By the end of this chapter, you will be able to:
Mastering these skills will help you solve real engineering problems with confidence and precision in your future trade.
Carry out binomial expansion is an essential skill for Mechanical Production Technicians in Kenya, enabling them to simplify and work with expressions involving powers of sums. It is particularly useful in calculations related to tolerances, stress-strain relationships, and approximations in mechanical design. This chapter develops the mathematical foundation for expanding binomial expressions, focusing on powers and coefficients necessary to handle expansions up to the fourth power and beyond.
Binomial powers refer to the exponent applied to a binomial expression such as \((a + b)^n\), where \(n\) is a non-negative integer. Expanding such powers into sums of terms allows Mechanical Production Technicians to estimate values and simplify complex expressions encountered in mechanical component design and analysis. The general expansion follows a pattern where the powers of the first term decrease, and the powers of the second term increase in each successive term.
The binomial theorem states that for any integer \(n \geq 0\),
$$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$
where \(\binom{n}{k}\) are the binomial coefficients.
Example 1: Expand \((x + 2)^2\) for a mechanical part dimension tolerance calculation.
Given: \(a = x\), \(b = 2\), \(n = 2\)
$$ (x + 2)^2 = \binom{2}{0} x^2 (2)^0 + \binom{2}{1} x^{1} (2)^{1} + \binom{2}{2} x^{0} (2)^2 $$
$$ = 1 \times x^2 \times 1 + 2 \times x \times 2 + 1 \times 1 \times 4 $$
$$ = x^2 + 4x + 4 $$
Answer: \(x^2 + 4x + 4\)
Example 2: Expand \((3 + y)^3\) where \(y\) represents a small deformation value in mm.
Given: \(a = 3\), \(b = y\), \(n = 3\)
$$ (3 + y)^3 = \binom{3}{0} 3^3 y^0 + \binom{3}{1} 3^2 y^1 + \binom{3}{2} 3^1 y^2 + \binom{3}{3} 3^0 y^3 $$
$$ = 1 \times 27 \times 1 + 3 \times 9 \times y + 3 \times 3 \times y^2 + 1 \times 1 \times y^3 $$
$$ = 27 + 27y + 9y^2 + y^3 $$
Answer: \(27 + 27y + 9y^2 + y^3\)
Example 3: Expand \((a + b)^4\) where \(a\) and \(b\) represent forces in kN applied on a machine component.
Given: \(a\), \(b\), \(n = 4\)
$$ (a + b)^4 = \sum_{k=0}^4 \binom{4}{k} a^{4-k} b^k = \binom{4}{0} a^4 + \binom{4}{1} a^3 b + \binom{4}{2} a^2 b^2 + \binom{4}{3} a b^3 + \binom{4}{4} b^4 $$
$$ = 1 \times a^4 + 4 \times a^3 b + 6 \times a^2 b^2 + 4 \times a b^3 + 1 \times b^4 $$
Answer: \(a^4 + 4a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4\)
Example 4: Expand \((2x + 5)^3\) where \(x\) is a dimension in cm.
Given: \(a = 2x\), \(b = 5\), \(n = 3\)
$$ (2x + 5)^3 = \binom{3}{0} (2x)^3 (5)^0 + \binom{3}{1} (2x)^2 (5)^1 + \binom{3}{2} (2x)^1 (5)^2 + \binom{3}{3} (2x)^0 (5)^3 $$
$$ = 1 \times 8x^3 \times 1 + 3 \times 4x^2 \times 5 + 3 \times 2x \times 25 + 1 \times 1 \times 125 $$
$$ = 8x^3 + 60x^2 + 150x + 125 $$
Answer: \(8x^3 + 60x^2 + 150x + 125\)
Example 5: Expand \((m - 3)^4\) where \(m\) is a measurement in mm, using the binomial theorem with subtraction.
Given: \(a = m\), \(b = -3\), \(n = 4\)
$$ (m - 3)^4 = \sum_{k=0}^4 \binom{4}{k} m^{4-k} (-3)^k = m^4 + 4 m^3 (-3) + 6 m^2 (-3)^2 + 4 m (-3)^3 + (-3)^4 $$
$$ = m^4 - 12 m^3 + 54 m^2 - 108 m + 81 $$
Answer: \(m^4 - 12 m^3 + 54 m^2 - 108 m + 81\)
Binomial coefficients are the numerical factors in the expansion of \((a + b)^n\). These coefficients determine the relative weight of each term in the expansion and can be found using Pascal's Triangle or the combination formula. In mechanical production calculations, coefficients help to quickly identify the magnitude of terms when approximating values or calculating tolerances.
The binomial coefficient \(\binom{n}{k}\) is given by the formula:
$$ \binom{n}{k} = \frac{n!}{k! (n-k)!} $$
where \(n!\) denotes the factorial of \(n\).
Example 1: Find the binomial coefficients for the expansion of \((a + b)^3\).
Given: \(n = 3\)
Using the formula:
$$ \binom{3}{0} = \frac{3!}{0! \times 3!} = \frac{6}{1 \times 6} = 1 $$
$$ \binom{3}{1} = \frac{3!}{1! \times 2!} = \frac{6}{1 \times 2} = 3 $$
$$ \binom{3}{2} = \frac{3!}{2! \times 1!} = \frac{6}{2 \times 1} = 3 $$
$$ \binom{3}{3} = \frac{3!}{3! \times 0!} = \frac{6}{6 \times 1} = 1 $$
Answer: The coefficients are 1, 3, 3, 1
Example 2: Calculate the coefficients for \((x + y)^5\).
Given: \(n = 5\)
$$ \binom{5}{0} = \frac{5!}{0! \times 5!} = 1 $$
$$ \binom{5}{1} = \frac{5!}{1! \times 4!} = \frac{120}{1 \times 24} = 5 $$
$$ \binom{5}{2} = \frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = 10 $$
$$ \binom{5}{3} = \frac{5!}{3! \times 2!} = \frac{120}{6 \times 2} = 10 $$
$$ \binom{5}{4} = \frac{5!}{4! \times 1!} = \frac{120}{24 \times 1} = 5 $$
$$ \binom{5}{5} = \frac{5!}{5! \times 0!} = 1 $$
Answer: The coefficients are 1, 5, 10, 10, 5, 1
Example 3: Find the coefficient of the fourth term in the expansion of \((a + b)^6\).
Given: \(n = 6\), term number \(k+1 = 4\) (so \(k = 3\))
$$ \binom{6}{3} = \frac{6!}{3! \times 3!} = \frac{720}{6 \times 6} = 20 $$
Answer: The coefficient of the fourth term is 20
Example 4: Determine the coefficients of \((2x + 3)^4\) using Pascal's Triangle or combination formula.
Given: \(n = 4\)
Coefficients from Pascal's Triangle for \(n=4\) are \(1, 4, 6, 4, 1\).
Using combination formula for confirmation:
$$ \binom{4}{0} = 1, \quad \binom{4}{1} = 4, \quad \binom{4}{2} = 6, \quad \binom{4}{3} = 4, \quad \binom{4}{4} = 1 $$
Answer: The coefficients are 1, 4, 6, 4, 1
Example 5: Calculate the coefficient of the term containing \(x^2 y^3\) in the expansion of \((x + y)^5\).
Given: \(n = 5\), powers of \(x\) and \(y\) are 2 and 3 respectively, so \(k = 3\).
$$ \binom{5}{3} = \frac{5!}{3! \times 2!} = \frac{120}{6 \times 2} = 10 $$
Answer: The coefficient of the term \(x^2 y^3\) is 10
Pascal's triangle is a triangular array of numbers that provides coefficients for expanding binomial expressions, which are common in mechanical production calculations such as tolerances and material strength estimations. In Kenyan mechanical workshops and manufacturing units, understanding Pascal's triangle helps technicians quickly expand powers of binomials without extensive multiplication, saving time and reducing errors in design computations. The triangle's structure reveals patterns useful for both small and moderately large powers of binomials, especially relevant in component stress analysis and machining parameter calculations.
