Question

An observer is \(20 \mathrm{m}\) above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is \(20 \mathrm{m}\) horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of \(5 \mathrm{m} / \mathrm{s}\), what is the rate of change of the angle of elevation when the elevator is \(10 \mathrm{m}\) above the ground? When the elevator is \(40 \mathrm{m}\) above the ground?

Short answer

Answer: In both cases, the rate of change of the angle of elevation is 1/4 radians/s.

Step-by-step solution

Set up and label a diagram

Draw a right triangle with the observer at one vertex, the elevator at another vertex, and the ground below the observer as the third vertex. We’ll label the angle between the horizontal and the line of sight to the elevator as θ, the distance from the observer to the elevator as r, and the height of the elevator above the ground as y.

Write an expression for the angle of elevation

In this right triangle, we can write an expression for the angle θ using the tangent function as: \[tan(\theta) = \frac{y-20}{20}\]

Differentiate both sides of the expression with respect to time

Differentiate both sides of the equation with respect to time t. Recall that the derivative of \(tan(\theta)\) with respect to θ is \(sec^2(\theta)\) and we need to apply the chain rule: \[\frac{d}{dt}(tan(\theta)) = \frac{d}{dt}\left(\frac{y-20}{20}\right)\] \[sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{20}\cdot\frac{dy}{dt}\]

Solve for the rate of change of the angle at given heights of the elevator

We know that \(\frac{dy}{dt}=5\). When the elevator is 10 meters above the ground (y = 10): In this case, we have: \[tan(\theta) = \frac{10 - 20}{20} = -\frac{1}{2}\] Now we can use the Pythagorean Theorem to find the length r: \[r^2 = 20^2 + (20 - 10)^2\] \[r^2 = 400 + 100\] \[r = 10\sqrt{5}\] And now, we find the secant of theta: \[sec(\theta) = \frac{r}{20} = \frac{10\sqrt{5}}{20} = \frac{\sqrt{5}}{2}\] Raise the secant value squared: \[sec^2(\theta) = \left(\frac{\sqrt{5}}{2}\right)^2 = \frac{5}{4}\] Now substitute the known values into the equation and solve for \(\frac{d\theta}{dt}\): \[\frac{5}{4}\cdot\frac{d\theta}{dt} = \frac{1}{20}\cdot5\] \[\frac{d\theta}{dt} = \frac{1}{4}\ radians/s\] Now, repeat this process for when the elevator is 40 meters above the ground (y = 40): \[tan(\theta) = \frac{40 - 20}{20} = 1\] \[r^2 = 20^2 + (40 - 20)^2 = 800\] \[r = 20\sqrt{2}\] \[sec(\theta) = \frac{r}{20} = \sqrt{2}\] \[sec^2(\theta) = 2\] \[\frac{d\theta}{dt} = \frac{1}{20}\cdot5\] \[\frac{d\theta}{dt} = \frac{1}{4} radians/s\] The rate of change of the angle of elevation is the same in both cases: \(\frac{d\theta}{dt} = \frac{1}{4} \mathrm{radians/s}\).

Key concepts

Related Rates

When we are dealing with situations where multiple variables are interconnected through equations, changes in one variable can affect the others. This concept is often explored in calculus problems, particularly within the realm of related rates. These problems require finding the rate of change of one quantity in relation to the change in another. To tackle these problems, we typically use the chain rule to relate the derivatives of the interconnected variables. For instance, in a scenario where an elevator's height changes over time, we can find the rate of change of an angle of elevation by relating the vertical speed of the elevator to the angle’s change rate.

Right Triangle Trigonometry

Right triangle trigonometry is essential in solving many geometry and calculus problems, especially when dealing with angles and lengths in right-angled triangles. The three main functions in trigonometry are sine, cosine, and tangent, often remembered by the mnemonic SOH-CAH-TOA. The tangent of an angle is particularly useful when we have a right triangle and want to relate the lengths of the opposite side to the adjacent side. By understanding these relationships, we can form equations that model scenarios like the one found in the aforementioned elevator problem, where the tangent of the angle of elevation is used to connect the height of the elevator and the horizontal distance from the observer.

Tangent Function

The tangent function is a ratio of the length of the opposite side to the length of the adjacent side in a right-angled triangle. For any angle \( \theta \), the tangent is defined as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). In the context of our problem, the angle of elevation’s tangent helps establish a relationship between the height of the elevator above a certain point and the fixed distance from the observer to the elevator’s base. This ratio is essential in setting up the equation that we'll differentiate to find related rates.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function that is not explicitly solved for one variable in terms of another. It allows us to differentiate both sides of an equation, maintaining the relationship between the two variables while applying the chain rule to solve for the derivative we are interested in. In our elevator scenario, we implicitly differentiate the equation involving the tangent of the angle to find how the angle of elevation changes over time without having to solve for the angle itself directly.

Pythagorean Theorem

The Pythagorean Theorem plays a pivotal role in many geometrical and trigonometrical calculations, expressing an essential relationship between the sides of a right-angled triangle. It asserts that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Represented as \( a^2 + b^2 = c^2 \), the theorem can be used, as seen in the elevator problem, to calculate unknown distances when given one side and the hypotenuse or to confirm the relationships between sides when considering the tangent and secant functions.