A rocket is being launched. After a while, it rises at a constant velocity of 15 miles per second. An observer is standing 84 miles from the launching site. How fast is the distance between the rocket and the observer changing when the rocket is at a height of 13 miles? Present your answer as an approximation, accurate up to three or more decimals.
Answer
Final Answer: The rate at which the distance between the rocket and the observer is changing when the rocket is at a height of 13 miles is approximately \(\boxed{2.294}\) miles per second.
Step 1 ：We are given a problem of related rates in calculus. We can model the situation as a right triangle, where the observer and the launching site form the base, the rocket's height is the perpendicular height, and the distance between the rocket and the observer is the hypotenuse.
Step 2 ：The rocket's height (h) is increasing at a constant rate of 15 miles per second (\(\frac{dh}{dt} = 15\)), and we are asked to find the rate at which the distance (d) between the rocket and the observer is changing (\(\frac{dd}{dt}\)) when h = 13.
Step 3 ：We can use the Pythagorean theorem to relate the sides of the triangle: \(d² = h² + 84²\). Differentiating both sides with respect to time (t), we get \(2d*(\frac{dd}{dt}) = 2h*(\frac{dh}{dt})\). We can solve this equation for \(\frac{dd}{dt}\), which is what we're asked to find.
Step 4 ：Substituting the given values into the equation: h = 13, \(\frac{dh}{dt} = 15\), base = 84, d = 85.0, we find that \(\frac{dd}{dt} = 2.2941176470588234\).
Step 5 ：Final Answer: The rate at which the distance between the rocket and the observer is changing when the rocket is at a height of 13 miles is approximately \(\boxed{2.294}\) miles per second.