Pascal's triangle starts with a single 1 at the top. Each number inside the triangle is found by adding the two numbers directly above it from the previous row. This pattern generates the coefficients used in binomial expansions.
$$ \begin{array}{ccccccc} &&&&1&&\\ &&&1&&1&\\ &&1&&2&&1\\ &1&&3&&3&&1\\ 1&&4&&6&&4&&1\\ \end{array} $$
The first and last number of each row is always 1. The rows correspond to the powers of the binomial expansion, starting from power 0 at the top.
Example 1: Find the coefficients for the expansion of \((a + b)^3\) using Pascal's triangle.
Given: Power \(n=3\)
Row 3 in Pascal's triangle is: \(1, 3, 3, 1\)
Answer: The coefficients are \(1, 3, 3, 1\).
Example 2: Determine the coefficients for \((x + y)^5\).
Given: Power \(n=5\)
Using Pascal's triangle rows:
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Answer: Coefficients are \(1, 5, 10, 10, 5, 1\).
Example 3: Find the coefficient of the 4th term in the expansion of \((m + n)^6\).
Given: Power \(n=6\), term number \(k=4\)
The coefficient corresponds to the 3rd index in row 6 (starting at index 0):
Row 6: \(1, 6, 15, 20, 15, 6, 1\)
Coefficient = 20
Answer: Coefficient of 4th term is \(20\).
Example 4: Using Pascal's triangle, find the coefficients for \((2a + 3b)^4\).
Given: Power \(n=4\)
Row 4 coefficients: \(1, 4, 6, 4, 1\)
Answer: Coefficients are \(1, 4, 6, 4, 1\).
Example 5: Verify the coefficient of the 5th term for \((p + q)^7\).
Given: Power \(n=7\), term number \(k=5\)
Row 7: \(1, 7, 21, 35, 35, 21, 7, 1\)
Coefficient at index 4 (term 5) = 35
Answer: Coefficient is \(35\).
The triangle exhibits symmetry and relationships between coefficients that simplify expansions:
Example 1: Show that the sum of coefficients for \((a + b)^4\) is \(2^4\).
Row 4 coefficients: \(1 + 4 + 6 + 4 + 1 = 16\)
Calculate:
$$2^4 = 16$$
Answer: Sum of coefficients is \(16\).
Example 2: Verify that the 3rd coefficient of \((x + y)^5\) equals \(\binom{5}{2}\).
Given: \(n=5\), \(k=2\)
Calculate combination:
$$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{120}{2 \times 6} = 10$$
From Pascal's triangle row 5: coefficients are \(1, 5, 10, 10, 5, 1\)
3rd coefficient = 10
Answer: Coefficient equals \(10\).
Example 3: Find the coefficient of the 2nd term in \((3m + 4n)^6\) using symmetry.
Row 6: \(1, 6, 15, 20, 15, 6, 1\)
2nd term coefficient = 6
Answer: Coefficient is \(6\).
Example 4: Calculate the sum of coefficients for \((p + q)^7\).
Sum = \(2^7 = 128\)
Row 7 coefficients: \(1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128\)
Answer: Sum of coefficients is \(128\).
Example 5: Using Pascal's triangle, find the coefficient of \(a^3b^2\) in \((a + b)^5\).
The term with \(a^{5-k} b^k\), \(k=2\):
Coefficient = \(\binom{5}{2} = 10\)
Answer: Coefficient is \(10\).
To find a specific term in \((a + b)^n\), use the coefficient from Pascal's triangle and powers of \(a\) and \(b\) that sum to \(n\).
The general term is:
$$T_{k+1} = \binom{n}{k} a^{n-k} b^{k}$$
where \(k=0, 1, 2, ..., n\).
Example 1: Find the 3rd term in \((x + y)^4\).
Given: \(n=4\), \(k=2\)
Coefficient \(\binom{4}{2} = 6\)
Term:
$$T_3 = 6 x^{4-2} y^{2} = 6 x^{2} y^{2}$$
Answer: \(6 x^{2} y^{2}\)
Example 2: Find the 5th term in \((2a + 3b)^6\).
Given: \(n=6\), \(k=4\)
Coefficient \(\binom{6}{4} = 15\)
Term:
$$T_5 = 15 (2a)^{6-4} (3b)^4 = 15 (2a)^2 (3b)^4$$
Calculate powers:
$$(2a)^2 = 4a^2$$
$$(3b)^4 = 81b^4$$
Multiply:
$$15 \times 4 \times 81 = 15 \times 324 = 4860$$
Term:
$$4860 a^{2} b^{4}$$
Answer: \(4860 a^{2} b^{4}\)
Example 3: Find the 2nd term in \((m, n)^5\).
Given: \(n=5\), \(k=1\), note the minus sign.
Coefficient \(\binom{5}{1} = 5\)
Term:
$$T_2 = 5 m^{5-1} (-n)^1 = 5 m^{4} (-n) = -5 m^{4} n$$
Answer: \(-5 m^{4} n\)
Example 4: Find the 4th term in \((x + 2y)^7\).
Given: \(n=7\), \(k=3\)
Coefficient \(\binom{7}{3} = 35\)
Term:
$$T_4 = 35 x^{7-3} (2y)^3 = 35 x^{4} (8 y^{3}) = 280 x^{4} y^{3}$$
Answer: \(280 x^{4} y^{3}\)
Example 5: Find the 6th term in \((3p - 4q)^6\).
Given: \(n=6\), \(k=5\)
Coefficient \(\binom{6}{5} = 6\)
Term:
$$T_6 = 6 (3p)^{6-5} (-4q)^5 = 6 (3p)^1 (-4q)^5$$
Calculate powers:
$$(3p)^1 = 3p$$
$$(-4q)^5 = (-4)^5 q^5 = -1024 q^5$$
Multiply:
$$6 \times 3 \times (-1024) = -18432$$
Term:
$$-18432 p q^{5}$$
Answer: \(-18432 p q^{5}\)
Pascal's triangle is practical for expansions up to about the 7th or 8th power due to the increasing size of coefficients and complexity of terms. In mechanical production tasks such as calculating load distributions or tolerances involving binomial expressions raised to moderate powers, Pascal's triangle provides a quick method for coefficient determination. For higher powers, factorial or combination formulas are more efficient or computer-aided techniques are used.
Example 1: Using Pascal's triangle, find coefficients for \((a + b)^8\).
Row 8 coefficients are obtained by adding adjacent pairs from row 7:
Row 7: \(1, 7, 21, 35, 35, 21, 7, 1\)
Add pairs:
\(1+7=8\), \(7+21=28\), \(21+35=56\), \(35+35=70\), \(35+21=56\), \(21+7=28\), \(7+1=8\)
Row 8:
\(1, 8, 28, 56, 70, 56, 28, 8, 1\)
Answer: Coefficients are \(1, 8, 28, 56, 70, 56, 28, 8, 1\).
Example 2: Find the coefficient of the 5th term in \((x + y)^8\).
Coefficient = 70 (from previous example)
Answer: Coefficient is \(70\).
Example 3: Find the 6th term in \((2m + 3n)^8\).
