Engineering Mathematics II for Land Survey Level 5 students requires mastery of algebraic tools essential for precise calculations in surveying tasks. The binomial expansion is a powerful mathematical technique that facilitates the expansion of expressions raised to a power without direct multiplication, saving time and reducing error in field computations. In land surveying, binomial expansion helps approximate values such as distance measurements, area calculations, and coordinate transformations where expressions involve powers of binomials. This chapter focuses on applying the binomial theorem to expand expressions, enabling surveyors to handle complex algebraic calculations with confidence.

1.1 Expression Expansion as per Binomial Theorem Expansion

The binomial theorem provides a systematic method to expand expressions of the form \((a + b)^n\), where \(n\) is a positive integer. This theorem is crucial in land surveying for simplifying calculations involving powers of sums, such as when adjusting coordinates or calculating error margins. The general formula for binomial expansion is:

$$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$

where \(\binom{n}{k} = \frac{n!}{k! (n-k)!}\) is the binomial coefficient.

1.1.1 Understanding the Binomial Theorem Formula

The binomial theorem expresses the expansion of \((a + b)^n\) as a sum of terms involving binomial coefficients. Each term in the expansion follows the pattern:

$$ \binom{n}{k} a^{n-k} b^k $$

where \(k\) ranges from 0 to \(n\).

Worked Examples

Example 1: Expand \((2 + x)^3\).

Given: \(a = 2\), \(b = x\), \(n = 3\)

$$ (2 + x)^3 = \sum_{k=0}^3 \binom{3}{k} 2^{3-k} x^k $$

Calculate each term:

For \(k=0\):

$$ \binom{3}{0} 2^{3} x^{0} = 1 \times 8 \times 1 = 8 $$

For \(k=1\):

$$ \binom{3}{1} 2^{2} x^{1} = 3 \times 4 \times x = 12x $$

For \(k=2\):

$$ \binom{3}{2} 2^{1} x^{2} = 3 \times 2 \times x^{2} = 6x^{2} $$

For \(k=3\):

$$ \binom{3}{3} 2^{0} x^{3} = 1 \times 1 \times x^{3} = x^{3} $$

Sum all terms:

$$ 8 + 12x + 6x^{2} + x^{3} $$

Answer: \(8 + 12x + 6x^{2} + x^{3}\)

Example 2: Expand \((1 + 3h)^4\).

Given: \(a = 1\), \(b = 3h\), \(n = 4\)

$$ (1 + 3h)^4 = \sum_{k=0}^4 \binom{4}{k} 1^{4-k} (3h)^k $$

Calculate each term:

For \(k=0\):

$$ \binom{4}{0} 1^{4} (3h)^0 = 1 \times 1 \times 1 = 1 $$

For \(k=1\):

$$ \binom{4}{1} 1^{3} (3h)^1 = 4 \times 1 \times 3h = 12h $$

For \(k=2\):

$$ \binom{4}{2} 1^{2} (3h)^2 = 6 \times 1 \times 9h^{2} = 54h^{2} $$

For \(k=3\):

$$ \binom{4}{3} 1^{1} (3h)^3 = 4 \times 1 \times 27h^{3} = 108h^{3} $$

For \(k=4\):

$$ \binom{4}{4} 1^{0} (3h)^4 = 1 \times 1 \times 81h^{4} = 81h^{4} $$

Sum all terms:

$$ 1 + 12h + 54h^{2} + 108h^{3} + 81h^{4} $$

Answer: \(1 + 12h + 54h^{2} + 108h^{3} + 81h^{4}\)

Example 3: Expand \((x - 2)^3\) using binomial theorem.

