Question

A lighthouse stands 500 m off a straight shore and the focused beam of its light revolves four times each minute. As shown in the figure, \(P\) is the point on shore closest to the lighthouse and \(Q\) is a point on the shore 200 m from \(P\). What is the speed of the beam along the shore when it strikes the point \(Q ?\) Describe how the speed of the beam along the shore varies with the distance between \(P\) and \(Q\). Neglect the height of the lighthouse.

Short answer

5-2-√29 right triangle, so: \[\theta = \tan^{-1}\left(\frac{200}{500}\right) = \tan^{-1}\left(\frac{2}{5}\right)\] With the angle \(\theta\) known, we can use the Pythagorean theorem to find \(AC\): \[AC = \sqrt{AB^2 + BC^2}\] \[AC = \sqrt{500^2 + 200^2} = 100\sqrt{29}\] #tag_title# Step 7: Evaluate the expression for the rate of change of AC, \(\frac{d(AC)}{dt}\), at the given values #tag_content# Plug in the known values for \(\theta\), \(AC\), and \(\frac{d\theta}{dt}\), and find the rate of change of AC when the beam is at point \(Q\): \[\frac{d(AC)}{dt} = -200 \sin \left(\tan^{-1}\left(\frac{2}{5}\right)\right) \frac{4\pi}{15} \frac{1}{100\sqrt{29}}\] Evaluating the expression, we get: \[\frac{d(AC)}{dt} \approx -13.28\, \text{m/s}\] #tag_title# Step 8: Discuss how the speed of the beam varies with the distance between P and Q #tag_content# The rate of change of AC (the speed of the beam) along the shore depends on the angle \(\theta\) and the distance AC. As the distance between P and Q increases, the angle of the beam will also increase and the beam will move faster along the shore. Conversely, as the distance between P and Q decreases, the angle will decrease, and the beam will move slower along the shore. **Short Answer:** The speed of the beam along the shore at point Q is approximately \(-13.28 \text{m/s}\). As the distance between P and Q increases, the speed of the beam also increases.

Step-by-step solution

Draw a diagram and label all given information

Draw a right triangle with the lighthouse at vertex \(A\), point \(P\) on the shore at vertex \(B\), and point \(Q\) on the shore at vertex \(C\). Label the distances: \(AB = 500 m\), and \(CB = 200 m\). The angle at vertex \(A\) (the lighthouse) is the angle we will use as the revolving angle, let's call it \(\theta\).

Set up the relationship between \(\theta\) and the revolving speed of the beam

We are given that the beam revolves four times per minute. Let's convert this to radians per second to make use of it later: \[4 \text{ rev/min} * 2\pi \frac{\text{ rad}}{\text{ rev}} * \frac{1 \text{ min}}{60 \text{ sec}} = \frac{4\pi}{15} \frac{\text{ rad}}{\text{ sec}}\] This is the rate at which the angle \(\theta\) changes, which we will denote \(\frac{d\theta}{dt}\): \[\frac{d\theta}{dt}=\frac{4\pi}{15}\]

Find an expression for the distance between points \(A\) and \(C\) using the Law of Cosines

Since \(ABC\) is a right triangle, we can use the Law of Cosines to find the distance \(AC\) in terms of \(\theta\): \[AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos \theta\] \[AC^2 = 500^2 + 200^2 - 2(500)(200) \cos \theta\]

Differentiate the distance expression with respect to time \(t\)

Now we will differentiate \(AC^2\) in terms of \(t\) to find the rate of change of AC: \[ \frac{d}{dt} (AC^2) = \frac{d}{dt} (500^2 + 200^2 - 2(500)(200) \cos \theta)\] Since \(AB\) and \(BC\) are constants, their derivatives are 0: \[\frac{d(AC^2)}{dt} = 0 + 0 - 2(500)(200) \frac{d}{dt}(\cos \theta)\] By using the chain rule, we get: \[\frac{d(AC^2)}{dt} = 2(500)(200) \sin \theta\cdot \frac{-d\theta}{dt}\]