Given: \(n=8\), \(k=5\)
Coefficient \(\binom{8}{5} = 56\)
Term:
$$T_6 = 56 (2m)^{8-5} (3n)^5 = 56 (2m)^3 (3n)^5$$
Calculate powers:
$$(2m)^3 = 8 m^{3}$$
$$(3n)^5 = 243 n^{5}$$
Multiply:
$$56 \times 8 \times 243 = 56 \times 1944 = 108864$$
Term:
$$108864 m^{3} n^{5}$$
Answer: \(108864 m^{3} n^{5}\)
Example 4: Why is Pascal's triangle less practical for powers above 8?
Example 5: Use combination formula to find coefficient for \((a + b)^{10}\) 4th term.
Given: \(n=10\), \(k=3\)
Calculate:
$$\binom{10}{3} = \frac{10!}{3! \times 7!} = \frac{3628800}{6 \times 5040} = \frac{3628800}{30240} = 120$$
Answer: Coefficient is \(120\).
The binomial theorem provides a general formula to expand powers of binomials \((a + b)^n\) for any positive integer \(n\). This theorem is crucial for mechanical production technicians when dealing with polynomial expressions in design calculations, stress analysis, and tolerance stacking. It allows quick and accurate expansion without repeated multiplication, which is essential in Kenyan manufacturing environments where precision and efficiency are valued.
The binomial theorem states that for any integer \(n \geq 0\),
$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} $$
where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient, giving the number of ways to choose \(k\) elements from \(n\).
Example 1: Expand \((x + y)^2\) using the binomial theorem.
Given: \(n=2\), \(a=x\), \(b=y\)
Formula:
$$(x + y)^2 = \binom{2}{0} x^{2} y^{0} + \binom{2}{1} x^{1} y^{1} + \binom{2}{2} x^{0} y^{2}$$
Calculate coefficients:
\(\binom{2}{0} = 1\), \(\binom{2}{1} = 2\), \(\binom{2}{2} = 1\)
Write terms:
$$1 \times x^{2} \times 1 + 2 \times x^{1} \times y^{1} + 1 \times 1 \times y^{2}$$
Simplify:
$$x^{2} + 2xy + y^{2}$$
Answer: \(x^{2} + 2xy + y^{2}\)
Example 2: Expand \((a + b)^3\).
Given: \(n=3\)
Formula:
$$(a + b)^3 = \sum_{k=0}^3 \binom{3}{k} a^{3-k} b^{k}$$
Calculate coefficients:
\(\binom{3}{0} = 1\), \(\binom{3}{1} = 3\), \(\binom{3}{2} = 3\), \(\binom{3}{3} = 1\)
Write terms:
$$1 a^{3} b^{0} + 3 a^{2} b^{1} + 3 a^{1} b^{2} + 1 a^{0} b^{3}$$
Simplify:
$$a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$$
Answer: \(a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}\)
Example 3: Expand \((2x + 3y)^2\).
Given: \(n=2\), \(a=2x\), \(b=3y\)
Formula:
$$(2x + 3y)^2 = \sum_{k=0}^2 \binom{2}{k} (2x)^{2-k} (3y)^k$$
Calculate coefficients:
\(\binom{2}{0} = 1\), \(\binom{2}{1} = 2\), \(\binom{2}{2} = 1\)
Write terms:
$$1 (2x)^2 (3y)^0 + 2 (2x)^1 (3y)^1 + 1 (2x)^0 (3y)^2$$
Calculate powers:
$$(2x)^2 = 4x^{2}$$
$$(3y)^0 = 1$$
$$(2x)^1 = 2x$$
$$(3y)^1 = 3y$$
$$(2x)^0 = 1$$
$$(3y)^2 = 9 y^{2}$$
Multiply:
$$1 \times 4x^{2} \times 1 = 4x^{2}$$
$$2 \times 2x \times 3y = 12 x y$$
$$1 \times 1 \times 9 y^{2} = 9 y^{2}$$
Sum:
$$4x^{2} + 12 x y + 9 y^{2}$$
Answer: \(4x^{2} + 12 x y + 9 y^{2}\)
Example 4: Expand \((m, n)^3\).
Given: \(n=3\), \(a=m\), \(b=-n\)
Formula:
$$(m, n)^3 = \sum_{k=0}^3 \binom{3}{k} m^{3-k} (-n)^k$$
Calculate coefficients:
\(\binom{3}{0} = 1\), \(\binom{3}{1} = 3\), \(\binom{3}{2} = 3\), \(\binom{3}{3} = 1\)
Write terms:
$$1 m^{3} (-n)^0 + 3 m^{2} (-n)^1 + 3 m^{1} (-n)^2 + 1 m^{0} (-n)^3$$
Calculate powers:
$$(-n)^0 = 1$$
$$(-n)^1 = -n$$
$$(-n)^2 = n^{2}$$
$$(-n)^3 = -n^{3}$$
Multiply:
$$1 \times m^{3} \times 1 = m^{3}$$
$$3 \times m^{2} \times (-n) = -3 m^{2} n$$
$$3 \times m \times n^{2} = 3 m n^{2}$$
$$1 \times 1 \times (-n^{3}) = -n^{3}$$
Sum:
$$m^{3} - 3 m^{2} n + 3 m n^{2} - n^{3}$$
Answer: \(m^{3} - 3 m^{2} n + 3 m n^{2} - n^{3}\)
Example 5: Expand \((x + 2y)^4\).
Given: \(n=4\), \(a=x\), \(b=2y\)
Formula:
$$(x + 2y)^4 = \sum_{k=0}^4 \binom{4}{k} x^{4-k} (2y)^k$$
Coefficients:
\(\binom{4}{0} = 1\), \(\binom{4}{1} = 4\), \(\binom{4}{2} = 6\), \(\binom{4}{3} = 4\), \(\binom{4}{4} = 1\)
Terms:
$$1 x^{4} (2y)^0 + 4 x^{3} (2y)^1 + 6 x^{2} (2y)^2 + 4 x^{1} (2y)^3 + 1 x^{0} (2y)^4$$
Calculate powers:
$$(2y)^0 = 1$$
$$(2y)^1 = 2 y$$
$$(2y)^2 = 4 y^{2}$$
$$(2y)^3 = 8 y^{3}$$
$$(2y)^4 = 16 y^{4}$$
Multiply terms:
$$1 \times x^{4} \times 1 = x^{4}$$
$$4 \times x^{3} \times 2 y = 8 x^{3} y$$
$$6 \times x^{2} \times 4 y^{2} = 24 x^{2} y^{2}$$
$$4 \times x \times 8 y^{3} = 32 x y^{3}$$
$$1 \times 1 \times 16 y^{4} = 16 y^{4}$$
Sum:
$$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + 16 y^{4}$$
Answer: \(x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + 16 y^{4}\)
To find any term in \((a + b)^n\), the general term \(T_{k+1}\) is:
$$ T_{k+1} = \binom{n}{k} a^{n-k} b^{k} $$
where \(k = 0, 1, 2, ..., n\).
This formula is used to find specific terms without expanding the entire expression, useful in mechanical engineering problems involving polynomial expressions.
Example 1: Find the 4th term of \((x + y)^5\).
Given: \(n=5\), \(k=3\)
Calculate coefficient:
$$\binom{5}{3} = \frac{5!}{3! 2!} = \frac{120}{6 \times 2} = 10$$
Term:
$$T_4 = 10 x^{5-3} y^{3} = 10 x^{2} y^{3}$$
Answer: \(10 x^{2} y^{3}\)
Example 2: Find the 3rd term in \((2m + 3n)^4\).
Given: \(n=4\), \(k=2\)
Calculate coefficient:
$$\binom{4}{2} = \frac{4!}{2! 2!} = \frac{24}{2 \times 2} = 6$$
Calculate powers:
$$(2m)^{4-2} = (2m)^2 = 4 m^{2}$$
$$(3n)^2 = 9 n^{2}$$
Multiply:
$$T_3 = 6 \times 4 m^{2} \times 9 n^{2} = 216 m^{2} n^{2}$$
Answer: \(216 m^{2} n^{2}\)
Example 3: Find the 5th term in \((3p - 2q)^6\).
