Related Rates
I am having trouble finding the equation to use to fit this all together.
At noon, ship A is 100km west of ship B. Ship A is sailing south at 35 km/s and ship B is sailing north at 25 km/s. How fast is the distance between the ships changing at 4:00pm.
To find the equation that relates the changing distance between the ships and their speeds, we can use the Pythagorean theorem.
Let's assume that the distance between the ships at any given time is represented by the variable "d." Since Ship A is sailing south and Ship B is sailing north, the distance between them can be visualized as a right-angled triangle, with d as the hypotenuse.
Since Ship A is 100 km west of Ship B, the horizontal distance between them can be represented as 100 km.
Let's denote the time at noon as t = 0, and the time at 4:00 pm as t = 4. The time elapsed from noon to 4:00 pm would be 4 - 0 = 4 hours.
Now, we need to relate the changing distance between the ships (dd/dt) with the speeds of Ship A and Ship B. At any given time, the change in distance (dd/dt) will be equal to the rate of change of the vertical distance (dy/dt) minus the rate of change of the horizontal distance (dx/dt).
To find dy/dt and dx/dt, we can use the following relations:
dy/dt = rate of change of distance covered by Ship A = 35 km/s
dx/dt = rate of change of distance covered by Ship B = -25 km/s (since it is sailing towards the north, which is considered negative in this context)
Now we can write the equation relating dd/dt, dy/dt, and dx/dt using the Pythagorean theorem:
(d/dt)^2 = (dy/dt)^2 + (dx/dt)^2
Differentiating both sides of the equation with respect to time t:
2(d/dt)(dd/dt) = 2(dy/dt)(d(dy/dt)/dt) + 2(dx/dt)(d(dx/dt)/dt)
Since we know the values of dy/dt, dx/dt, and d(dy/dt)/dt = 0 because the speed is constant, we can substitute them into the equation:
2(d/dt)(dd/dt) = 2(35 km/s)(0) + 2(-25 km/s)(0)
Simplifying further:
2(d/dt)(dd/dt) = 2(0) + 2(0)
2(d/dt)(dd/dt) = 0
From this equation, we can deduce that the rate of change of the distance between the ships (dd/dt) is 0. Therefore, at 4:00 pm, the distance between the ships is not changing.
So, at 4:00 pm, the rate of change of the distance between the ships is 0 km/s.
To solve this problem, we need to find the rate at which the distance between the two ships is changing. Let's call this rate "d". To do this, we can use the concept of related rates.
Let's break down the problem step by step:
1. Start by visualizing the scenario. At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km/s, and ship B is sailing north at 25 km/s. We want to find the rate at which the distance between the ships is changing at 4:00 pm.
2. Let's assume that ship A is moving along the x-axis, and ship B is moving along the y-axis. We can create a coordinate system where the position of ship A at any given time t is given by (x(t), 0), and the position of ship B is given by (0, y(t)).
3. We can use the Pythagorean theorem to find the distance between the two ships at any given time t. The distance, which we'll call "D," is given by:
D^2 = (x(t) - 0)^2 + (0 - y(t))^2
4. To relate the position of each ship to its respective rate, we can take the derivative of both sides of the equation with respect to time:
2D * dD/dt = 2(x(t) * dx/dt - y(t) * dy/dt)
Simplifying, we get:
D * dD/dt = x(t) * dx/dt - y(t) * dy/dt
5. We're given the speeds of the ships, which we can substitute into the equation. At any given time t:
dx/dt = 35 km/s (positive because ship A is moving south)
dy/dt = -25 km/s (negative because ship B is moving north)
6. We also know that at 4:00 pm, which is 4 hours after noon, t = 4 hours. We need to find the values of x(t) and y(t) at that time.
Given that ship A is moving south at a constant rate of 35 km/s, at 4:00 pm, it would have traveled a total distance of:
x(4) = 35 km/s * 4 hours = 140 km
Ship B, on the other hand, is moving north at 25 km/s, so at 4:00 pm, it would have traveled a total distance of:
y(4) = -25 km/s * 4 hours = -100 km
7. Plugging all the values into the equation, we can solve for dD/dt:
D * dD/dt = x(t) * dx/dt - y(t) * dy/dt
D * dD/dt = (140 km) * (35 km/s) - (-100 km) * (25 km/s)
Note: Don't forget to use consistent units (km and km/s).
Simplifying:
D * dD/dt = 4900 km^2/s - (-2500 km^2/s)
D * dD/dt = 4900 km^2/s + 2500 km^2/s
D * dD/dt = 7400 km^2/s
8. Finally, we need to find D, the distance between the ships at 4:00 pm. Using the Pythagorean theorem equation:
D^2 = (x(4) - 0)^2 + (0 - y(4))^2
Plugging in the values:
D^2 = (140 km)^2 + (-100 km)^2
D^2 = 19600 km^2 + 10000 km^2
D^2 = 29600 km^2
Taking the square root of both sides:
D ≈ √29600 km
D ≈ 172.05 km
9. Now that we have the value of D, we can find dD/dt by dividing our previous result by D:
dD/dt = (7400 km^2/s) / (172.05 km)
dD/dt ≈ 43 km/s
Therefore, the distance between the ships is changing at a rate of approximately 43 km/s at 4:00 pm.