(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.

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

Sure! Let's set up the problem and label the variables.

Let's label the variables as follows:
- The height of the observer's head from the ground: h (in feet)
- The distance from the observer's head to the balloon: x (in feet)
- The angle of elevation of the observer's head: θ (in radians)

Based on the problem, we have the following information:
- The observer is 3 feet tall, so the height of the observer's head from the ground is h = 3 feet.
- The observer holds the balloon 1 foot in front of her face, so the distance from the observer's head to the balloon is x = 1 foot.
- The balloon rises vertically at a rate of 5 feet per second.

We need to find the rate of change of the angle of elevation of the observer's head 2 seconds after releasing the balloon, which is expressed as dθ/dt.

To find this rate, we can use the tangent function, since the tangent of the angle of elevation is equal to the ratio of the height of the observer's head to the distance from the observer's head to the balloon:

tan(θ) = h / x

Taking the derivative of both sides of the equation with respect to time (t), we get:

sec^2(θ) * dθ/dt = (dh/dt * x - h * dx/dt) / x^2

To solve for dθ/dt, we need to find dh/dt (the rate of change of the height of the observer's head) and dx/dt (the rate of change of the distance from the observer's head to the balloon).

Since the balloon rises vertically at a rate of 5 feet per second, the rate of change of the height of the observer's head is equal to the rate of change of the height of the balloon. Therefore, dh/dt = 5 feet per second.

The rate of change of the distance from the observer's head to the balloon is zero because the balloon is released and no longer moves away from the observer. Therefore, dx/dt = 0 feet per second.

Substituting these values into the equation, we have:

sec^2(θ) * dθ/dt = (5 * 1 - 3 * 0) / 1^2

Simplifying further, we get:

sec^2(θ) * dθ/dt = 5 / 1

Therefore, we can conclude that:

sec^2(θ) * dθ/dt = 5

Hence, the rate of change of the angle of elevation of the observer's head at the moment when she realizes the balloon is gone forever is 5 / sec^2(θ).