Given: \(n=6\), \(k=4\), \(b = -2q\)
Calculate coefficient:
$$\binom{6}{4} = \frac{6!}{4! 2!} = \frac{720}{24 \times 2} = 15$$
Calculate powers:
$$(3p)^{6-4} = (3p)^2 = 9 p^{2}$$
$$(-2q)^4 = 16 q^{4}$$
Multiply:
$$T_5 = 15 \times 9 p^{2} \times 16 q^{4} = 2160 p^{2} q^{4}$$
Answer: \(2160 p^{2} q^{4}\)
Example 4: Find the 2nd term in \((x + 5y)^7\).
Given: \(n=7\), \(k=1\)
Calculate coefficient:
$$\binom{7}{1} = 7$$
Calculate powers:
$$(x)^{7-1} = x^{6}$$
$$(5y)^1 = 5 y$$
Multiply:
$$T_2 = 7 \times x^{6} \times 5 y = 35 x^{6} y$$
Answer: \(35 x^{6} y\)
Example 5: Find the coefficient of the 7th term in \((2a - 3b)^9\).
Given: \(n=9\), \(k=6\), \(b = -3b\)
Calculate coefficient:
$$\binom{9}{6} = \frac{9!}{6! 3!} = \frac{362880}{720 \times 6} = 84$$
Calculate powers:
$$(2a)^{9-6} = (2a)^3 = 8 a^{3}$$
$$(-3b)^6 = (-3)^6 b^{6} = 729 b^{6}$$
Multiply:
Coefficient term:
$$84 \times 8 \times 729 = 84 \times 5832 = 489,888$$
Term:
$$489,888 a^{3} b^{6}$$
Answer: \(489,888 a^{3} b^{6}\)
For mechanical production technicians, expansions of binomials with negative or fractional powers occur in advanced calculations, such as stress-strain relationships or fluid flow equations. The binomial theorem extends to these cases using infinite series, but expansions are approximated to a finite number of terms.
The general expansion for \((1 + x)^n\) where \(n\) is any real number is:
$$ (1 + x)^n = 1 + n x + \frac{n(n-1)}{2!} x^{2} + \frac{n(n-1)(n-2)}{3!} x^{3} + \cdots $$
valid for \(|x| < 1\).
Example 1: Expand \((1 + x)^{-1}\) up to \(x^{3}\).
Terms:
\(1\)
\(n x = (-1) x = -x\)
\(\frac{n(n-1)}{2!} x^{2} = \frac{-1 \times -2}{2} x^{2} = 1 x^{2} = x^{2}\)
\(\frac{n(n-1)(n-2)}{3!} x^{3} = \frac{-1 \times -2 \times -3}{6} x^{3} = \frac{-6}{6} x^{3} = -x^{3}\)
Sum:
$$1 - x + x^{2} - x^{3}$$
Answer: \(1 - x + x^{2} - x^{3}\)
Example 2: Expand \((1 + x)^{1/2}\) up to \(x^{3}\).
Terms:
\(1\)
\(n x = \frac{1}{2} x\)
\(\frac{n(n-1)}{2!} x^{2} = \frac{\frac{1}{2} \times -\frac{1}{2}}{2} x^{2} = -\frac{1}{8} x^{2}\)
\(\frac{n(n-1)(n-2)}{3!} x^{3} = \frac{\frac{1}{2} \times -\frac{1}{2} \times -\frac{3}{2}}{6} x^{3} = \frac{3}{16} x^{3}\)
Sum:
$$1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{3}{16} x^{3}$$
Answer: \(1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{3}{16} x^{3}\)
Example 3: Expand \((1 - 2x)^{-2}\) up to \(x^{2}\).
Rewrite as:
$$(1 + (-2x))^{-2}$$
Terms:
\(1\)
\(n x = -2 \times (-2x) = 4x\)
\(\frac{n(n-1)}{2!} x^{2} = \frac{-2 \times (-3)}{2} (-2x)^2 = \frac{6}{2} \times 4 x^{2} = 3 \times 4 x^{2} = 12 x^{2}\)
Sum:
$$1 + 4 x + 12 x^{2}$$
Answer: \(1 + 4 x + 12 x^{2}\)
Example 4: Approximate \(\sqrt{1 + 0.1}\) using binomial expansion up to \(x^{2}\).
Given: \(n = \frac{1}{2}\), \(x = 0.1\)
Approximation:
$$1 + \frac{1}{2} (0.1) - \frac{1}{8} (0.1)^2 = 1 + 0.05 - 0.00125 = 1.04875$$
Answer: Approximate value is \(1.04875\)
Example 5: Expand \((1 + 3x)^{-1}\) up to \(x^{2}\).
Rewrite as:
$$(1 + 3x)^{-1} = 1 + (-1) \times 3x + \frac{-1 \times (-2)}{2} (3x)^2$$
Calculate terms:
\(1\)
\(-3x\)
\(\frac{2}{2} \times 9 x^{2} = 9 x^{2}\)
Sum:
$$1 - 3 x + 9 x^{2}$$
Answer: \(1 - 3 x + 9 x^{2}\)
Binomial expansions are used in mechanical production for estimating powers of sums in load calculations, material deformation, and machining parameters. They simplify polynomial expressions, allowing technicians to predict component behavior under varying conditions efficiently.
Example 1: A shaft radius increases by \(x = 0.02\) mm from nominal \(r = 50\) mm. Estimate \((r + x)^3\) using binomial expansion.
Given: \(r=50\) mm, \(x=0.02\) mm, \(n=3\)
Expand:
$$(r + x)^3 = r^{3} + 3 r^{2} x + 3 r x^{2} + x^{3}$$
Calculate:
$$50^{3} = 125,000 \text{ mm}^3$$
$$3 \times 50^{2} \times 0.02 = 3 \times 2,500 \times 0.02 = 150$$
$$3 \times 50 \times (0.02)^2 = 3 \times 50 \times 0.0004 = 0.06$$
$$(0.02)^3 = 0.000008$$
Sum:
$$125,000 + 150 + 0.06 + 0.000008 = 125,150.060008 \text{ mm}^3$$
Answer: Volume estimate is approximately \(125,150.06\) mm\(^3\).
Example 2: Calculate \((1.05)^4\) using binomial expansion.
Given: \(a=1\), \(b=0.05\), \(n=4\)
Expand:
$$(1 + 0.05)^4 = 1^{4} + 4 \times 1^{3} \times 0.05 + 6 \times 1^{2} \times (0.05)^2 + 4 \times 1 \times (0.05)^3 + (0.05)^4$$
Calculate terms:
$$1$$
$$4 \times 0.05 = 0.2$$
$$6 \times 0.0025 = 0.015$$
$$4 \times 0.000125 = 0.0005$$
$$0.00000625$$
Sum:
$$1 + 0.2 + 0.015 + 0.0005 + 0.00000625 = 1.21550625$$
Answer: \((1.05)^4 \approx 1.2155\)
Example 3: Approximate \((1 - 0.03)^5\) using binomial expansion.
Given: \(a=1\), \(b=-0.03\), \(n=5\)
Terms:
$$1 + 5(-0.03) + 10(-0.03)^2 + 10(-0.03)^3 + 5(-0.03)^4 + (-0.03)^5$$
Calculate powers:
\(-0.03)^2 = 0.0009\)
\(-0.03)^3 = -0.000027\)
\(-0.03)^4 = 0.00000081\)
\(-0.03)^5 = -0.0000000243\)
Calculate terms:
$$1 - 0.15 + 0.009 - 0.00027 + 0.00000405 - 0.0000000243$$
Sum:
$$1 - 0.15 + 0.009 - 0.00027 + 0.00000405 - 0.0000000243 = 0.858734$$
Answer: Approximate value is \(0.8587\)
Example 4: The diameter of a rod increases by 0.01 m from 0.5 m. Estimate the volume change using \((d + x)^3\).
