A train is heading due west from St. Louis. At noon, a plane flying horizontally due north at a fixed altitude of 4 miles passes directly over the train. When the train has

traveled another mile, it is going 80 mph, and the plane has traveled another 5 miles and is going 500 mph. At that moment, how fast is the distance between the train and the plane increasing?

Is the answer 105 mph.

Measuring time (t) in hours from that moment, the vector distance R between train and plane at later times is

R = (1+80t) i + (5+ 500t) j + 4 k
where i, j and k are unit vectors west, north and vertical.

dR/dt = 80 i + 500 j

The magnitude of that rate of change is
sqrt[(80)^2 + (500)^2] = 506 mph.

This could also have been solved simply by taking the magnitude of the relative velocity vector, 80 i + 500 j mph

To find the rate at which the distance between the train and the plane is increasing, we can use the concept of related rates. Let's break down the problem into smaller parts.

1. Firstly, let's consider the horizontal motion of the train and the plane separately. Since the train is traveling due west and the plane is flying due north, their horizontal motions are independent of each other. Therefore, we can focus on their vertical motions and find the rate of change of their distances from a fixed point, say the starting point of the train.

2. Let's denote the horizontal distance traveled by the train as x (in miles) and the horizontal distance traveled by the plane as y (in miles).

3. Given that the train is traveling at 80 mph, we know that dx/dt (the rate of change of x with respect to time) is 80 mph.

4. Also, given that the plane is flying at 500 mph, dy/dt (the rate of change of y with respect to time) is 500 mph.

5. Now, let's find an expression for the distance between the train and the plane at any given time t. Let D represent this distance.

Using the Pythagorean theorem, we have:
D^2 = x^2 + y^2

6. Differentiating both sides of the equation implicitly with respect to time t, we get:
2D * dD/dt = 2x * dx/dt + 2y * dy/dt

7. Rearranging the equation, we get:
dD/dt = (x * dx/dt + y * dy/dt) / D

8. We are interested in finding the rate at which the distance D changes when x has increased by 1 mile. So, we need to find the values of x, y, and D when x has increased by 1 mile.

At the given moment, when the train has traveled another mile, x = 1 mile, and y = 5 miles (since the plane has traveled another 5 miles).

9. To find D, we substitute the values of x and y into the equation D^2 = x^2 + y^2:
D^2 = 1^2 + 5^2
D^2 = 1 + 25
D^2 = 26
D ≈ √26 miles

10. Now, we can substitute the values of x, dx/dt, y, and dy/dt into the equation for dD/dt to find the rate at which D is changing at that moment:
dD/dt = (1 * 80 + 5 * 500) / √26 ≈ 105.77 mph

Therefore, the rate at which the distance between the train and the plane is increasing at the given moment is approximately 105.77 mph, not exactly 105 mph as you mentioned.