Problem 5 - Calculus Early Transcendentals

Question

Expanding square The sides of a square increase in length at a rate of \(2 \mathrm{m} / \mathrm{s}\). a. At what rate is the area of the square changing when the sides are \(10 \mathrm{m}\) long? b. At what rate is the area of the square changing when the sides are 20 m long? c. Draw a graph that shows how the rate of change of the area varies with the side length.

Step-by-Step Solution

Answer

Based on the given information about the square and the step-by-step solution provided, the rate at which the area of the square is changing can be determined for a square with side length s. a) When the sides are 10 m long, the area is increasing at a rate of 40 m²/s. b) When the sides are 20 m long, the area is increasing at a rate of 80 m²/s. c) The graph of the rate of change of area with respect to the side length can be represented by the equation dA/dt = 4s, where dA/dt is the rate of change of area and s is the side length. This graph will be a straight line with a slope of 4, passing through the origin (0,0).

Step 1: Area of square

A = s^2 As we know, the side length is increasing at a rate of 2 m/s:

Step 2: Rate of side length

ds/dt = 2 m/s Now we want to find the rate at which the area is changing, which is dA/dt.

Step 3: Differentiation of area

First, let's differentiate the area equation with respect to time: (dA/dt) = d(s^2)/dt By applying the chain rule, we get: (dA/dt) = 2s (ds/dt) Now let's substitute the values for each case: (a) s = 10 m

Step 4: Calculating dA/dt for s = 10 m

(dA/dt) = 2(10) (2) = 40 m^2/s The area is increasing at a rate of 40 m²/s when the sides are 10 m long. (b) s = 20 m

Step 5: Calculating dA/dt for s = 20m

(dA/dt) = 2(20) (2) = 80 m^2/s The area is increasing at a rate of 80 m²/s when the sides are 20 m long. (c) Drawing the graph: As we have observed, the rate of change of area (dA/dt) is proportional to the side length: (dA/dt) = 2s (ds/dt) The graph will be a straight line with a slope equal to 2(ds/dt), which is 2(2) = 4. The graph will pass through the origin (0,0). The graph represents the relation between the side length (s) and the rate of change of area (dA/dt) as follows:

Step 6: Graph Equation

dA/dt = 4s

Key Concepts

Understanding Related Rates Problems

Related rates problems in calculus revolve around finding the rate at which one quantity changes in relation to another. This type of problem is commonplace in real-life situations where multiple variables are interconnected and change over time.

For instance, consider the expanding square from our exercise. The situation sets up a dynamic relationship where the length of the sides of the square (denoted as 's') is increasing, which in turn affects the square's area (denoted as 'A'). The crux of related rates problems is to determine how fast 'A' changes (expressed as \(\frac{dA}{dt}\)) when 's' changes at a known rate (expressed as \(\frac{ds}{dt}\)).

In the provided exercise, it's given that the sides of the square are increasing at \(2 \text{m/s}\). This rate, \(\frac{ds}{dt}\), is a known value, and we are asked to find \(\frac{dA}{dt}\) when the sides are of certain lengths. The key to solving these problems is recognizing the relationship between 's' and 'A' and then applying appropriate calculus principles, such as differentiation, to find the unknown rate.

Applying Chain Rule Differentiation

The chain rule is a fundamental differentiation technique in calculus, used when dealing with composite functions – functions made up by combining other functions. In the context of related rates and our expanding square, the chain rule helps us link the rate of change of the area with the rate of change of the side length of the square.

When we differentiate the area of the square (\(A = s^2\)) with respect to time (\(t\)), we treat 's' as an implicit function of time – that is, 's' changes as time changes. The chain rule then provides us with the formula \(\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}\), where \(\frac{ds}{dt}\) is the rate at which the side length increases. The coefficient '2s' comes from the derivative of \(s^2\), which is straightforward to differentiate without the time component.

Understanding and applying the chain rule is critical in finding related rates, as it allows us to formally connect various rates of change that are happening simultaneously but are perhaps not immediately apparent.

Square Area Calculation and Rates of Change

The calculation of a square's area is simple: it is the side length squared (\(A = s^2\)). However, in related rates problems like our exercise, we are interested in how the area changes over time, not just the area itself.

The rate of change of the area of the square (\(\frac{dA}{dt}\)) can be visualized as how fast the square