Given: \(a = x\), \(b = -2\), \(n = 3\)

$$ (x - 2)^3 = \sum_{k=0}^3 \binom{3}{k} x^{3-k} (-2)^k $$

Calculate each term:

For \(k=0\):

$$ \binom{3}{0} x^{3} (-2)^0 = 1 \times x^{3} \times 1 = x^{3} $$

For \(k=1\):

$$ \binom{3}{1} x^{2} (-2)^1 = 3 \times x^{2} \times (-2) = -6x^{2} $$

For \(k=2\):

$$ \binom{3}{2} x^{1} (-2)^2 = 3 \times x \times 4 = 12x $$

For \(k=3\):

$$ \binom{3}{3} x^{0} (-2)^3 = 1 \times 1 \times (-8) = -8 $$

Sum all terms:

$$ x^{3} - 6x^{2} + 12x - 8 $$

Answer: \(x^{3} - 6x^{2} + 12x - 8\)

1.1.2 Expansion for Surveying Measurements with Small Quantities

In land surveying, measurements often involve small increments or corrections expressed as binomials with small variables (e.g., \((1 + x)^n\) where \(x\) is small). Using binomial expansion helps approximate these expressions by truncating higher-order terms, reducing calculation complexity without significant loss of accuracy.

The binomial expansion for \((1 + x)^n\) is:

$$ (1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k $$

For small \(x\), terms with higher powers of \(x\) become negligible.

Worked Examples

Example 1: Approximate \((1 + 0.02)^4\) by expanding up to the second power of \(x\).

Given: \(x = 0.02\), \(n = 4\)

$$ (1 + 0.02)^4 = \sum_{k=0}^4 \binom{4}{k} (0.02)^k $$

Calculate terms up to \(k=2\):

For \(k=0\):

$$ \binom{4}{0} (0.02)^0 = 1 \times 1 = 1 $$

For \(k=1\):

$$ \binom{4}{1} (0.02)^1 = 4 \times 0.02 = 0.08 $$

For \(k=2\):

$$ \binom{4}{2} (0.02)^2 = 6 \times 0.0004 = 0.0024 $$

Sum these terms:

$$ 1 + 0.08 + 0.0024 = 1.0824 $$

Answer: \(1.0824\) (approximate value of \((1 + 0.02)^4\))

Example 2: Calculate \((1 - 0.01)^5\) by expanding terms up to the third power.

Given: \(x = -0.01\), \(n = 5\)

$$ (1 - 0.01)^5 = \sum_{k=0}^5 \binom{5}{k} (-0.01)^k $$

Calculate terms for \(k=0\) to \(k=3\):

For \(k=0\):

$$ \binom{5}{0} (-0.01)^0 = 1 \times 1 = 1 $$

For \(k=1\):

$$ \binom{5}{1} (-0.01)^1 = 5 \times (-0.01) = -0.05 $$

For \(k=2\):

$$ \binom{5}{2} (-0.01)^2 = 10 \times 0.0001 = 0.001 $$

For \(k=3\):

$$ \binom{5}{3} (-0.01)^3 = 10 \times (-0.000001) = -0.00001 $$

Sum these terms:

$$ 1 - 0.05 + 0.001 - 0.00001 = 0.95099 $$

Answer: \(0.95099\) (approximate value of \((1 - 0.01)^5\))

Example 3: Approximate \((1 + 0.005)^6\) up to the second power term.

Given: \(x = 0.005\), \(n = 6\)

$$ (1 + 0.005)^6 = \sum_{k=0}^6 \binom{6}{k} (0.005)^k $$

Calculate terms for \(k=0\) to \(k=2\):

For \(k=0\):

$$ \binom{6}{0} (0.005)^0 = 1 \times 1 = 1 $$

For \(k=1\):

$$ \binom{6}{1} (0.005)^1 = 6 \times 0.005 = 0.03 $$

For \(k=2\):

$$ \binom{6}{2} (0.005)^2 = 15 \times 0.000025 = 0.000375 $$

Sum these terms:

$$ 1 + 0.03 + 0.000375 = 1.030375 $$

Answer: \(1.030375\) (approximate value of \((1 + 0.005)^6\))

1.1.3 Using Binomial Expansion to Calculate Surveying Areas and Volumes

Land surveyors often calculate areas and volumes where dimensions include small variations or expansions expressed as binomial expressions. Binomial expansion helps in estimating these values by expanding powers of sums, facilitating measurements of irregular plots or earthworks.

$$ (a + b)^2 = a^2 + 2ab + b^2 $$

This principle extends to higher powers for volume or other measurements.

