tanTheta= h/1 where h is 3ft(at t=0)
d/dt (tanTheta)=1/1ft*dh/dt
sec^2 (Theta) dTheta/dt=dh/dt
Now, at t=2, tanTheta=10ft/1=10
sec^2 Theta(dTheta/dt)=1/ft*dh/dt
dh/dt=5,
so at t=2, then secTheta=1/sqrt(1^2+10^2)=1/sqrt101
sec^2 theta=1/101
so dTheta/dt=5*101 rad/sec
yep, the balloon is lost.
check my work
(A poignant problem) A 3-foot-tall observer holds a balloon 1 foot in front of her face. Experimentally, the observer releases the balloon and watches as it rises vertically at 5 feet per second. 2 seconds after releasing the the balloon, the observer realizes that the balloon is gone forever. Find the rate of change of the angle of elevation of the observer's head at this moment.
I'm really having problems setting the example up, can you please list and label the variables also.
1 answer
1 answer