of ship B.
Step 2: Use the Pythagorean theorem to calculate the distance between the ships at 4:00 PM.
The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the distance between ship A and ship B at 4:00 PM is the hypotenuse, and the east-west distance is one side, while the north-south distance is the other side.
Using the Pythagorean theorem, we have:
(distance)^2 = (east-west distance)^2 + (north-south distance)^2
Let d be the distance between the ships at 4:00 PM.
Therefore:
d^2 = (30 km)^2 + (0 km)^2
d^2 = 900 km^2
d = √900 km
d = 30 km
So at 4:00 PM, the distance between the ships is 30 km.
Step 3: Calculate how fast the distance between the ships is changing at 4:00 PM.
We are given the speeds of each ship: ship A is sailing east at 25 km/h, and ship B is sailing north at 15 km/h.
To find how fast the distance is changing, we need to calculate the rate of change of the distance with respect to time. This is a derivative problem.
Let's assume t represents time in hours, and d represents the distance between the ships in kilometers at time t.
Using the Pythagorean theorem mentioned earlier, we have the equation:
d^2 = (30 km)^2 + (0 km)^2
d^2 = 900 km^2
Differentiating both sides of the equation with respect to t, we get:
2d * (dd/dt) = 0
2d * (dd/dt) = 0
Plugging in the known value of d = 30 km, we have:
2 * 30 km * (dd/dt) = 0
60 km * (dd/dt) = 0
(dd/dt) = 0 km/h
Therefore, the distance between the ships is not changing at 4:00 PM.