Worked Examples

Example 1: Calculate the approximate area of a square plot with side length \(20 + 0.5\) m.

Given: \(a = 20\) m, \(b = 0.5\) m

$$ (20 + 0.5)^2 = 20^2 + 2 \times 20 \times 0.5 + 0.5^2 $$

Calculate each term:

$$ 20^2 = 400 $$

$$ 2 \times 20 \times 0.5 = 20 $$

$$ 0.5^2 = 0.25 $$

Sum terms:

$$ 400 + 20 + 0.25 = 420.25 \text{ m}^2 $$

Answer: \(420.25 \text{ m}^2\)

Example 2: Estimate the volume of earthwork for a cube with side length \((5 + 0.1)\) m.

Given: \(a = 5\) m, \(b = 0.1\) m, volume \(V = (a + b)^3\)

$$ (5 + 0.1)^3 = \sum_{k=0}^3 \binom{3}{k} 5^{3-k} (0.1)^k $$

Calculate terms:

For \(k=0\):

$$ \binom{3}{0} 5^{3} (0.1)^0 = 1 \times 125 \times 1 = 125 $$

For \(k=1\):

$$ \binom{3}{1} 5^{2} (0.1)^1 = 3 \times 25 \times 0.1 = 7.5 $$

For \(k=2\):

$$ \binom{3}{2} 5^{1} (0.1)^2 = 3 \times 5 \times 0.01 = 0.15 $$

For \(k=3\):

$$ \binom{3}{3} 5^{0} (0.1)^3 = 1 \times 1 \times 0.001 = 0.001 $$

Sum terms:

$$ 125 + 7.5 + 0.15 + 0.001 = 132.651 \text{ m}^3 $$

Answer: \(132.651 \text{ m}^3\)

Example 3: Calculate the approximate volume of a cuboid with dimensions \((10 + 0.2)\) m, \((5 + 0.1)\) m, and height 3 m, using binomial expansion for the base area.

Given: Length \(L = 10 + 0.2\) m, Width \(W = 5 + 0.1\) m, Height \(H = 3\) m

First, calculate base area \(A = L \times W = (10 + 0.2)(5 + 0.1)\)

Use expansion:

$$ (10 + 0.2)(5 + 0.1) = 10 \times 5 + 10 \times 0.1 + 0.2 \times 5 + 0.2 \times 0.1 $$

Calculate each term:

$$ 10 \times 5 = 50 $$

$$ 10 \times 0.1 = 1 $$

$$ 0.2 \times 5 = 1 $$

$$ 0.2 \times 0.1 = 0.02 $$

Sum terms:

$$ 50 + 1 + 1 + 0.02 = 52.02 \text{ m}^2 $$

Now calculate volume:

$$ V = A \times H = 52.02 \times 3 = 156.06 \text{ m}^3 $$

Answer: \(156.06 \text{ m}^3\)

Chapter Summary

This chapter focused on applying the binomial theorem to expand algebraic expressions raised to a power, enabling the simplification of complex polynomial forms. The process of expression expansion was demonstrated step-by-step, highlighting the use of binomial coefficients and powers of terms. Following this, the chapter addressed the calculation of roots of numbers, explaining methods to approximate square roots and other roots using binomial expansion techniques. The concept of error analysis was introduced to understand how small changes in input values affect the output, emphasizing the importance of estimating errors in practical engineering calculations. Throughout the chapter, the relationship between small variations and their propagation through mathematical functions was explored, reinforcing the accuracy of approximations. The coverage of these topics equips students with essential tools to handle expansions, root approximations, and error estimation in engineering contexts.