A balloon is rising at a constant speed of 5 ft/sec....?

Question

A balloon is rising at a constant speed of 5ft/sec. A boy is cycling along a straight road at a speed of 15 ft/sec. when he passes under the balloon, it is 45 ft above him. How fast is the distance between the boy and the balloon increasing 3 seconds later?

A balloon is rising at a constant speed of 5ft/sec. A boy is cycling along a straight road at a speed of 15 ft/sec. when he passes under the balloon, it is 45 ft above him. How fast is the distance between the boy and the balloon increasing 3 seconds later?


Can anyone help me at least get started?

Answer 1

At t=3, the height of the balloon is 60 ft and
y(t)=460+5*t

The boy is at x=45 ft and
x(t)=45+15*t

the distance is
r=sqrt(x^2+y^2)
and the rate of change of the distance is
dr/dt

r=sqrt((60+5*t)^2+(45+15*t)^2)
r=sqrt(5625+1950*t+250*t^2)

dr/dt=(1950+500*t)/(2*sqrt(5625+1950*t+250*t^2))
at t=3
dr/dt=14.7 ft/s

j

Answer 2

hey doofus 14.7 isnt the answer....
you know at t=3, x=45 and y=60 so z=75.
then xdx/dt+ydy/dt=zdz/dt
45(15)+60(5)=75 dz/dt
13=dz/dt