Problem 38 - Calculus of a Single Variable

Question

A balloon rises at a rate of 4 meters per second from a point on the ground 50 meters from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 50 meters above the ground.

Step-by-Step Solution

Answer

The rate of change of the angle of elevation when the balloon is 50 meters above the ground is 0.04 rad/s.

Step 1: Understand the Problem and Draw the Diagram

Illustrate the problem. The observer, balloon and the point on the ground (right below the balloon) form a right triangle. The observer's line of sight to the balloon is the hypotenuse of this triangle. The vertical upwards distance represents the opposite site, and the ground distance between observer and the point below the balloon is the adjacent side of this right-angled triangle.

Step 2: Set up the Relationship

Write down the trigonometric relationship that's relevant in this context. Because we will be dealing with the angle of elevation and the sides of a right triangle, the tangent function (opposite/adjacent) is most appropriate. Let \( h \) represent the height of the balloon and \( \theta \) represents the angle of elevation, we then get the relationship \( \tan(\theta) = h/50 \).

Step 3: Differentiate both Sides

With respect to time (t), the equation becomes \( \frac{d}{dt} \tan(\theta) = \frac{d}{dt}(h/50) \). When simplifying, this leads to \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{dh}{dt} / 50 \)

Step 4: Substitute Known Values

Given that \( \frac{dh}{dt} = 4 \), the height \( h = 50 \), and \( \theta \) is such that \( \tan(\theta) = 50/50 = 1 \) (which means \( \theta = 45 \) degrees or \( \theta = \pi/4 \) radians), from this the value \( \sec^2(\theta) \) can be found and substituted into the equation, giving \( \frac{d\theta}{dt} = \frac{4}{50 \cdot \sec^2(\theta)} \). \(\sec^2(45^\circ)\) = 2, so \( \frac{d\theta}{dt} = \frac{4}{50 \cdot 2} = 0.04 \) rad/sec.