Question

A rocket is being launched. After a while, it rises at a constant velocity of 40 miles per second. An observer is standing 55 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 48 miles? Present your answer as an approximation, accurate up to three or more decimals.

Answer 1

distance between the rocket and the observer is changing at a rate of approximately 3904.874 miles per second when the rocket is at a height of 48 miles.

Step-by-step explanation:

To find the rate at which the distance between the rocket and the observer is changing, we can use the concept of related rates. Let's denote the distance between the rocket and the observer as D(t) at time t, where t is measured in seconds, and the height of the rocket as h(t) at time t.

From the given information, we know that the rocket is rising at a constant velocity of 40 miles per second. Therefore, we have:

dh(t)/dt = 40 miles/second.

We also know that the observer is standing 55 miles from the launching site. So, the distance between the rocket and the observer can be expressed as:

D(t) = √(h(t)^2 + 55^2).

To find how fast D(t) is changing with respect to time, we differentiate both sides of the equation with respect to t:

d(D(t))/dt = d(√(h(t)^2 + 55^2))/dt.

Using the chain rule, we have:

d(D(t))/dt = (1/2) * (h(t)^2 + 55^2)^(-1/2) * 2 * h(t) * dh(t)/dt.

Substituting the given values into the equation, we have:

d(D(t))/dt = (1/2) * (√(h(t)^2 + 55^2)) * 2 * h(t) * 40.

We are given that the height of the rocket is 48 miles. Substituting h(t) = 48 into the equation, we have:

d(D(t))/dt = (1/2) * (√(48^2 + 55^2)) * 2 * 48 * 40.

Calculating the expression, we find:

d(D(t))/dt ≈ 3904.874.

Therefore, the distance between the rocket and the observer is changing at a rate of approximately 3904.874 miles per second when the rocket is at a height of 48 miles.