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?
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 denote the distance between the train and the plane as 'd' (in miles) and the time as 't' (in hours).
First, let's find an equation that relates the distances traveled by the train and the plane to the distance 'd'.
Since the train is going due west and the plane is flying due north, we can consider the paths of the train and the plane as the legs of a right triangle. The distance 'd' represents the hypotenuse of this triangle.
By using the Pythagorean theorem, the equation for 'd' becomes:
d^2 = (x + 1)^2 + (5 + 4t)^2
where x is the distance traveled by the train (in miles).
Next, we need to differentiate both sides of this equation with respect to time 't' to find how 'd' is changing with respect to time. This gives us:
2d * (dd/dt) = 2(x + 1) * (dx/dt) + 2(5 + 4t) * (29)
Now, let's plug in the given values:
dx/dt = 80 mph (rate at which the train is traveling when the train has traveled another mile)
dt/dt = 500 mph (rate at which the plane is traveling when the plane has traveled another 5 miles)
x = 1 mile (distance traveled by the train when the train has traveled another mile)
Plugging these values into the equation above, we get:
2d * (dd/dt) = 2(1 + 1) * 80 + 2(5 + 4t) * (500)
Simplifying further:
dd/dt = (2(2) * 80 + 2(5 + 4t) * (500)) / (2d)
At the moment when the train has traveled another mile, x = 1. This means d = sqrt((1+1)^2 + (5+4t)^2) can be simplified to:
d = sqrt(4 + (5 + 4t)^2)
Plugging this value into the equation for dd/dt, we get:
dd/dt = (2(2) * 80 + 2(5 + 4t) * (500)) / (2 * sqrt(4 + (5 + 4t)^2))
The expression inside the square root can be simplified as:
dd/dt = (4 * 80 + (5 + 4t) * (500)) / sqrt(4 + (5 + 4t)^2)
Thus, to find the rate at which the distance between the train and the plane is increasing at that moment, we can substitute t = 0 (since it is not specified in the question) into the equation for dd/dt:
dd/dt = (4 * 80 + (5 + 4 * 0) * (500)) / sqrt(4 + (5 + 4 * 0)^2)
dd/dt = 320 + 5 * 500 / sqrt(4 + 5^2) = 320 + 2500 / sqrt(29)
Simplifying further, we get:
dd/dt = 320 + 2500 / sqrt(29)
So, at that moment, the distance between the train and the plane is increasing at a rate of 320 + 2500 / sqrt(29) miles per hour.