Problem 12 - Calculus Volume 1

Question

Sketch the situation if necessary and used related rates to solve for the quantities. A 5-ft-tall person walks toward a wall at a rate of 2 ft/ sec. A spotlight is located on the ground 40 ft from the wall. How fast does the height of the person’s shadow on the wall change when the person is 10 ft from the wall?

Step-by-Step Solution

Answer

The shadow's height changes at \( \frac{4}{9} \) ft/sec.

Step 1: Setup the Scenario

Imagine a right triangle formed by the person, their shadow on the wall, and the distance from the spotlight to the wall. Let the distance of the person from the wall be denoted as \( x \) ft. The spotlight is 40 ft from the wall, so the distance from the spotlight to the person is \( 40 - x \) ft. Let \( y \) represent the height of the shadow on the wall.

Step 2: Establish the Relationship

Use similar triangles to establish the relationship between the height of the person, their distance from the spotlight, and the height of the shadow.The ratio of corresponding sides is:\[ \frac{5}{40-x} = \frac{y}{40} \]Solve for \( y \):\[ y = \frac{5(40)}{40-x} \]

Step 3: Differentiate with Respect to Time

Since we want the rate of change of the height of the shadow, differentiate \( y \) with respect to \( t \):\[ \frac{dy}{dt} = \frac{d}{dt} \left( \frac{200}{40-x} \right) \]Use the quotient rule or note that you can apply the chain rule due to the form given:

Step 4: Apply the Chain Rule

Differentiating \( y = \frac{200}{40-x} \) with respect to \( t \), we apply the chain rule:\[ \frac{dy}{dt} = -200 \cdot \frac{1}{(40-x)^2} \cdot \left( -\frac{dx}{dt} \right) \]Simplify:\[ \frac{dy}{dt} = \frac{200}{(40-x)^2} \cdot \frac{dx}{dt} \]

Step 5: Substitute Known Values

We know \( \frac{dx}{dt} = 2 \) ft/sec and \( x = 10 \) ft.Substitute these values into the differentiated equation:\[ \frac{dy}{dt} = \frac{200}{(40-10)^2} \cdot 2 \]Calculate:\[ \frac{dy}{dt} = \frac{200}{900} \cdot 2 = \frac{400}{900} \]Simplify:\[ \frac{dy}{dt} = \frac{4}{9} \]

Step 6: Conclude the Solution

The rate at which the height of the shadow on the wall changes is \( \frac{4}{9} \) ft/sec.

Key Concepts

Similar Triangles

In the world of geometry, similar triangles are a powerful tool. These triangles have the same shape but may differ in size. Their corresponding angles are equal, and their corresponding sides are in proportion. This is useful in related rates problems.
In this exercise, think of two setups forming triangles: one involving the person and the spotlight, and the other involving the person's shadow and the wall.

• The first triangle has sides formed by the height of the person and distance to the wall.
• The second triangle is formed by the height of the shadow and the distance from the spotlight to the wall.

We used similar triangles to express the relationship between the height of the person and the height of their shadow. This relationship allows us to set up a proportion and find unknown quantities.

Chain Rule

The chain rule is a fundamental concept in calculus. It lets us differentiate composite functions. That means functions inside other functions. In related rates problems, we often mix variables that depend on time.When you differentiate these sorts of functions, you "chain" the derivatives together. For instance, if you have a function inside another function, the derivative takes into account both. For example, consider differentiating the function given by:\[ y = \frac{200}{40-x} \]Using the chain rule, we ensure to multiply by the derivative of the inside expression, which is in this case, as the rate of change of distance \(-\frac{dx}{dt}\). It’s like peeling layers of an onion- working from the outside in.

Differentiation with Respect to Time

This step is crucial in related rates problems. Here, we differentiate variables with influence from time. When a variable changes over time, we express it in terms of its rate of change. This involves differentiating it concerning time. That means, using calculus skills to understand how something changes each second.This method is used when we differentiate both sides of our equation about time. For example, here we want the derivative of the shadow’s height \(y\) over time \(t\): \[ \frac{dy}{dt} \] reflects how quickly the height of the shadow changes as time goes by.

Related Rates Problems

Related rates problems are real-world scenarios where related quantities change over time. It's an area of interest in calculus that deals with these dynamic changes. In these problems, quantities are linked through relationships, usually by equations. We need to find out how quickly one variable changes concerning another. Consider the exercise: two quantities—the position of a person and their shadow's height—are related because as the person moves, their shadow's length also changes. To solve these problems:

• First, understand the geometry or physical layout.
• Write an equation relating the quantities.
• Differentiation is then done concerning time to find the rate of change.
• Substitute known values to solve.

This process helps us understand the dynamic relationships in the movement and changes of real-world entities.