Given: \(d=0.5\) m, \(x=0.01\) m, \(n=3\)
Expand:
$$(d + x)^3 = d^{3} + 3 d^{2} x + 3 d x^{2} + x^{3}$$
Calculate:
$$0.5^{3} = 0.125 \text{ m}^3$$
$$3 \times 0.5^{2} \times 0.01 = 3 \times 0.25 \times 0.01 = 0.0075$$
$$3 \times 0.5 \times 0.0001 = 0.00015$$
$$(0.01)^3 = 0.000001$$
Sum:
$$0.125 + 0.0075 + 0.00015 + 0.000001 = 0.132651 \text{ m}^3$$
Answer: Volume estimate is approximately \(0.13265\) m\(^3\).
Example 5: Use binomial expansion to approximate \((1.08)^{3}\).
Given: \(a=1\), \(b=0.08\), \(n=3\)
Expand:
$$1 + 3 \times 0.08 + 3 \times (0.08)^2 + (0.08)^3$$
Calculate powers:
$$(0.08)^2 = 0.0064$$
$$(0.08)^3 = 0.000512$$
Calculate terms:
$$1 + 0.24 + 0.0192 + 0.000512 = 1.259712$$
Answer: \((1.08)^3 \approx 1.2597\)
Expanding binomial expressions with positive integer powers is a fundamental skill in mechanical production, especially when calculating forces, material properties, or dimensions raised to powers. The binomial expansion allows efficient computation without multiplying the expression repeatedly. The expansion of \((a + b)^n\) for positive integer \(n\) uses coefficients from Pascal's triangle and follows the formula:
$$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^{k} $$
where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) are the binomial coefficients.
Example 1:
Expand \((x + 2)^3\) and find the term when \(x = 4\).
Given: \(n=3\), \(a = x\), \(b=2\), \(x=4\)
$$ (x + 2)^3 = \sum_{k=0}^3 \binom{3}{k} x^{3-k} 2^k $$
$$ = \binom{3}{0} x^3 2^0 + \binom{3}{1} x^2 2^1 + \binom{3}{2} x^1 2^2 + \binom{3}{3} x^0 2^3 $$
Calculate coefficients:
\(\binom{3}{0} = 1\), \(\binom{3}{1} = 3\), \(\binom{3}{2} = 3\), \(\binom{3}{3} = 1\)
$$ = 1 \times x^3 + 3 \times x^2 \times 2 + 3 \times x \times 4 + 1 \times 8 $$
Substitute \(x=4\):
$$ = 1 \times 4^3 + 3 \times 4^2 \times 2 + 3 \times 4 \times 4 + 8 $$
Calculate powers and products:
$$ = 64 + 3 \times 16 \times 2 + 3 \times 4 \times 4 + 8 $$
$$ = 64 + 96 + 48 + 8 $$
Sum all terms:
$$ = 216 $$
Answer: 216
Example 2:
Find the expansion of \((3 + y)^4\) and evaluate for \(y = 1\).
Given: \(n=4\), \(a=3\), \(b=y\), \(y=1\)
$$ (3 + y)^4 = \sum_{k=0}^4 \binom{4}{k} 3^{4-k} y^k $$
Coefficients:
\(\binom{4}{0} = 1\), \(\binom{4}{1} = 4\), \(\binom{4}{2} = 6\), \(\binom{4}{3} = 4\), \(\binom{4}{4} = 1\)
Expansion:
$$ = 1 \times 3^4 + 4 \times 3^3 y + 6 \times 3^2 y^2 + 4 \times 3 y^3 + 1 \times y^4 $$
Substitute \(y=1\):
$$ = 81 + 4 \times 27 \times 1 + 6 \times 9 \times 1 + 4 \times 3 \times 1 + 1 $$
Calculate:
$$ = 81 + 108 + 54 + 12 + 1 $$
Sum:
$$ = 256 $$
Answer: 256
Example 3:
A mechanical part has dimensions expressed as \((2m + 0.5)^5\). Find the expanded form and calculate the value when \(m=1.2\).
Given: \(n=5\), \(a=2m\), \(b=0.5\), \(m=1.2\)
$$ (2m + 0.5)^5 = \sum_{k=0}^5 \binom{5}{k} (2m)^{5-k} (0.5)^k $$
Coefficients:
\(\binom{5}{0} = 1\), \(\binom{5}{1} = 5\), \(\binom{5}{2} = 10\), \(\binom{5}{3} = 10\), \(\binom{5}{4} = 5\), \(\binom{5}{5} = 1\)
Expand terms:
$$ = 1 \times (2m)^5 + 5 \times (2m)^4 \times 0.5 + 10 \times (2m)^3 \times (0.5)^2 + 10 \times (2m)^2 \times (0.5)^3 + 5 \times (2m) \times (0.5)^4 + 1 \times (0.5)^5 $$
Calculate powers of 0.5:
\[ (0.5)^2 = 0.25, \quad (0.5)^3 = 0.125, \quad (0.5)^4 = 0.0625, \quad (0.5)^5 = 0.03125 \]
Substitute \(m=1.2\):
Calculate powers of \(2m = 2 \times 1.2 = 2.4\):
\[
(2.4)^5 = 2.4 \times 2.4 \times 2.4 \times 2.4 \times 2.4
\]
Calculate stepwise:
\[
2.4^2 = 5.76
\]
\[
2.4^3 = 5.76 \times 2.4 = 13.824
\]
\[
2.4^4 = 13.824 \times 2.4 = 33.1776
\]
\[
2.4^5 = 33.1776 \times 2.4 = 79.62624
\]
Similarly,
\[ (2.4)^4 = 33.1776, \quad (2.4)^3 = 13.824, \quad (2.4)^2 = 5.76, \quad (2.4)^1 = 2.4 \]
Now substitute:
$$ = 1 \times 79.62624 + 5 \times 33.1776 \times 0.5 + 10 \times 13.824 \times 0.25 + 10 \times 5.76 \times 0.125 + 5 \times 2.4 \times 0.0625 + 1 \times 0.03125 $$
Calculate each term:
$$ = 79.62624 + 5 \times 16.5888 + 10 \times 3.456 + 10 \times 0.72 + 5 \times 0.15 + 0.03125 $$
$$ = 79.62624 + 82.944 + 34.56 + 7.2 + 0.75 + 0.03125 $$
Sum all:
$$ = 205.11149 $$
Answer: 205.11
Example 4:
Calculate the coefficient of the \(x^3 y^2\) term in the expansion of \((x + y)^5\).
Given: \(n=5\), term with \(x^{3}\), so \(k=2\) since powers add to 5.
Coefficient:
$$ \binom{5}{2} = \frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = 10 $$
Term:
$$ 10 \times x^{3} \times y^{2} $$
Answer: Coefficient is 10
Example 5:
A tensile force \(F\) is given by \((100 + 5x)^4\), where \(x\) is a small extension in mm. Find the term independent of \(x\) and the coefficient of the \(x^2\) term.
