Derivatives - chain rule

An airplane, flying horizontally at an altitude of 1 mile, passes directly over an observer. If the constant speed of the airplane is 400 miles per hour, how fast is its distance from the observer increasing 45 seconds later? Hint: note that in 45 seconds (3/4 * 1/60 = 1/80), the airplane goes 5 miles.

the distance z is found using

z^2 = 1^2 + x^2
at x=5, z^2 = 26

z dz/dt = x dx/dt

so at the moment in question

√26 dz/dt = 5*400

note that z will be in mi/hr

To solve this problem, we can use the chain rule in calculus to find the rate of change of the distance between the observer and the airplane.

Let's define the following variables:
- y: The distance between the observer and the airplane (in miles)
- t: Time (in seconds)

We are given that the airplane is flying horizontally at a constant speed of 400 miles per hour. Therefore, the rate at which the airplane is changing its position along the x-axis (horizontal distance) is 400 miles per hour.

We want to find how fast the distance between the observer and the airplane (y) is changing, which can be represented as dy/dt.

Given that dy/dt represents the rate of change of y (distance) with respect to t (time), we need to find dy/dt when t = 45 seconds.

Now, let's derive an equation that connects y and t to help us find dy/dt using the chain rule.

Since the airplane is flying horizontally, we can use the Pythagorean theorem to relate y and t:

y^2 = x^2 + h^2,

where h represents the altitude of the airplane (1 mile) and x represents the horizontal distance traveled by the airplane, which is changing with time.

Differentiating both sides of the equation with respect to t (time), we get:

2y * (dy/dt) = 2x * (dx/dt) + 0,

since h is constant (its derivative with respect to time is 0).

Now, let's substitute the given values into the equation to find dx/dt when t = 45 seconds:

When t = 0, the airplane is directly above the observer. Hence, x = 0.
When t = 45 seconds, the airplane has traveled 5 miles horizontally, as given in the hint.

So, when t = 45 seconds, x = 5 miles.

Now, substituting these values into the equation, we have:

2 * y * (dy/dt) = 2 * 5 * (dx/dt) + 0,
2 * y * (dy/dt) = 10 * (dx/dt).

Since the airplane is flying horizontally at a constant speed, dx/dt = 400 miles per hour.

Substituting this value into the equation, we get:

2 * y * (dy/dt) = 10 * 400,
2 * y * (dy/dt) = 4000.

Now, solving for dy/dt:

dy/dt = 4000 / (2 * y).

We are asked to find dy/dt when t = 45 seconds, so we need to find y when t = 45 seconds. Since the airplane is flying horizontally, y will remain constant at 1 mile (altitude of the airplane) throughout the flight.

Therefore, substituting y = 1 mile into the equation:

dy/dt = 4000 / (2 * 1),
dy/dt = 2000 miles per hour.

Hence, the distance between the observer and the airplane is increasing at a rate of 2000 miles per hour 45 seconds later.