Question

A drawbridge \(30 \mathrm{ft}\). long is being slowly raised by chains passing overa windlass and being drawn in at the rate of \(8 \mathrm{ft}\). a minute. \(\mathrm{A}\) distant electric light sends out horizontal rays and the bridge thus casts a shadow on a vertical wall, consisting of the other half of the bridge, which has been already raised. Find how fast the shadow is ereeping up the wall when half the chain has been drawn in.

Short answer

Answer: The shadow is moving up the wall at a rate of \(-\frac{120}{169}\) ft/min. The negative sign indicates that the shadow is actually moving down the wall.

Step-by-step solution

Set up similar triangles

Let x be the length of the shadow on the wall. When the bridge has been lifted by 4 ft, the height of the triangle formed by the shadow and the wall is (30-4 = 26) ft long. Since the rays are horizontal, we can see that the triangles formed are similar. The ratio of the sides is: \(\frac{4}{26}=\frac{30}{x}\)

Solve for x (the shadow length)

To solve for x, we can cross-multiply and simplify: \(4x = 30 \times 26\) \(x=\frac{30\times 26}{4}=195\)

Differentiate the equation with respect to time

Now, we will differentiate the equation concerning time (t) to find the rate at which the shadow is moving up the wall: \(\frac{d}{dt}(\frac{4}{26})=\frac{d}{dt}(\frac{30}{x})\)

Plug in the given rate and solve for dx/dt

We are given that the rate at which the chains are being drawn is 8 ft/min. That means \(\frac{d(4)}{dt}=8\) ft/min. Plugging the given rate and the value of x into the equation, we get: \(\frac{8}{26}=\frac{30 \times (-\frac{dx}{dt})}{195}\) Now, we will solve for \(\frac{dx}{dt}\), which is the rate we are looking for.

Calculate the rate at which the shadow is moving up the wall

Solving for \(\frac{dx}{dt}\), we get: \(\frac{dx}{dt}=\frac{-30\times 8}{195\times 26}\) \(\frac{dx}{dt}=-\frac{120}{169}\) Now, we have found that the shadow is moving up the wall at a rate of \(-\frac{120}{169}\) ft/min when half the chain has been drawn in. The negative sign indicates that the shadow is actually moving down the wall, as expected.

Key concepts

Related Rates

Related rates problems are a fascinating application of derivatives in calculus. They examine how different quantities change together over time. In our case, as the drawbridge is being raised, we're interested in how quickly the shadow on the wall is changing. The term 'related rates' comes from the idea that the rate at which the bridge's length is changing is directly connected to the rate at which the shadow's length changes.

To solve such problems, we typically identify the quantities that change together and then relate their rates of change. This is often done by deriving a relationship between the quantities and using differentiation to include the time factor, allowing us to find the rate of change for one quantity when we know the rate of change for another. The intricacy often arises not from the calculus itself, but from setting up the relationships correctly.

Similar Triangles

Similar triangles play a crucial role in solving many geometry-related calculus problems, including this one involving a drawbridge and its shadow. Cognizant of the properties of similar triangles can simplify the process of creating equations that relate different variables.

Triangles are similar when they have corresponding angles that are equal and the lengths of their corresponding sides are in proportion, which is the case with our drawbridge scenario. The drawbridge and its shadow create one triangle, while the standing part of the bridge and its height form another. By recognizing that these two triangles are similar, we can construct a proportionality relationship. This allows us to create an equation that relates the height of the bridge to the length of the shadow, vital for finding the rate at which the shadow changes.

Differentiation

Differentiation is the core mathematical concept used to solve calculus problems related to rates of change. In the context of our drawbridge problem, differentiation gives us the means to find the answer by analytically tackling how one variable changes in relation to another over time.

Once we have our equation from the similar triangles, we take its derivative with respect to time. This is where the calculus magic happens: we're transforming static geometric relationships into dynamic ones, unveiling the rate at which the shadow moves up the wall. The act of differentiation turns our spatial puzzle into a quantifiable change over time, providing the exact value we're after. Understanding how to differentiate equations effectively is an essential skill in calculus that unlocks the door to solving related rates problems.

Problem-Solving in Calculus

Problem-solving in calculus, especially with related rates problems, requires a step-by-step approach to dissect the situation, construct relationships between variables, and use calculus techniques such as differentiation to find the solution.

In this drawbridge problem, the method involves setting up an equation using similar triangles, differentiating with respect to time to find the relationship between the rates of change, and then substituting known values to solve for the desired rate. This systematic process aids in making complex problems more manageable and highlights calculus' power in providing real-world solutions. By cultivating a strong foundation in problem-solving strategies and practicing different types of problems, one can develop the skill set needed to tackle a variety of challenging calculus questions.