Problem 38 - Calculus Single Variable

Question

A lighthouse is 100 feet tall. It keeps its beam focused on a boat that is sailing away from the lighthouse at the rate of 300 feet per minute. If \(\theta\) denotes the acute angle between the beam of light and the surface of the water, then how fast is \(\theta\) changing at the moment the boat is 1000 feet from the lighthouse?

Step-by-Step Solution

Answer

\(\frac{d\theta}{dt} \approx -0.00297 \text{ radians per minute}\).

Step 1: Draw and Label the Diagram

Visualize the scenario by drawing a right triangle where the lighthouse is the vertical side (\(100\) feet), the distance from the lighthouse to the boat is the horizontal (\(x\)), and the hypotenuse is the light beam. The angle \(\theta\) is between the vertical (lighthouse) and the hypotenuse.

Step 2: Set Up the Trigonometric Relationship

In this setup, we use tangent because we know the opposite side (lighthouse height) and the adjacent side (distance from lighthouse to boat). The relation is: \[ \tan(\theta) = \frac{100}{x} \] where \(100\) is the height of the lighthouse and \(x\) is the horizontal distance from the lighthouse to the boat.

Step 3: Differentiate with Respect to Time

Differentiate the trigonometric equation \( \tan(\theta) = \frac{100}{x} \) with respect to time \(t\). This involves implicit differentiation. \[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{100}{x^2} \cdot \frac{dx}{dt} \]

Step 4: Substitute Known Values

We know \(\frac{dx}{dt} = 300 \text{ ft/min}\) as the boat sails away. At the moment the boat is \(1000\) feet away, use this to substitute: \[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{100}{1000^2} \cdot 300 \] Also, calculate \(\sec^2(\theta)\) when \(x = 1000\): \[ \tan(\theta) = \frac{1}{10}, \quad \sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left(\frac{1}{10}\right)^2 = \frac{101}{100} \]

Step 5: Solve for \(\frac{d\theta}{dt}\)

Plug the values back into the differentiated equation to find \(\frac{d\theta}{dt}\): \[ \frac{101}{100} \cdot \frac{d\theta}{dt} = -\frac{100}{1000000} \cdot 300 \] Simplify and solve for \( \frac{d\theta}{dt} \): \[ \frac{d\theta}{dt} = -\frac{3}{1010} \] This approximates to \( \frac{d\theta}{dt} \approx -0.00297 \text{ radians/minute} \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the variables are intertwined. This means you can't easily solve for one variable in terms of another before differentiating. It’s like pulling apart a tangled knot of yarn. When a relationship between variables is given implicitly, differentiating both sides requires treating every part as a function of time or another variable.
For example, in our exercise with the lighthouse and the boat, we have the trigonometric equation \( \tan(\theta) = \frac{100}{x} \). Here, \(\theta\) is a function of time, but it's not isolated. Instead, \(x\) (the distance from the lighthouse to the boat) is also a function of time. Thus, while differentiating, we also consider how \(x\) changes with time.

• The general rule for implicit differentiation mandates applying the chain rule for all derivative terms.
• In practice, this means you'll hear about derivatives like \(\sec^2(\theta) \cdot \frac{d\theta}{dt}\), showing the dependency on time.

By understanding this, you gain a powerful tool for attacking calculus problems that can't be easily pieced apart.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent play a crucial role in solving many calculus problems because they relate angles to side lengths in right triangles. In the context of our lighthouse problem, we used the tangent function because we knew the height of the lighthouse and the distance to the boat—making it a classic right triangle situation.
Here's a quick refresher to strengthen your grasp:

• The tangent function, \( \tan(\theta) = \frac{{\text{opposite}}}{\text{adjacent}} \), fits perfectly when you know two sides of a triangle.
• Understanding how each trigonometric function changes helps when you're solving dynamic (related rates) problems.
• In this problem, once we had \( \tan(\theta)=\frac{100}{x}\), it unlocked the pathway to relate \(\theta\) to \(x\) and ultimately how they both change over time.

Getting comfortable with these functions is essential not only for homework but for any calculus or physics scenario involving lengths and angles.

Calculus Problems

Calculus problems cover a wide spectrum, but one common type is the *related rates* problem. These involve rates of change and often use implicit differentiation. The essence is to relate two or more functions and determine how their rates change with respect to each other. The lighthouse problem is a classic related rates scenario, illustrating how if one quantity changes (in this case, the distance), it impacts another (the angle \(\theta\)).
To tackle such problems:

• Begin by identifying which quantities are changing and try to express their relationship mathematically.
• Set up an equation involving these quantities. Use derivatives to express how the respective rates of these quantities are related.
• Substitute any known values. This simplifies the equation to allow solving for the unknown rate, such as \(\frac{d\theta}{dt}\).

These steps are split into understanding the physical situation, converting it into a mathematical model, and then solving by leveraging calculus tools. Once you grasp related rates, they will unlock a deeper understanding of how interconnected change occurs in numerous scenarios.