Given: \(n=4\), \(a=100\), \(b=5x\)
The general term is:
$$ T_{k+1} = \binom{4}{k} 100^{4-k} (5x)^k $$
$$ T_1 = \binom{4}{0} 100^4 (5x)^0 = 1 \times 100000000 \times 1 = 100000000 $$
$$ T_3 = \binom{4}{2} 100^{2} (5)^2 x^{2} = 6 \times 10000 \times 25 \times x^{2} $$
Calculate coefficient:
$$ 6 \times 10000 \times 25 = 6 \times 250000 = 1500000 $$
Answer:
Expanding binomial expressions with negative integer powers is crucial in mechanical production calculations involving inverse relations such as stiffness, compliance, or stress-strain inverses. The binomial series for negative powers is an infinite series and converges when \(|b/a| < 1\). The general formula for \((1 + x)^n\) where \(n\) is negative integer is:
$$ (1 + x)^n = 1 + n x + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots $$
The expansion continues infinitely but in practical engineering, only the first few terms are used for approximations.
Example 1:
Expand \((1 + x)^{-1}\) up to \(x^3\) terms and evaluate for \(x = 0.1\).
Given: \(n = -1\)
$$ (1 + x)^{-1} = 1 + (-1) x + \frac{(-1)(-2)}{2!} x^2 + \frac{(-1)(-2)(-3)}{3!} x^3 $$
Calculate coefficients:
$$ = 1 - x + \frac{2}{2} x^2 - \frac{6}{6} x^3 = 1 - x + x^2 - x^3 $$
Substitute \(x=0.1\):
$$ = 1 - 0.1 + 0.01 - 0.001 = 0.909 $$
Answer: 0.909
Example 2:
Find the first four terms of \((1 - 0.2)^{-2}\) using binomial expansion.
Given: \(n = -2\), \(x = -0.2\)
Formula:
$$ (1 + x)^n = 1 + n x + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 $$
Calculate terms:
$$ = 1 + (-2)(-0.2) + \frac{-2 \times -3}{2} (-0.2)^2 + \frac{-2 \times -3 \times -4}{6} (-0.2)^3 $$
Calculate each term:
Sum:
$$ = 1 + 0.4 + 0.12 + 0.032 = 1.552 $$
Answer: 1.552
Example 3:
A mechanical part's compliance varies as \((1 + x)^{-3}\). Find the expansion up to \(x^3\) for \(x=0.05\).
Given: \(n = -3\), \(x=0.05\)
Formula:
$$ (1 + x)^{-3} = 1 + n x + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 $$
Calculate coefficients:
$$ = 1 - 3 x + \frac{-3 \times -4}{2} x^2 - \frac{-3 \times -4 \times -5}{6} x^3 $$
Calculate terms:
Sum:
$$ = 1 - 0.15 + 0.015 + 0.00125 = 0.86625 $$
Answer: 0.86625
Example 4:
Expand \((1 + 2x)^{-2}\) up to the \(x^3\) term and evaluate for \(x = 0.1\).
Given: \(n = -2\), \(x = 2x\), substitute \(2x\) as variable.
Formula:
$$ (1 + 2x)^{-2} = 1 + n(2x) + \frac{n(n-1)}{2!} (2x)^2 + \frac{n(n-1)(n-2)}{3!} (2x)^3 $$
Calculate coefficients and powers:
$$ = 1 - 2 \times 2x + \frac{-2 \times -3}{2} (2x)^2 - \frac{-2 \times -3 \times -4}{6} (2x)^3 $$
$$ = 1 - 4x + 3 \times 4x^2 - 4 \times 8 x^3 = 1 - 4x + 12 x^2 - 32 x^3 $$
Substitute \(x=0.1\):
$$ = 1 - 4 \times 0.1 + 12 \times 0.01 - 32 \times 0.001 $$
Calculate:
$$ = 1 - 0.4 + 0.12 - 0.032 = 0.688 $$
Answer: 0.688
Example 5:
A machine's efficiency factor is given by \(\frac{1}{(1 + 0.03x)^4}\). Expand using binomial expansion up to \(x^3\) and calculate for \(x=2\).
Given: \(n = -4\), \(x = 0.03x\), \(x=2\)
Formula:
$$ (1 + 0.03x)^{-4} = 1 + n (0.03x) + \frac{n(n-1)}{2!} (0.03x)^2 + \frac{n(n-1)(n-2)}{3!} (0.03x)^3 $$
Calculate coefficients:
$$ = 1 - 4 (0.06) + \frac{-4 \times -5}{2} (0.06)^2 - \frac{-4 \times -5 \times -6}{6} (0.06)^3 $$
Calculate powers:
\[
0.03 \times 2 = 0.06
\]
\[
(0.06)^2 = 0.0036
\]
\[
(0.06)^3 = 0.000216
\]
Calculate terms:
Sum:
$$ = 1 - 0.24 + 0.036 + 0.00432 = 0.80032 $$
Answer: 0.80032
Fractional powers represent roots and fractional exponents of numbers, which are common in mechanical production calculations such as determining dimensions based on volume or stress-strain relations. In Kenyan workshops and manufacturing settings, understanding fractional powers is essential for accurate measurements and design calculations where quantities like square roots or cube roots appear frequently.
The general form for fractional powers is:
$$ a^{\frac{m}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m $$
where \(a\) is a positive real number, \(m\) and \(n\) are integers, and \(n > 0\).
Example 1:
Calculate the square root of 81 raised to the power 3, i.e., \(81^{\frac{3}{2}}\), for a metal rod's cross-sectional area calculation.
Given: \(a = 81\), \(m = 3\), \(n = 2\)$$a^{\frac{m}{n}} = \sqrt[n]{a^m}$$$$81^{\frac{3}{2}} = \sqrt{81^3}$$ Calculate \(81^3\):$$81^3 = 81 \times 81 \times 81 = 531441$$ Calculate \(\sqrt{531441}\):$$\sqrt{531441} = 729$$ Answer: 729
Example 2:
Find the cube root of 125 raised to the power 2, \(125^{\frac{2}{3}}\), used in volume-related machining operations.
Given: \(a = 125\), \(m = 2\), \(n = 3\)$$125^{\frac{2}{3}} = \sqrt[3]{125^2}$$
Calculate \(125^2\):$$125^2 = 15625$$
Calculate \(\sqrt[3]{15625}\):
Since \(25^3 = 15625\),$$\sqrt[3]{15625} = 25$$
Answer: 25
Example 3:
Evaluate \(( \frac{16}{81} )^{\frac{3}{4}}\), relevant in stress analysis where material properties involve fractional exponents.
Given: \(a = \frac{16}{81}\), \(m = 3\), \(n = 4\)$$(\frac{16}{81})^{\frac{3}{4}} = (\sqrt[4]{\frac{16}{81}})^3$$ Calculate \(\sqrt[4]{\frac{16}{81}}\):$$\sqrt[4]{16} = 2$$$$\sqrt[4]{81} = 3$$ So,$$\sqrt[4]{\frac{16}{81}} = \frac{2}{3}$$ Raise to the power 3:$$(\frac{2}{3})^3 = \frac{8}{27}$$ Answer: \(\frac{8}{27}\)
Example 4:
Calculate \(49^{-\frac{1}{2}}\), a fractional negative power used in inverse square root scaling in tool wear rate estimations.
Given: \(a = 49\), \(m = -1\), \(n = 2\)$$49^{-\frac{1}{2}} = \frac{1}{\sqrt{49}}$$ Calculate \(\sqrt{49}\):$$\sqrt{49} = 7$$$$49^{-\frac{1}{2}} = \frac{1}{7}$$ Answer: \(\frac{1}{7}\)
Example 5:
A gear ratio calculation involves \(27^{\frac{4}{3}}\). Find the value.
