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 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 5.65 miles away.

Something is not right with this question.

Your train is going 500 mph ?????

I think you want that to be the speed of the plane. How fast is the train going?

your answer is a "distance", it asked "how fast is the distance between them increasing" which is velocity.

Please check the wording of this great Calculus question.

To determine how fast the distance between the train and the plane is increasing at that moment, we need to apply some basic principles of geometry and related rates.

Let's set up a coordinate system, with the initial position of the train being the origin (0,0), and the plane passing over it at time t=0.

Let x represent the horizontal distance traveled by the train (in miles) and y represent the vertical distance of the plane from the origin (in miles). Then, the position of the plane at time t can be described by the equation y = 4t.

To find the distance between the train and the plane at any given time, we can use the distance formula for two points in a 2D plane.

The distance, D, between the train and the plane is given by:

D = sqrt(x^2 + y^2)

Substituting the equation for y, we have:

D = sqrt(x^2 + (4t)^2)

Now, let's differentiate the equation with respect to time (t) to determine how the distance is changing:

dD/dt = (1/2)*(x^2 + (4t)^2)^(-1/2)*(2x*dx/dt + 2*(4t)*(4*dt/dt))
dD/dt = (1/2)*(x^2 + 16t^2)^(-1/2)*(2x*dx/dt + 32t)
dD/dt = (1/sqrt(x^2 + 16t^2))*(x*dx/dt + 16t)

We are given that dx/dt (the speed of the train) is 500 mph. At the moment in question, x = 5 miles.

dD/dt = (1/sqrt(25 + 16t^2))*(5*500 + 16t)

To find the rate at the moment when x = 5 miles, we need to determine the value of t at that time.
Given that the train is traveling at 500 mph, after traveling an additional 5 miles, it takes t = 0.01 hours = 0.01*60 minutes = 0.6 minutes.

Now, let's calculate the rate:

dD/dt = (1/sqrt(25 + 16(0.01)^2))*(5*500 + 16(0.01))
dD/dt = (1/sqrt(25 + 0.025))*(2500 + 0.16)
dD/dt = (1/sqrt(25.025))*(2500.16)
dD/dt = (1/5.002)*2500.16
dD/dt = 499.975

Therefore, at the moment when the train has traveled another 5 miles and is moving at 500 mph, the distance between the train and the plane is increasing at a rate of approximately 499.975 miles per hour.

Thus, the answer is not 5.65 miles away, but rather the distance is increasing at a rate of approximately 499.975 miles per hour.