Alicia watches as a rocket is being launched from the Morgan Valley Space Center. After a while, the rocket rises at a constant velocity of 10 miles per second. Alicia is standing 45 miles from the launching site. How fast is the distance between the rocket and Alicia changing when the rocket is at a height of 28 miles? Round answer to at least 3 decimal places.
The distance is changing at a rate of:
Select an answer $\checkmark$

Step 1 ：We can model the situation as a right triangle, where the rocket's height is one leg, Alicia's distance from the launch site is the other leg, and the hypotenuse is the distance between Alicia and the rocket.

Step 2 ：We can use the Pythagorean theorem to relate these quantities: \(d^2 = h^2 + 45^2\), where d is the distance between Alicia and the rocket, and h is the height of the rocket.

Step 3 ：Differentiating both sides with respect to time t, we get: \(2d \frac{dd}{dt} = 2h \frac{dh}{dt}\).

Step 4 ：We know that \(\frac{dh}{dt}\) (the rate of change of the rocket's height) is 10 miles per second, and we want to find \(\frac{dd}{dt}\) (the rate of change of the distance between Alicia and the rocket) when h = 28 miles.

Step 5 ：We can solve for \(\frac{dd}{dt}\) in terms of d, h, and \(\frac{dh}{dt}\), and then substitute the given values to find the answer.

Step 6 ：Substituting the given values, we get: h = 28, dh_dt = 10, distance_from_launch = 45, d = 53.0, dd_dt = 5.283.

Step 7 ：Final Answer: The rate at which the distance between the rocket and Alicia is changing when the rocket is at a height of 28 miles is approximately \(\boxed{5.283}\) miles per second.