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Chapter 4: Problem 10

If the side \(e\) of a square is increased by \(1 \%,\) then the area is increased approximately (A) \(0.02 e\) (B) \(0.02 e^{2}\) (C) \(0.01 e^{2}\) (D) \(0.01 e\)

Short Answer

Expert verified
The approximate increase in area is \( 0.02e^{2} \), so the answer is (B).

Step by step solution

01

Understand the Problem

We need to find the approximate increase in the area of a square when its side length is increased by 1%. This increase must be expressed in terms of the original side length \( e \).
02

Original Area Calculation

If the original side length is \( e \), then the original area of the square is given by \( A = e^2 \).
03

New Side Length

If the side \( e \) is increased by 1%, then the new side length is \( e + 0.01e = 1.01e \).
04

Calculate New Area

The new area, using the increased side length, is \( A_{ ext{new}} = (1.01e)^{2} = 1.0201e^{2} \).
05

Calculate Increase in Area

The increase in area is the new area minus the original area: \( \Delta A = 1.0201e^{2} - e^{2} = 0.0201e^{2} \).
06

Approximate the Increase

Simplifying the increase, we see that it approximately equals \( 0.02e^{2} \) because \( 0.0201e^{2} \approx 0.02e^{2} \). This makes use of the approximation that 0.0201 is close to 0.02.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area of a Square
The concept of the area of a square is quite straightforward. The square is a geometric shape that has four equal sides. To calculate the area, you simply square the length of one of its sides.

The formula to find the area of a square is given by:
  • Let the side of the square be denoted by \( e \).
  • The area \( A \) of the square can then be calculated as \( A = e^2 \).
This formula comes from multiplying the side by itself, since both the length and the width are equal in a square.

Calculating the area of a square helps in many real-life problems, such as determining the space available for a garden or the material needed for flooring.
Percentage Increase
Understanding percentage increase is essential when evaluating how a number grows relative to its original value. To find the percentage increase, you determine how much larger one value is compared to another.

Here's a basic breakdown of how percentage increase works:
  • First, find the absolute increase in value by subtracting the original amount from the new amount.
  • Then, divide this increase by the absolute original value.
  • Finally, multiply the result by 100 to get a percentage.
Using the problem's context:
  • The side of the square, \( e \), is increased by 1%, meaning the length is adjusted by \( e + 0.01e = 1.01e \).
  • This small adjustment in side increases the area significantly, showcasing the compound effect of percentage increases on squared quantities.
Problem-Solving Steps
Solving calculus problems often involves a methodical approach, and following specific steps can help break down the process. Let's discuss the problem-solving steps, which can be applied to problems like the one in this exercise.

Here are the essential steps to tackle any problem efficiently:
  • Understand the Problem: Identify what is being asked. In this scenario, you needed to find the increase in area after a side length change.
  • Original Calculation: Begin by determining what you know. Calculate the original measure using known formulas (e.g., the original area \( A = e^2 \)).
  • New Value Determination: Adjust the known value according to given changes (e.g., new side length \( 1.01e \)).
  • Calculate New Outcome: Use the new values to find the new result (e.g., \( A_{new} = (1.01e)^2 \)).
  • Determine the Change: Subtract the old from the new to find the increase (e.g., \( \Delta A = A_{new} - A \)).
  • Approximate if Needed: Sometimes results can be approximated for simplification (e.g., \( 0.0201e^2 \approx 0.02e^2 \)).
By consistently applying these steps, solving similar problems becomes straightforward, reinforcing your problem-solving skills across different contexts.

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Most popular questions from this chapter

A tangent to the curve \(y^{2}-x y+9=0\) is vertical when (A) \(y=0\) (B) \(y=\pm \sqrt{3}\) (C) \(y=\frac{1}{2}\) (D) \(y=\pm 3\)

The displacement from the origin of a particle moving on a line is given by \(s=t^{4}-4 t^{3}\). The maximum displacement during the time interval \(-2 \leqq t \leqq 4\) is (A) 27 (B) 3 (C) 48 (D) 16

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