A balloon is rising vertically above a level, straight road at a constant rate of 5 ft divided by sec. Just when the balloon is 106 ft above the ground, a bicycle moving at a constant rate of 17 ft divided by sec passes under it. How fast is the distance s (t) between the bicycle and balloon increasing 6 seconds later?

db = 14.2 ft/sec

Explanation:

from the question we are given the following values:

height of the balloon (h) = 106 ft

rate of increase in height (speed) = 5ft/sec

rate of movement of the bicycle (speed) = 17 ft/sec

tine (t) = 6 sec

The height of the balloon, the horizontal distance of the bicycle and the distance of the balloon from the bicycle after three seconds all form three sides of a triangle The distance between the balloon and the bicycle is the hypotenuse (b) The horizontal distance is the adjacent side (a) The height of the balloon is the opposite side (h)

Therefore

b^{2} = a^{2} + h^{2} ... equation 1

the height (h) after 6 seconds becomes = 106 + (5 x 6) = 136 ft

the horizontal distance (a) after 6 seconds = 17 x 6 = 102 ft

therefore

b^{2} = 102^{2} + 136^{2}

b = 170 ft

differentiating equation 1 in terms of time it becomes

2b/frac{db}[dt} = 2a/frac{da}[dt} + 2h/frac{dh}[dt}

where da = rate of change of horizontal distance

db = rate of change of vertical height

2 x 170 x / frac{db}[dt} = (2 x 102 x / frac{17}[dt}) + (2 x 136 x / frac{5}[dt})

2 x 170 x db = (2 x 102 x 17) + (2 x 136 x 5)

db = 14.2 ft/sec