Problem 52 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Each side of a square is increasing at the uniform rate of \(1 \mathrm{~m} / \mathrm{sec}\). If after some time the area of the square is increasing at the rate of \(8 \mathrm{~m}^{2} / \mathrm{sec}\), then the area of square at that time in sq. meters is: (A) 4 (B) 9 (C) 16 (D) 25

Step-by-Step Solution

Answer

The area of the square is 16 square meters.

Step 1: Understanding the Problem

We have a square whose sides are increasing at a rate of \(1 \text{ m/s}\). We need to find the area of the square when the area is increasing at a rate of \(8 \text{ m}^2/\text{s}\).

Step 2: Establish Known Equations

Let \(s\) be the side length of the square, which is increasing at \(\frac{ds}{dt} = 1 \text{ m/s}\). The area \(A\) of the square is given by \(A = s^2\), and the rate of change of the area is \(\frac{dA}{dt}\).

Step 3: Differentiate Area with Respect to Time

Differentiate the area formula with respect to time to find \(\frac{dA}{dt}\). Using the chain rule: \(\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}\).

Step 4: Substitute Known Values into the Differentiated Equation

We know \(\frac{ds}{dt} = 1 \text{ m/s}\) and \(\frac{dA}{dt} = 8 \text{ m}^2/\text{s}\). Substitute these values into the differentiated equation: \[ 8 = 2s \cdot 1 \] which simplifies to \(s = 4\).

Step 5: Calculate the Area of the Square

Now that we know \(s = 4\), substitute this back into the area formula to find the area of the square: \[ A = s^2 = 4^2 = 16 \text{ square meters}\].

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It helps us to understand how a quantity changes concerning another variable, often time. In the context of this square problem, we are interested in how the area of the square changes as the side length increases over time. The goal is to find the derivative of the area with respect to time, written as \( \frac{dA}{dt} \). This can be done using the chain rule.The chain rule in calculus states that if you have a function of a function, you differentiate the outer function and multiply it by the derivative of the inner function. For our square, we have \( A = s^2 \). When we differentiate with respect to time, we get:

• Differentiate the area formula \( A = s^2 \)
• Apply the chain rule: \( \frac{dA}{dt} = 2s \cdot \frac{ds}{dt} \)

This tells us how fast the area is changing in relation to how fast the side length is growing. Understanding this rate of change is crucial for solving related rate problems.

Related Rates

Related rates are a common type of problem in calculus where you have two or more quantities that change over time, and you want to find the rate of change of one of these quantities in terms of the other. The idea is to relate these rates using differentiation based on the known relationship between the quantities.In our square problem, the side length \( s \) and the area \( A \) are inter-related. We know the rate at which the side length grows, \( \frac{ds}{dt} = 1 \text{ m/s} \), and we want to find when the rate of the area increase, \( \frac{dA}{dt} \), reaches 8 \( \text{ m}^2/\text{s} \).Since these two rates are connected through the differentiation of the area formula, we use:

• Substitute known values into the differentiated equation: \( 8 = 2s \cdot 1 \)
• Solve for \( s \) to discover \( s = 4 \)

By understanding how different variables change concerning one another, we can determine crucial information like the side length at a given rate of area change.

Geometry

Geometry provides the visual and structural understanding of sizes, shapes, and relative positions of figures. For problems involving geometric shapes like squares, it's important to understand how each part of the shape behaves and changes.With a square, the relationship between the side length and the area is defined by the formula \( A = s^2 \). Here, as the side length changes, so does the area in a very predictable manner thanks to this geometric relationship. Knowing this, when we find that the side length of our square is \( s = 4 \), determining the area becomes straightforward:

• Substitute \( s \) back into the area formula: \( A = s^2 = 4^2 \)
• Calculate the area: \( A = 16 \text{ square meters} \)

Through geometric intuition and algebra, you can easily handle and solve these types of problems by connecting the properties of the shapes with calculus tools.