Ahot air is attached to a spool of rope that is 125 feet away from the balloon when it is on the ground. the hot air balloon rises straight up in such a way that the length of rope increases at a rate of 15 ft/sec. how fast is the hot air balloon rising 20 seconds after it lifts off ?

Let the length of the rope =rheight of the balloon = hwe are given dr/dt = 15ft/secat any time t we know: r =125 + 15*tby pythagoras theorem we know: h^2 + 125^2 = r^2we have to find dh/dt at time t= 20secd/dt(h^2 + 125^2) =d/dt(r^2)by chain rule: d/dh(h^2 + 125^2)* dh/dt = d/dr(r^2)*dr/dt2h*dh/dt = 2r* dr/dth*dh/dt = r*dr/dtdh/dt = (r/h) * dr/dt = [(125 + 15*20)/(425^2 -125^2)^1/2]*15 > at time t= 20sec = 15.69 ft/sec