26. 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?

Let's measure position from the ground location at noon, where the train was when the plane was overhead. It will make the math simpler but will not affect the answer. North position from there will have a "j" unit vector and west poeitiom from there will have a "-i" unit vector. The vertical unit vector is k.

At the time in question, the airplane position vector is
R2 = (5 + 500 t) j + 4 k
and the train position vector is
R1 = -(1 + 80 t)i

The separation vector is
D = R2 - R1 = (1 + 80 t)i + (5 + 500 t) j + 4 k
The rate of change of this vector is
dD/dt = 80 i + 500j mph
Take the square root of the sum of the squares for the rate the distance in increasing.
That would be 506.4 mph

Note that the altitude and the separation at that time do not matter, only the speeds.

To find how fast the distance between the train and the plane is increasing, we need to use the concept of related rates. We can use the Pythagorean theorem to represent the distance between the train and the plane at any given time. Let's denote the distance between them as "d" and the time as "t".

From the information given, we know that the train is traveling at a speed of 80 mph, and the plane is traveling at a speed of 500 mph. At the moment in question, the train has traveled 1 mile (because it traveled another mile when the plane traveled another 5 miles), and the plane has traveled 5 miles.

Using the Pythagorean theorem, the equation to represent the distance between the train and the plane is:

d^2 = (80t)^2 + (500t)^2

To find the rate at which the distance d is changing with respect to time, we can differentiate both sides of the equation with respect to time (t):

2d * dd/dt = 2(80t)(80) + 2(500t)(500)

Simplifying the equation, we have:

dd/dt = [(80t)(80) + (500t)(500)] / d

Now, we need to find the value of d. At the moment in question, the train has traveled another mile (so t = 1) and the plane has traveled another 5 miles. Using these values, we can solve for d:

d^2 = (80*1)^2 + (500*1)^2

d^2 = 6400 + 250000

d^2 = 256400

d ≈ 506.34 miles (rounded to two decimal places)

Now we can substitute this value of d back into the equation for dd/dt:

dd/dt = [(80*1)(80) + (500*1)(500)] / 506.34

dd/dt ≈ (6400 + 250000) / 506.34

dd/dt ≈ 313.35 mph (rounded to two decimal places)

Therefore, at that moment, the distance between the train and the plane is increasing at a rate of approximately 313.35 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 define:
x = distance traveled by the train (in miles)
y = distance traveled by the plane (in miles)
D = distance between the train and the plane (in miles)

Given:
The rate at which the train is traveling, dx/dt = 80 mph
The rate at which the plane is traveling, dy/dt = 500 mph
The altitude of the plane, which is the same as the height of the plane from the train, h = 4 miles

We need to find dD/dt, the rate at which the distance between the train and the plane is changing.

By Pythagoras' theorem, we know that:
D^2 = x^2 + (y - h)^2

Differentiating both sides of the equation with respect to time t, we get:
2D * dD/dt = 2x * dx/dt + 2(y - h) * dy/dt

Substituting the given values for dx/dt, dy/dt, and h, and the known values for x and y when the train has traveled another mile and the plane has traveled another 5 miles:
dx/dt = 80 mph
dy/dt = 500 mph
h = 4 miles
x = 1 mile
y = 5 miles

2 * D * dD/dt = 2 * 1 * 80 + 2 * (5 - 4) * 500

Simplifying this equation gives:
2 * D * dD/dt = 160 + 2 * 500

Now we can solve for dD/dt:
2 * D * dD/dt = 1160

Dividing both sides by 2D:
dD/dt = 1160 / (2 * D)

To find the exact value of dD/dt, we need to know the current value of D. Unfortunately, that information is not provided in the question.