At noon, ship A is 130 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

To find the speed at which the distance between the ships is changing at 4:00 PM, we can use the concept of relative motion.

Given:
- Ship A is initially 130 km west of Ship B.
- Ship A is sailing east at 30 km/h.
- Ship B is sailing north at 20 km/h.

Let's break down the problem into two components: one along the horizontal axis (east-west direction) and the other along the vertical axis (north-south direction).

1. Horizontal Component:
Since Ship A is sailing east at 30 km/h, the horizontal distance between Ship A and Ship B decreases by 30 km/h.

2. Vertical Component:
Since Ship B is sailing north at 20 km/h, the vertical distance between Ship A and Ship B increases by 20 km/h.

Since the horizontal and vertical components are perpendicular to each other, we can use the Pythagorean theorem to find the distance between the ships.

At any given time, the distance D between Ship A and Ship B can be represented as:

D^2 = (130 - 30t)^2 + (20t)^2

Now, let's differentiate both sides of the equation with respect to time t to find the rate at which the distance is changing over time.

2D * dD/dt = 2(130 - 30t)(-30) + 2(20t)(20)

Simplifying:

2D * dD/dt = -60(130 - 30t) + 40t^2

At 4:00 PM, t = 4. Substitute this value into the equation:

2D * dD/dt = -60(130 - 30(4)) + 40(4)^2

2D * dD/dt = -60(130 - 120) + 40(16)

2D * dD/dt = -60(10) + 640

2D * dD/dt = -600 + 640

2D * dD/dt = 40

Since we are interested in finding dD/dt, the rate at which the distance is changing, we divide both sides of the equation by 2D:

dD/dt = 40 / 2D

Now substitute the distance D at 4:00 PM into the equation. To find this, we need to calculate the distance using the given information:

D = sqrt((130 - 30t)^2 + (20t)^2)

D = sqrt((130 - 30(4))^2 + (20(4))^2)

D = sqrt(50^2 + 80^2)

D = sqrt(2500 + 6400)

D = sqrt(8900)

D = 94.34 km (approx.)

Now substitute D into the equation:

dD/dt = 40 / 2(94.34)

dD/dt = 40 / 188.68

dD/dt ≈ 0.212 km/h

Therefore, the distance between the ships is changing at a rate of approximately 0.212 km/h at 4:00 PM.

To find the rate at which the distance between the ships is changing, we need to use the concept of related rates. Let's break down the problem into different components.

Let x represents the horizontal distance ship A has traveled (east), and y represents the vertical distance ship B has traveled (north). At noon, ship A is 130 km west of ship B, so x = -130 km.

Given:
dx/dt = 30 km/h (Rate at which ship A is traveling east)
dy/dt = 20 km/h (Rate at which ship B is traveling north)

To find the rate at which the distance between the ships is changing (dz/dt), we need to find dz/dt when t = 4:00 PM.

We can use the Pythagorean theorem to relate x, y, and z:
z^2 = x^2 + y^2

Differentiating both sides with respect to time (t):
2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt)

We want to find dz/dt when t = 4:00 PM, which means we want to find z, x, and y at that time.

At 4:00 PM, t = 4 hours since noon. Since ship A is sailing east at a constant speed of 30 km/h, the horizontal distance traveled by ship A is:
x = dx/dt * t = 30 km/h * 4 hours = 120 km

Since ship B is sailing north at a constant speed of 20 km/h, the vertical distance traveled by ship B is:
y = dy/dt * t = 20 km/h * 4 hours = 80 km

Using the Pythagorean theorem, we can find the distance between the ships at 4:00 PM:
z^2 = x^2 + y^2
z^2 = (120 km)^2 + (80 km)^2
z^2 = 14400 km^2 + 6400 km^2
z^2 = 20800 km^2
z ≈ 144.22 km

Now, we can substitute the values we have into the equation we derived earlier to find dz/dt:
2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt)
2 * 144.22 km * (dz/dt) = 2 * 120 km * 30 km/h + 2 * 80 km * 20 km/h

Simplifying further:
288.44 km * (dz/dt) = 7200 km * h + 3200 km * h
288.44 km * (dz/dt) = 10400 km * h

Solving for dz/dt:
(dz/dt) = 10400 km * h / 288.44 km
(dz/dt) ≈ 36.01 km/h

Therefore, the distance between the ships is changing at a rate of approximately 36.01 km/h at 4:00 PM.

well, you know the distance z at time t is

z^2 = (130-30t)^2 + (20t)^2
= 1300t^2 - 7800t + 16900
so, at 4:00, z = 10√145

z dz/dt = 2600t - 7800 = 2600(t-3)

Now plug in t=4