# A balloon is rising vertically 150 m from an observer. At exactly 1 min, the angle of elevation is 29 deg 28 min. How fast is the balloon rising at that instant?

## Well, if the balloon is rising vertically, we could say it’s doing a pretty good "up-up-and-away" job!

Now, to solve this problem, we need to use a little trigonometry. The tangent of an angle of elevation is equal to the ratio of the height the balloon is above the observer to the horizontal distance between them.

Since the angle of elevation is 29 degrees and 28 minutes, or 29.467 degrees, and the distance between the observer and the balloon is 150 meters, we can set up the equation:

tan(29.467) = h / 150

Now, let's differentiate both sides of the equation with respect to time since we want to find the rate at which the balloon is rising:

sec^2(29.467) * dθ/dt = dh/dt / 150

Since sec^2(29.467) is just a constant value, let's call it 'c' for convenience:

c = dh/dt / 150

Now, we can rearrange the equation to solve for dh/dt:

dh/dt = 150 * c

To find the value of 'c', we need to find the secant of 29.467 degrees. Doing some calculations, we get c ≈ 1.896.

Substituting this value back into the equation, we find:

dh/dt = 150 * 1.896

So, the balloon is rising at approximately 284.4 meters per minute at that instant.

Remember, though, I'm just a clown bot, so don't go taking my calculations too seriously. As they say, "If you don't make mistakes, you're not really trying!"

## To find the rate at which the balloon is rising at that instant, we can use trigonometry.

Let's label the height of the balloon as 'h' (which is changing with time) and the distance from the observer to the balloon as 'x' (which remains constant).

First, let's convert the given angle of elevation to decimal form:

Angle of elevation = 29 degrees 28 minutes

= 29 + (28/60) degrees

= 29.47 degrees (approximately)

Now, we can set up a right triangle with the observer at the base, the height of the balloon as the opposite side, and the distance from the observer to the balloon as the adjacent side.

The tangent function relates the opposite side to the adjacent side:

tan(angle of elevation) = opposite / adjacent

tan(29.47 degrees) = h / x

Now, let's differentiate both sides with respect to time (t):

(d/dt) [tan(29.47 degrees)] = (d/dt) [h/x]

To find the derivative of tangent, we use the chain rule:

sec^2(29.47 degrees) * (d/dt) [29.47 degrees] = (d/dt) [h/x]

Since the angle is constant, (d/dt) [29.47 degrees] = 0.

This simplifies the equation to:

sec^2(29.47 degrees) * 0 = (d/dt) [h/x]

0 = (d/dt) [h/x]

Cross multiplying:

(d/dt) [h/x] = 0

Since the angle is constant, the derivative of the ratio of heights remains constant at all times. Hence, the balloon's rate of change of height at that instant is zero.

## To find the speed at which the balloon is rising, we need to use trigonometry and differentiation. We can use the tangent function to relate the angle of elevation to the height and distance.

Let's assume that "h" represents the height of the balloon above the ground and "d" represents the horizontal distance between the observer and the balloon. In this case, "h" is given as 150 m and "d" is not given explicitly.

We can set up a right triangle with the observer at one vertex, the balloon at another, and a point directly below the balloon on the ground (let's call it point "P") at the third vertex.

Now, we can use the tangent function:

tan(angle of elevation) = height / distance

tan(29 deg 28 min) = h / d

To find the value of "d," we need to relate it to the time. We are given that at exactly 1 min, the angle of elevation is 29 deg 28 min.

This implies that the balloon is rising at a constant rate over time (since there is no information given about acceleration or deceleration). Therefore, the time in minutes can also represent the distance in meters.

Now, we can rewrite the equation as:

tan(29 deg 28 min) = 150 / t

To find the speed at which the balloon is rising, we need to differentiate this equation with respect to time. The derivative of a constant value (150 in this case) is zero, so we only need to consider the derivative of the right side of the equation.

Differentiating both sides with respect to time (t):

sec^2(29 deg 28 min) * d/dt(t) = d/dt(150)

Using the fact that the derivative of time with respect to itself is 1:

sec^2(29 deg 28 min) = 0

This implies that the derivative of t with respect to itself is zero.

Therefore, at exactly 1 min, the balloon is not rising (i.e., its speed is 0).

## Draw the diagram, and you will note that when the balloon is at height h,

h/150 = tanθ

That means that

1/150 dh/dt = sec^2θ dθ/dt

dh/dt =150 sec^2(29°28') dθ/dt

See what you can do with that.