A​ hot-air balloon is 180 ft above the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes directly beneath it going 45 mi/hr ​(66 ft/s​). If the balloon rises vertically at a rate of 9 ft/s​, what is the rate of change of the distance between the motorcycle and the balloon 9 seconds ​later?

4 answers

at time t, the distance z is
z^2 = (180+9t)^2 + (66t)^2 = 4437t^2 + 3240t + 32400
z(9) = 648.81
2z dz/dt = 8874t + 3240
Now plug in z and t to find dz/dt
horizontal distance = x
height = h
z is the distance between,
right triangle so
z^2=h^2+x^2
2 z dz = 2 h dh + 2 x dx
and thus
z dz/dt = h dh/dt + x dx/dt
plug and chug
cool oobleck, never seen it done that way before :)
Charity was probably taught along these lines, same answer as oobleck

At a time of t seconds after the bike passed directly beneath the balloon,
let the height of the balloon be 180+9t ft
let the distance covered by the bike be 66t ft

you have a right-angled triangle.
Let the distance between balloon and bike be d ft (the hypotenuse)

(180+9t)^2 + (66t)^2 = d^2
differentiate with respect to t
2(180+9t)(9) + 2(66t)(66) = 2d dd/dt
9(180+9t) + 4356t = d dd/dt

when t = 9
(180+81)^2 + (594)^2 = d^2
d = √420957

into 9(180+9t) + 4356t = d dd/dt
9(180+81) + 4356(9) = √420957 dd/dt

solve for dd/dt, I got appr 64 ft/sec

check my work