Problem 5 - Calculus - AP Edition

Question

Let \(A\) be the area of a square whose sides have length \(x,\) and assume that \(x\) varies with the time \(t\) (a) Draw a picture of the square with the labels \(A\) and \(x\) placed appropriately. (b) Write an equation that relates \(A\) and \(x\) (c) Use the equation in part (b) to find an equation that relates \(d A / d t\) and \(d x / d t\) (d) At a certain instant the sides are 3 ft long and increasing at a rate of \(2 \mathrm{ft} / \mathrm{min}\). How fast is the area increasing at that instant?

Step-by-Step Solution

Answer

The area is increasing at 12 square feet per minute.

Step 1: Draw the square

Sketch a square and label each side with length \(x\). Inside the square, write the letter \(A\) to indicate the area of the square. This visual represents that \(A\), the area, is dependent on the side length \(x\).

Step 2: Relate area and side length

Write the formula for the area of the square in terms of its side length. The area \(A\) of a square with side length \(x\) is given by the equation: \(A = x^2\).

Step 3: Differentiate with respect to time

Differentiate the equation \(A = x^2\) with respect to time \(t\) to find the relationship between \(\frac{dA}{dt}\) and \(\frac{dx}{dt}\). Using implicit differentiation, the derivative is: \(\frac{dA}{dt} = 2x \frac{dx}{dt}\).

Step 4: Substitute known values

At the given instant, \(x = 3\) ft and \(\frac{dx}{dt} = 2\) ft/min. Substitute these values into the differentiated equation: \(\frac{dA}{dt} = 2 \cdot 3 \cdot 2\). Calculate to find \(\frac{dA}{dt}\).

Step 5: Calculate the rate of change

Perform the multiplication: \(\frac{dA}{dt} = 2 \cdot 3 \cdot 2 = 12\). Thus, the area is increasing at the rate of 12 square feet per minute when the side length is 3 feet.

Key Concepts

Derivatives

In calculus, derivatives represent the rate at which a quantity changes. They are fundamental for solving problems involving changing rates, such as those in related rates exercises. When considering the change in the area of a square over time, we employ derivatives to express how fast the area is growing as a function of how fast the sides of the square are lengthening. If the side length of a square is given by function \(x(t)\), where \(t\) represents time, then the area \(A(t)\) can be expressed as \(A = x^2\). Derivatives allow us to determine \(\frac{dA}{dt}\), the rate at which area changes, in relation to \(\frac{dx}{dt}\), the rate at which the side length changes.

• Derivative of \(x^2\): Using the power rule in calculus, where the derivative of \(x^n\) is \(nx^{n-1}\), the derivative of \(x^2\) with respect to \(x\) is \(2x\).
• Using derivatives with time: We express the change over time, \(\frac{dA}{dt} = 2x\frac{dx}{dt}\), by multiplying the derivative result by the rate of change of \(x\) with time.

Understanding and calculating derivatives are crucial to solving related rates problems, as they help us determine dependencies between varying quantities.

Implicit Differentiation

Implicit differentiation is a technique in calculus used to find derivatives of equations that are not explicitly solved for one variable. In situations like our square example, where the area \(A\) is a function of the side length \(x\), both changing over time, we use implicit differentiation to determine the relationship between \(\frac{dA}{dt}\) and \(\frac{dx}{dt}\).
To apply implicit differentiation, follow these steps:

• Differentiate both sides of the equation with respect to time \(t\). For our square, start with the area formula \(A = x^2\).
• The derivative of \(A\) with respect to \(t\) is \(\frac{dA}{dt}\), and the derivative of \(x^2\), with respect to \(x\) followed by the chain rule, involves multiplying by \(\frac{dx}{dt}\), resulting in \(2x\frac{dx}{dt}\).

Implicit differentiation is often easier than first solving for the dependent variable, making it invaluable for handling more complex related rates problems, especially those involving geometric figures.

Area of a Square

The area of a square, one of the most basic geometric shapes, is calculated easily using a simple formula: the length of one side squared. The formula is \(A = x^2\), where \(x\) represents the length of a side. This formula reflects the proportional relationship between the side length and the area.
In real-world applications, the area of a square can change over time if our side length \(x\) changes. For example, during an exercise exploring related rates, calculating the area’s rate of change involves understanding how variations in side length over time affect overall area. This uses the formula we derived through differentiation: \(\frac{dA}{dt} = 2x\frac{dx}{dt}\).
The concept of squaring ensures that any increase in side length results in a quadratic increase in area, showcasing how even simple geometric shapes can present interesting relationships in calculus. Knowing these formulas and their implications simplifies problem-solving in multivariate calculus scenarios.