# A hot air balloon rising vertically is tracked by an observer located 4 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is \frac{\pi}{3} , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?

## Hey, I just did one very much like this. Your turn.

## by the way,

d/dt (tan A) = sec^2 A * dA/dt

= (1/cos A)^2 * dA/dt

## To find how fast the balloon is rising at the given moment, we need to use trigonometry and differentiation.

Let's consider the situation and set up a right triangle with the observer, the balloon, and the horizontal distance between them.

Let H be the height of the balloon, and let x be the horizontal distance between the observer and the lift-off point. We are given that x = 4 miles.

We can represent the angle between the observer's line-of-sight and the horizontal as θ. We are given that θ = π/3 (60 degrees).

Using trigonometry, we know that the tangent of an angle is equal to the ratio of the opposite side to the adjacent side. In this case, the tangent of θ is H/x.

So, we have the equation: tan(θ) = H/x

To solve for H, we multiply both sides of the equation by x: x * tan(θ) = H

Differentiating both sides of the equation with respect to time t, we get: (dH/dt) = (dx/dt) * tan(θ) + x * sec^2(θ) * (dθ/dt)

Since the balloon is rising vertically, the horizontal distance x between the observer and the balloon is not changing. Therefore, dx/dt = 0.

We are also given that (dθ/dt) = 0.1 rad/min.

Plugging in the known values, we get: (dH/dt) = 0 * tan(π/3) + 4 * sec^2(π/3) * 0.1

The secant of an angle is the reciprocal of the cosine. We know that cos(π/3) = 1/2, so sec(π/3) = 2.

Simplifying the equation, we have: (dH/dt) = 0 + 4 * (2^2) * 0.1

Calculating further, we get: (dH/dt) = 4 * 4 * 0.1

Therefore, the rate at which the balloon is rising at this moment is: (dH/dt) = 1.6 miles per minute.