# A balloon leaves the ground 500 feet away from an observer and rises vertically at the rate of 140 feet per minute. At what rate is the angle of inclination of the observer's line of sight increasing at the instant when the balloon is exactly 500 feet above the ground?

## tan A = h/500

d tan A/ dh = 500(-1)/500^2 = -1/500

d tan A/dA = sec^2 A
so d tan A = sec^2 A dA

sec^2 A dA/dh = -1/500
dA/dh = cos^2 A /500
dA/dt = dh/dt * dA/dh
dA/dt = 140 cos^2 A / 500
at 500 ft. cos A = cos pi/4 = .707
so cos^2 A = .5
dA/dt = 140 * .5 /500
or 8 degrees/minute

## tan A = h/500

d tan A/ dh = 500(1)/500^2 = 1/500

d tan A/dA = sec^2 A
so d tan A = sec^2 A dA

sec^2 A dA/dh = 1/500
dA/dh = cos^2 A /500
dA/dt = dh/dt * dA/dh
dA/dt = 140 cos^2 A / 500
at 500 ft. cos A = cos pi/4 = .707
so cos^2 A = .5
dA/dt = 140 * .5 /500
or 8 degrees/minute

## To solve this problem, we'll need to use trigonometry and related rates. Let's define some variables:

Let x represent the horizontal distance between the balloon and the observer.
Let y represent the height of the balloon above the ground.
Let θ represent the angle of inclination of the observer's line of sight.

We are given that the balloon leaves the ground 500 feet away from the observer (x = 500 ft) and rises vertically at a rate of 140 feet per minute (dy/dt = 140 ft/min). We need to find the rate at which the angle θ is changing when the balloon is exactly 500 feet above the ground (y = 500 ft).

First, let's draw a diagram to help us understand the situation:

B (balloon)
|
|
| O (observer)
--------------- 500 ft ---------------

From the diagram, we can see that tan(θ) = y / x.

Differentiating both sides of this equation with respect to time t, we get:

sec²(θ) * dθ/dt = (dy/dt)/x [using the chain rule]

Now we can solve for dθ/dt by substituting the given values: dy/dt = 140 ft/min, x = 500 ft, and y = 500 ft:

sec²(θ) * dθ/dt = (140 ft/min) / 500 ft
dθ/dt = (140 ft/min) / (500 ft * sec²(θ))

To find the angle θ, we can use the trigonometric identity:

sin(θ) = y / √(x² + y²)

Substituting the given values: x = 500 ft and y = 500 ft, we have:

sin(θ) = 500 ft / √(500 ft)²
sin(θ) = 500 ft / √(500 ft)²
sin(θ) = 500 ft / 500 ft
sin(θ) = 1

Since θ is the angle of a right triangle, sin(θ) = 1 when θ = 90°.

Plugging this back into the equation for dθ/dt, we have:

dθ/dt = (140 ft/min) / (500 ft * sec²(90°))
= (140 ft/min) / (500 ft * 1)
= (140 ft/min) / 500 ft

Finally, simplifying gives us:

dθ/dt = 0.28°/min

Therefore, the angle of inclination of the observer's line of sight is increasing at a rate of approximately 0.28° per minute when the balloon is exactly 500 feet above the ground.

## To find the rate at which the angle of inclination of the observer's line of sight is increasing, we can use trigonometry and related rates.

Let's denote:
- The distance between the observer and the balloon as "x" (in feet).
- The height of the balloon above the ground as "y" (in feet).

We are given that the balloon is 500 feet away from the observer (x = 500 feet) and that it is rising vertically at a rate of 140 feet per minute (dy/dt = 140 ft/min). We want to find the rate at which the angle of inclination (θ) of the observer's line of sight is increasing when the balloon is exactly 500 feet above the ground (dy/dt when y = 500 ft).

We can form a right triangle with the observer at its vertex, the ground directly below the balloon, and the line of sight as the hypotenuse. The height of the triangle is y, the base is x, and the hypotenuse is the line of sight. By using trigonometry, we can relate these quantities:

tan(θ) = y / x

Now, we need to differentiate this equation implicitly with respect to time (t):

d(tan(θ)) = (d(y) / dt * x - y * d(x) / dt) / x^2

The right side simplifies to:

d(tan(θ)) = (dy - y * dx) / x^2

Now, we substitute the given values and known rates:

d(tan(θ)) = (140 - 500 * 0) / 500^2