A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?

Answer: To determine how fast the distance between the balloon and the automobile is changing after 1 second, we can use the Pythagorean theorem to model the situation.

Let \( h \) be the height of the balloon above the ground, and let \( d \) be the distance of the car from a point on the ground directly beneath the balloon. Let \( s \) be the distance between the balloon and the car. After 1 second, the height of the balloon will be \( h + 15 \) ft because it is rising at 15 ft/sec, and the position of the car along the road will be \( d + 66 \) ft since it is traveling at 66 ft/sec.

Since the balloon and the car are moving along perpendicular paths, we have a right triangle where \( h \) and \( d \) are the legs, and \( s \) is the hypotenuse. The relationship between these is given by Pythagoras' theorem:

\( s^2 = h^2 + d^2 \).

We are interested in finding the rate of change of \( s \) with respect to time, which is \( ds/dt \). To do this, we differentiate both sides of the equation with respect to time \( t \):

\( 2s \cdot ds/dt = 2h \cdot dh/dt + 2d \cdot dd/dt \).

Now, we simplify by dividing every term by 2:

\( s \cdot ds/dt = h \cdot dh/dt + d \cdot dd/dt \).

We know the rates of change \( dh/dt \) (the rate at which the balloon's height is increasing) and \( dd/dt \) (the rate at which the car's distance from the point directly below the balloon on the ground is increasing):

\( dh/dt = 15 \) ft/sec, \( dd/dt = 66 \) ft/sec.

Initially, the balloon is 200 ft off the ground, so \( h = 200 \) ft, and at 1 second later \( d = 66 \) ft because the car has moved 66 feet along the road.

Plugging in these values:

\( s \cdot ds/dt = 200 \cdot 15 + 66 \cdot 66 \).

First, we need to find \( s \), the initial distance between the balloon and the car. Thus:

\( s^2 = 200^2 + 66^2 \), \( s^2 = 40000 + 4356 \), \( s^2 = 44356 \), \( s \approx 210.61 \) ft (taking the positive square root since distance is positive),

Now, substituting \( s \) and the rates of changes into the equation:

\( 210.61 \cdot ds/dt = 200 \cdot 15 + 66 \cdot 66 \), \( 210.61 \cdot ds/dt = 3000 + 4356 \), \( 210.61 \cdot ds/dt = 7356 \), \( ds/dt = 7356 / 210.61 \), \( ds/dt \approx 34.93 \) ft/sec.

The distance between the balloon and the automobile is increasing at an approximate rate of 34.93 feet per second after 1 second.