Let's say a train is traveling at 0.8km/min along a straight track. A movie camera 0.5km away from the track

Question

is focused on the train.

The diagram that comes with the problem is a right triangle with the distance from the camera to the train as the hypotenuse. The question is:

How fast is the camera rotating (in radians/min) at the moment when the train is 1 km from the camera?

It isn't a trick question. The train and the camera are 0.5km away from each other. The train heads off at 0.8km/min whil the camera rotates to focus on it.

Direction the train is moving is always positive starting at 0.

Answer 1

Pythagreon theorum - a^2 + b^2 = c^2

a, distance from track to camera = 0.5 km
b, distance train has to travel until it is directly in front of camera
c, distance from train to camera = 1 km

So, b = Sqrt(1-.25) = 0.866 km

So, the angle (with this condition being t=0) is theta(t) = sin ((.866 km - 8 km/hr * t)/.5 km)

Now, unfortunately, you need calculus.

The derivative of this with respect to t is
theta'(t) = -1.6*cos(8t/5-.1732) (I did this with my calculator, I can't remember how to derivitize trig functions).

So, theta'(0) = -1.6 cos(-.1732) = -1.5761 rad/hr = -.0263 rad/min

Since it doesn't specify direction, the negative can be removed:

0.026 rad/min

Answer 2

I remember a similar problem from first year calculus, only that was a lighthouse offshore with its beam sweeping along a straight shore line and they gave to rotation rate and we had to find the speed of the beam on the shore line. I can't think exactly how to do it now, but you need to write an expression for the angle of the camera with position of the train and then differentiate. The expression would be something like TAN(A)=D/.5, where .5 is the distance of the camera from the track, A is the pointing angle of the camera and D is the distance of the train from the point directly opposite the camera, which is where the right angle of the triangle is.

Answer 3

Sounds like a trick question. The camera should be stationary viewing the train from 1km away...it will not rotate until the train starts to leave the frame much closer.

Answer 4

The hypotenuse is 5.064km.
I don't remember how to calculate radians.