Question

A balloon is rising vertically above a​ level, straight road at a constant rate of 2 ft divided by sec. Just when the balloon is 84 ft above the​ ground, a bicycle moving at a constant rate of 12 ft divided by sec passes under it. How fast is the distance s (t )between the bicycle and balloon increasing 6 seconds​ later? A coordinate plane has a horizontal x-axis and a vertical y-axis. The angle between the axes is marked with a small square. The location of a balloon on the vertical axis is labeled y(t). An arrow pointing upward lies on the y-axis just below the point y(t). The location of a bicycle on the horizontal axis is labeled x(t). An arrow pointing to the right lies on the x-axis to the right of the point x(t). A segment falling from left to right begins at the balloon and ends at the bicycle and is labeled s(t).

Answer 1

Answer:

Step-by-step explanation:

At time t seconds after the bike passed, the distance z between the balloon and the bike is

z^2 = (84+2t)^2 + (12t)^2

So, at t=6,

z^2 = (84+2*6)^2 + (12*6)^2 = 14400

So, z = 120

2z dz/dt = 2(84+2t)*2 + 2(12t)*12

At time t=6,

240 dz/dt = 2(84+2*6)*2 + 2(12*6)*12 = 2112

dz/dt = 2112/240 = 44/5 ft/s