If the balloon continues to go up 10 feet per second, how fast is the distance between the motorcycle and the balloon changing 10 seconds later?

Question

A hot air balloon is 150 ft above the ground while a motorcycle passes directly beneath it heading North at a speed of 40 miles per hour also known as 58.67 feet per second. If the balloon continues to go up 10 feet per second, how fast is the distance between the motorcycle and the balloon changing 10 seconds later?

Answer 1

Let x be the distance from the starting point to the position of the motorcycle.

Let y be the vertical distance of the balloon.

After 10 seconds, x = 586.7 feet and y = 160 feet.

dx/dt = 58.67 f/s and dy/dt = 10 f/s

Let s be the distance between the balloon and the motorcycle.

s² = x²+y²

2s ds/dt = 2x dx/dt + 2y dy/dt

ds/dt = (x dx/dt + y dy/dt)/s

s = √(x²+y²)

ds/dt = (x dx/dt + y dy/dt)/√(x²+y²)

Just plug in the known values to get ds/dt.