Find the rate of change of AC, \(\frac{d(AC)}{dt}\)

Since we have the derivative of \(AC^2\), we need to find the derivative of \(AC\). This can be done by using the chain rule again: \[\frac{d(AC)}{dt} = \frac{1}{2} \frac{1}{\sqrt{AC^2}} \frac{d(AC^2)}{dt}\] Now, we plug in the expression from step 4: \[\frac{d(AC)}{dt} = \frac{-1}{2(500)(\sqrt{AC^2})} \cdot 2(500)(200) \sin \theta \frac{d\theta}{dt}\] We can simplify this expression to: \[\frac{d(AC)}{dt} = -200 \sin{\theta} \frac{d\theta}{dt} \frac{1}{\sqrt{AC^2}}\]

Plug in the given values for BC, dθ/dt, and find the angle θ and distance AC

Now, we plug in the given values for \(BC = 200m\) and \(\frac{d\theta}{dt}= \frac{4\pi}{15}\). To find the angle \(\theta\) and distance \(AC\) at the moment when the beam is at \(Q\), we can use the triangle properties. In this case, we see that \(\triangle ABC\) is a 5-2-0ffsettriangle1047-RIGHT_WHITE_SMALL_SQUARE

Key concepts

Related Rates

Understanding related rates in calculus is crucial when analyzing how one quantity changes in relation to another, especially in real-world scenarios like motion. A common example involves the rate at which the angle between a lighthouse beam and a point on the shore changes as the light rotates.

In the presented exercise, by utilizing the concept of related rates, we can examine the speed of the lighthouse's beam along the shoreline -- a dynamic scenario where the rates of change of distance and angles are interconnected. It starts with the recognition that the angle of the beam and the point it hits on the shore are related through time—hence, 'related rates'. The key is to set up an equation that describes this relationship and then differentiate with respect to time to find out how fast one quantity is changing in terms of the other.

Trigonometry in Calculus

Incorporating trigonometry into calculus is essential when dealing with problems involving angles and their relationships to lengths. For example, in our motion-related problem, we need trigonometry to describe the positions of the points relative to the lighthouse beam.

Trigonometric functions like sine and cosine describe the relationships between the angles and side lengths of a triangle. When the problem involves a revolving lighthouse beam, trigonometric functions help relate the rotation angle to the position on the shore. The proper application of these relationships allows us to proceed to calculating derivatives which are the foundation of solving related rates problems. In the exercise, trigonometry has been used to establish the initial relationship that is differentiated to find the speed of the beam.

Law of Cosines

The Law of Cosines is a powerful tool in solving problems that involve measuring distances in non-right triangles. In the context of calculus problems concerning motion, such as determining the path of a lighthouse beam along a shore, the Law of Cosines can be used to relate the sides of a triangle to the cosine of its angles.

For the given exercise, we employ the Law of Cosines to express the distance between the lighthouse and a point on the shore as a function of the changing angle. This relationship sets the stage for finding the rate at which this distance changes, a derivative calculation we need for the solution. The Law of Cosines formula, in its differentiated form, helps us understand how the change in angle affects the distance 'AC' over time.

Derivatives of Trigonometric Functions

The derivatives of trigonometric functions are frequently encountered in calculus, especially when solving problems involving periodic motion or waves. These derivatives enable us to understand how functions like sine and cosine change over time, which is essential when calculating related rates.

In our example featuring the rotating lighthouse beam, we must differentiate the cosine function to describe the rate at which the angle between the lighthouse and a point on the shore, 'Q', is changing with respect to time. Knowing that the derivative of cosine is negative sine, and taking the derivative of a constant yields zero, we can simplify the expression from the problem to find the rate at which the distance 'AC' changes as the beam moves. These fundamental principles of differentiating trigonometric functions are key to solving not only this particular problem but also a plethora of related rates problems involving angles and time.