Given: \(a = 27\), \(m = 4\), \(n = 3\)$$27^{\frac{4}{3}} = (\sqrt[3]{27})^4$$ Calculate \(\sqrt[3]{27}\):$$\sqrt[3]{27} = 3$$ Calculate \(3^4\):$$3^4 = 81$$ Answer: 81
In mechanical production, small changes in dimensions or measurements can significantly affect the performance of components. Estimating errors due to these small changes is crucial for quality control and ensuring mechanical parts meet tolerance requirements. The differential calculus approach is used to approximate errors for functions of one or more variables.
For a function \(y = f(x)\), the approximate change in \(y\) due to a small change \(\Delta x\) in \(x\) is given by:
$$ \Delta y \approx \frac{dy}{dx} \Delta x $$
where \(\frac{dy}{dx}\) is the derivative of the function.
For functions involving powers such as \(y = x^n\), the error estimation formula becomes:
$$ \Delta y \approx n x^{n-1} \Delta x $$
Example 1:
Estimate the error in the volume of a cube if the side length \(x = 10\,cm\) has an error \(\Delta x = 0.1\,cm\). Volume \(V = x^3\).
Given: \(x = 10\,cm\), \(\Delta x = 0.1\,cm\), \(V = x^3\)
Derivative:$$\frac{dV}{dx} = 3x^2$$
Error in volume:$$\Delta V \approx 3x^2 \Delta x$$
Substitute values:$$\Delta V \approx 3 \times (10)^2 \times 0.1$$$$\Delta V \approx 3 \times 100 \times 0.1 = 30\,cm^3$$
Answer: 30 cm³
Example 2:
A shaft diameter is \(d = 50\,mm\) with a possible measurement error of \(\Delta d = 0.2\,mm\). Calculate the approximate error in cross-sectional area \(A = \frac{\pi}{4} d^2\).
Given: \(d = 50\,mm\), \(\Delta d = 0.2\,mm\),$$A = \frac{\pi}{4} d^2$$ Derivative:$$\frac{dA}{dd} = \frac{\pi}{2} d$$ Error in area:$$\Delta A \approx \frac{\pi}{2} d \Delta d$$ Substitute values:$$\Delta A \approx \frac{3.1416}{2} \times 50 \times 0.2$$$$\Delta A \approx 1.5708 \times 50 \times 0.2 = 15.708\,mm^2$$ Answer: 15.708 mm²
Example 3:
Estimate the error in the length of a cylindrical rod \(L = k t^{\frac{1}{2}}\) where \(k = 5\), thickness \(t = 16\,mm\) and error \(\Delta t = 0.1\,mm\).
Given: \(L = 5 t^{\frac{1}{2}}\), \(t = 16\,mm\), \(\Delta t = 0.1\,mm\)
Derivative:$$\frac{dL}{dt} = 5 \times \frac{1}{2} t^{-\frac{1}{2}} = \frac{5}{2 \sqrt{t}}$$
Calculate derivative at \(t=16\):$$\frac{dL}{dt} = \frac{5}{2 \times 4} = \frac{5}{8} = 0.625$$
Error in length:$$\Delta L \approx 0.625 \times 0.1 = 0.0625\,mm$$
Answer: 0.0625 mm
Example 4:
A rectangular plate has length \(l = 2\,m\) and width \(w = 1\,m\) with possible errors \(\Delta l = 0.01\,m\), \(\Delta w = 0.005\,m\). Estimate the error in area \(A = l \times w\).
Given: \(l = 2\,m\), \(w = 1\,m\), \(\Delta l = 0.01\,m\), \(\Delta w = 0.005\,m\)
Total differential:$$\Delta A \approx w \Delta l + l \Delta w$$
Substitute values:$$\Delta A \approx 1 \times 0.01 + 2 \times 0.005 = 0.01 + 0.01 = 0.02\,m^2$$
Answer: 0.02 m²
Example 5:
Estimate the error in the volume \(V = \pi r^2 h\) of a cylinder where \(r = 0.3\,m\), \(h = 1\,m\), \(\Delta r = 0.002\,m\), \(\Delta h = 0.005\,m\).
Given: \(V = \pi r^2 h\), \(r=0.3\,m\), \(h=1\,m\), \(\Delta r=0.002\,m\), \(\Delta h=0.005\,m\)
Partial derivatives:$$\frac{\partial V}{\partial r} = 2 \pi r h$$$$\frac{\partial V}{\partial h} = \pi r^2$$
Error estimation:$$\Delta V \approx | \frac{\partial V}{\partial r} | \Delta r + | \frac{\partial V}{\partial h} | \Delta h$$
Calculate derivatives:$$\frac{\partial V}{\partial r} = 2 \times 3.1416 \times 0.3 \times 1 = 1.88496$$$$\frac{\partial V}{\partial h} = 3.1416 \times (0.3)^2 = 3.1416 \times 0.09 = 0.282744$$
Calculate error:$$\Delta V \approx 1.88496 \times 0.002 + 0.282744 \times 0.005$$$$\Delta V \approx 0.00376992 + 0.00141372 = 0.00518364\,m^3$$
Answer: 0.00518 m³
This chapter began by exploring powers in binomial expressions, establishing the foundation for understanding terms raised to various exponents. The concept of coefficients was then introduced, explaining their role as multipliers in each term of the expansion. Pascal's triangle was presented as a practical tool to quickly determine these coefficients without complex calculations. The binomial theorem was developed to provide a systematic method for expanding binomials raised to any power. Attention then shifted to expansions involving positive powers of n, demonstrating straightforward polynomial forms. More advanced cases included negative powers of n, where infinite series expansions were necessary. The chapter also covered fractional powers of n, involving roots and their corresponding series representations. Finally, methods for the estimation of errors caused by small changes in variables were discussed, highlighting the practical importance of approximations in engineering calculations.
