Question
A hot-air balloon is 120 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 30 mi/hr (44 ft/s). If the balloon rises vertically at a rate of 15 ft/s, what is the rate of change of the distance between the motorcycle and the balloon 6 seconds later?
Answers
as usual, draw a diagram. If the bike is x ft away, then the distance z is
z^2 = x^2 + 120^2
So that means that
z dz/dt = x dx/dt
at the moment in question, z^2 = (6*44)^2 + 120^2, so z = 290 ft
Now to find dz/dt,
290 dz/dt = (6*44) * 15
dz/dt = 396/29 ≈ 13.66 ft/s
z^2 = x^2 + 120^2
So that means that
z dz/dt = x dx/dt
at the moment in question, z^2 = (6*44)^2 + 120^2, so z = 290 ft
Now to find dz/dt,
290 dz/dt = (6*44) * 15
dz/dt = 396/29 ≈ 13.66 ft/s
it was 43.77
oops. do you see my mistake? I was considering the balloon at a constant height of 120 ft. So, in reality, at time t,
z^2 = (44t)^2 + (120+15t)^2 = 2161t^2 + 3600t + 14400
at t=6, we have z^2 = 113796
z =337.34 and
2z dz/dt = 4322t + 3600
674.68 dz/dt = 29532
dz/dt = 43.77
z^2 = (44t)^2 + (120+15t)^2 = 2161t^2 + 3600t + 14400
at t=6, we have z^2 = 113796
z =337.34 and
2z dz/dt = 4322t + 3600
674.68 dz/dt = 29532
dz/dt = 43.77