Question

a hot air balloon rising straight up from a level field is being tracked by a range finder 150 meters from the liftoff point. [see image] . at the moment the range finder's elevation angle is , the angle is increasing at a rate of 0.14 radians per minute. how can we relate the elevation angle and the height? [mark all that apply]

Answer 1

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Explanation:

To relate the elevation angle (\(\theta\)) and the height of the hot air balloon, we can use trigonometry, specifically the tangent function.

Let \(h\) be the height of the hot air balloon above the level field.

1. **Tangent Function:**

- The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side.

- In this case, the opposite side is the height (\(h\)), and the adjacent side is the horizontal distance from the range finder to the balloon, which is 150 meters.

- Therefore, we can write the tangent function as:

\[ \tan(\theta) = \frac{h}{150} \]

2. **Differential Relation:**

- To relate the rate of change of the angle (\(\frac{d\theta}{dt}\)) with the rate of change of the height (\(\frac{dh}{dt}\)), we can differentiate the tangent function with respect to time:

\[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{150} \cdot \frac{dh}{dt} \]

- Rearranging this equation gives the relationship between the rate of change of the angle and the rate of change of the height.

These relationships help connect the elevation angle with the height of the hot air balloon and describe how changes in the angle correspond to changes in the height.