Calculate \((1 + x)^3\) using binomial expansion for \(x = 0.2\). (2 marks)
Find the coefficient of \(x^2\) in the expansion of \((2 + x)^4\). (2 marks)
Using Pascal’s triangle, write down the coefficients for the expansion of \((a + b)^5\). (2 marks)
Expand \((1 + 3x)^3\) using the binomial theorem and find the value when \(x = 0.1\). (3 marks)
Calculate the positive power term \(\binom{6}{2} x^2\) in the expansion of \((1 + x)^6\) for \(x = 0.05\). (2 marks)
Use binomial expansion to approximate \((1 - 0.02)^5\) up to the third term. (3 marks)
Using fractional powers, expand \((1 + x)^{\frac{1}{2}}\) up to the second term for \(x = 0.04\). (3 marks)
Estimate the percentage error in the volume of a cylindrical shaft if the radius is measured as 10 cm with an error of +0.1 cm, using binomial expansion for volume formula \(V = \pi r^2 h\) with \(h = 50\) cm. (4 marks)
For a metal rod of length 2 m, the length changes by +0.005 m due to temperature. Use binomial expansion to estimate the change in cross-sectional area if the radius 1 cm expands similarly. (4 marks)
A gear tooth thickness is measured as 20 mm with a possible error of -0.1 mm. Using binomial expansion, estimate the resulting percentage change in the moment of inertia \(I = \frac{bt^3}{12}\), where \(b = 50\) mm, \(t\) is thickness. (5 marks)
Formula: $$ (1 + x)^3 = \sum_{k=0}^3 \binom{3}{k} x^k = 1 + 3x + 3x^2 + x^3 $$
Substitute \(x=0.2\):$$ = 1 + 3(0.2) + 3(0.2)^2 + (0.2)^3 $$ Calculate stepwise:$$ = 1 + 0.6 + 3(0.04) + 0.008 $$$$ = 1 + 0.6 + 0.12 + 0.008 = 1.728 $$ Answer: \(1.728\)
Formula:$$ (2 + x)^4 = \sum_{k=0}^4 \binom{4}{k} 2^{4-k} x^k $$
Coefficient of \(x^2\) is:$$ \binom{4}{2} 2^{4-2} = 6 \times 2^2 = 6 \times 4 = 24 $$
Answer: Coefficient of \(x^2\) is 24
Row 5 coefficients are:$$ 1, 5, 10, 10, 5, 1 $$
Answer: \(1, 5, 10, 10, 5, 1\)
Formula:$$ (1 + 3x)^3 = \sum_{k=0}^3 \binom{3}{k} (3x)^k = 1 + 3(3x) + 3(3x)^2 + (3x)^3 $$
Calculate terms:$$ = 1 + 9x + 27x^2 + 27x^3 $$
Substitute \(x=0.1\):$$ = 1 + 9(0.1) + 27(0.1)^2 + 27(0.1)^3 $$ Calculate stepwise:$$ = 1 + 0.9 + 27(0.01) + 27(0.001) $$$$ = 1 + 0.9 + 0.27 + 0.027 = 2.197 $$
Answer: \(2.197\)
Calculate \(\binom{6}{2}\):$$ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 $$
Calculate term:$$ 15 \times (0.05)^2 = 15 \times 0.0025 = 0.0375 $$
Answer: \(0.0375\)
Binomial expansion:$$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2} x^2 + \ldots $$
Here, \(n=5\), \(x = -0.02\):
Calculate terms:$$ 1 + 5(-0.02) + \frac{5 \times 4}{2} (-0.02)^2 = 1 - 0.1 + 10 \times 0.0004 $$$$ = 1 - 0.1 + 0.004 = 0.904 $$
Answer: \(0.904\)
Binomial expansion for fractional power:$$ (1 + x)^{\alpha} \approx 1 + \alpha x + \frac{\alpha(\alpha -1)}{2} x^2 $$
Only up to second term means:$$ 1 + \frac{1}{2} x $$
Calculate:$$ 1 + \frac{1}{2} \times 0.04 = 1 + 0.02 = 1.02 $$
Answer: \(1.02\)
Relative change in volume:$$ \frac{\Delta V}{V} \approx 2 \frac{\Delta r}{r} $$
Calculate:$$ 2 \times \frac{0.1}{10} = 2 \times 0.01 = 0.02 = 2\% $$
Volume increases by approximately 2%.
Answer: \(2\%\) increase
Fractional change in radius:$$ \frac{\Delta r}{r} = \frac{0.01}{1} = 0.01 $$
Cross-sectional area \(A = \pi r^2\), relative change:$$ \frac{\Delta A}{A} \approx 2 \frac{\Delta r}{r} = 2 \times 0.01 = 0.02 = 2\% $$
Answer: Cross-sectional area increases by approximately 2%
Moment of inertia:$$ I = \frac{b t^3}{12} $$
Relative change in \(I\):$$ \frac{\Delta I}{I} \approx 3 \frac{\Delta t}{t} = 3 \times \frac{-0.1}{20} = 3 \times -0.005 = -0.015 = -1.5\% $$
Answer: Moment of inertia decreases by approximately 1.5%
Question 11 (Compulsory - 20 marks)
A mechanical workshop at Kenya Railways is modifying a cylindrical piston. The piston radius increases by 2% due to thermal expansion, and the length increases by 1.5%.
a) Use binomial expansion to estimate the percentage increase in the piston volume up to the second order terms. (10 marks)
b) If the piston volume originally is 500 cm³, calculate the approximate new volume after expansion using your expansion results. (10 marks)
Question 12 (20 marks)
A steel rod length \(L\) depends on temperature and is given by \(L = L_0 (1 + \alpha \Delta T)^n\), where \(\alpha = 0.000012 / ^\circ C\), \(\Delta T\) is temperature change, and \(n = \frac{1}{2}\).
a) Use binomial expansion to approximate \(L\) for small \(\Delta T = 30 ^\circ C\) up to the third term. (10 marks)
b) Explain how fractional powers in the expansion affect the accuracy of length estimation. (10 marks)
Question 13 (20 marks)
A metal sheet thickness is reduced by 3%, and its width is increased by 2%.
a) Using binomial expansion, estimate the approximate percentage change in the cross-sectional area of the sheet. (10 marks)
b) Discuss the implications of ignoring higher-order terms in such mechanical tolerance calculations. (10 marks)
Question 14 (20 marks)
The force \(F\) in a mechanical spring is given by \(F = kx^n\), where \(k\) is a constant, \(x\) is displacement, and \(n = -2\). If the displacement \(x\) changes slightly by \(\Delta x = 0.05x\),
a) Use binomial expansion to estimate the percentage change in force \(F\). (10 marks)
b) Calculate the approximate new force if the original force is 1000 N. (10 marks)
Question 11
a) Volume \( V = \pi r^2 h \), percentage increases in \(r\) and \(h\) are 2% and 1.5% respectively. Using binomial expansion:$$ (1 + 0.02)^2 = 1 + 2(0.02) + (0.02)^2 = 1 + 0.04 + 0.0004 = 1.0404 $$$$ (1 + 0.015) = 1 + 0.015 $$
Combined volume increase factor:$$ 1.0404 \times 1.015 = 1.056006 $$
Percentage increase:$$ (1.056006 - 1) \times 100 = 5.6\% $$
b) New volume:$$ 500 \times 1.056006 = 528.0 \text{ cm}^3 $$ Answer: Approximate new volume is 528.0 cm³
Question 12
a) Given \( L = L_0 (1 + \alpha \Delta T)^{\frac{1}{2}} \), with \(\alpha \Delta T = 0.000012 \times 30 = 0.00036\)
Binomial expansion up to third term:$$ (1 + x)^{\frac{1}{2}} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} x^3 $$
Substitute \( x = 0.00036 \):$$ L = L_0 (1 + \frac{1}{2}(0.00036) - \frac{1}{8}(0.00036)^2 + \frac{1}{16}(0.00036)^3 ) $$
Calculate terms:$$ \frac{1}{2} \times 0.00036 = 0.00018 $$$$ \frac{1}{8} \times (0.00036)^2 = \frac{1}{8} \times 0.0000001296 = 0.0000000162 $$$$ \frac{1}{16} \times (0.00036)^3 = \frac{1}{16} \times 0.0000000467 = 0.0000000029 $$
So:$$ L = L_0 (1 + 0.00018 - 0.0000000162 + 0.0000000029) \approx L_0 (1.00018) $$
b) Fractional powers reduce the magnitude of changes for small \(x\), improving estimation accuracy for small temperature changes.
Question 13
a) Cross-sectional area \( A = \text{thickness} \times \text{width} \)
Thickness decreases by 3% → factor \( (1 - 0.03) \)
Width increases by 2% → factor \( (1 + 0.02) \)
Area change factor:$$ (1 - 0.03)(1 + 0.02) = 1 - 0.03 + 0.02 - 0.0006 = 0.9894 $$
Percentage change:$$ (0.9894 - 1) \times 100 = -1.06\% $$
b) Ignoring higher-order terms (like the product \( -0.0006 \)) can lead to slight errors in critical tolerance calculations, affecting quality control.
Question 14
a) Given \( F = k x^{-2} \), small change \( \Delta x = 0.05 x \), so:$$ F = k x^n = k x^{-2} $$
Using binomial expansion for \( (1 + \frac{\Delta x}{x})^n \):$$ (1 + 0.05)^{-2} = 1 - 2(0.05) + \frac{(-2)(-3)}{2} (0.05)^2 = 1 - 0.1 + 0.0075 = 0.9075 $$
Percentage change:$$ (0.9075 - 1) \times 100 = -9.25\% $$
b) New force:$$ 1000 \times 0.9075 = 907.5 \text{ N} $$
Answer: Force decreases by approximately 9.25% to 907.5 N
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