Final answer:
When the train is 2 miles away from the camera, the distance between the train and the camera is increasing at a rate of 3/sqrt(5) miles per minute.
Step-by-step explanation:
Rate of Change in Distance From the Camera to the Train
Let's visualize the scenario where a train is moving away from a camera on a straight track at 1.5 miles per minute, and we want to find out how fast the distance from the camera to the train is changing when the train is 2 miles away from the camera. To solve this problem, we can apply the Pythagorean theorem to a right triangle formed by the camera, the point on the track closest to the camera, and the train's position when it is 2 miles away from the camera.
The distance of the camera from the track is 1 mile, and the distance of the train from the point of closest approach is now 2 miles. Using derivatives and the Pythagorean theorem, we get:
- Let x be the distance of the train from the point on the track closest to the camera.
- Let y be the distance from the camera to the train.
- Pythagorean theorem: x^2 + 1^2 = y^2.
- When x is 2 miles, y can be found by calculating y = sqrt(2^2 + 1^2) = sqrt(5), so the distance is sqrt(5) miles.
- We differentiate both sides of the equation with respect to time t to find dy/dt when x is 2 miles.
- 2x(dx/dt) = 2y(dy/dt) after differentiation.
- Solving for dy/dt when x = 2 miles and dx/dt = 1.5 miles/minute, we find that dy/dt = (2 * 2 * 1.5) / (2 * sqrt(5)) = 3/sqrt(5) miles/minute, which is the rate at which the distance from the camera to the train is changing.
Therefore, when the train is 2 miles away from the camera, the distance between them is increasing at a rate of 3/sqrt(5) miles per minute.