Engineering mathematics is fundamental for quantity surveyors in Kenya, enabling precise analysis and problem-solving in construction cost estimation, project management, and design evaluation. Calculus operations, especially differentiation, are essential for modeling rates of change such as material usage over time, cost variations, and structural behavior under load. This chapter focuses on differentiation of algebraic, trigonometric, and hyperbolic functions, providing practical techniques and worked examples relevant to quantity surveying applications.
Differentiation of algebraic functions allows quantity surveyors to determine how quantities such as volume, cost, or work rates change with respect to variables like time, length, or quantity of materials. Mastery of differentiation rules enables efficient calculation of marginal costs, optimization of resources, and evaluation of project dynamics.
The power rule is used to differentiate functions of the form \( f(x) = x^n \), where \( n \) is any real number. It states that the derivative of \( x^n \) with respect to \( x \) is given by:
$$ \frac{d}{dx} x^n = n x^{n-1} $$
Example 1: A quantity surveyor models the cost \( C \) in thousands of Ksh for materials as \( C(x) = 5x^3 \), where \( x \) is the volume of concrete in cubic meters. Find the rate of change of cost with respect to volume.
Given: \( C(x) = 5x^3 \)
$$ \frac{dC}{dx} = \frac{d}{dx} (5x^3) $$
$$ = 5 \times \frac{d}{dx} x^3 $$
$$ = 5 \times 3x^{2} $$
$$ = 15x^{2} $$
Answer: \(\frac{dC}{dx} = 15x^{2}\) (thousands of Ksh per cubic meter)
Example 2: The length \( L \) of a beam varies with time \( t \) in hours as \( L(t) = 2t^{4} \). Find the instantaneous rate of change of length at \( t = 3 \) hours.
Given: \( L(t) = 2t^{4} \)
$$ \frac{dL}{dt} = 2 \times \frac{d}{dt} t^{4} = 2 \times 4t^{3} = 8t^{3} $$
At \( t = 3 \):
$$ \frac{dL}{dt} = 8 \times 3^{3} = 8 \times 27 = 216 $$
Answer: \(\frac{dL}{dt} = 216\) units per hour
Example 3: The volume \( V \) of soil excavated is modeled by \( V(x) = 7x^{2.5} \), where \( x \) is the excavation depth in meters. Find the rate of change of volume with respect to depth when \( x = 4 \) meters.
Given: \( V(x) = 7x^{2.5} \)
$$ \frac{dV}{dx} = 7 \times \frac{d}{dx} x^{2.5} = 7 \times 2.5 x^{1.5} = 17.5 x^{1.5} $$
At \( x = 4 \):
$$ \frac{dV}{dx} = 17.5 \times 4^{1.5} = 17.5 \times (4 \times \sqrt{4}) = 17.5 \times (4 \times 2) = 17.5 \times 8 = 140 $$
Answer: \(\frac{dV}{dx} = 140\) cubic meters per meter
The product rule is applied when differentiating the product of two functions, say \( u(x) \) and \( v(x) \). It states:
$$ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) $$
This rule is useful when cost or volume is modeled as a product of two varying quantities.
Example 1: The cost function is \( C(x) = x^2 (3x + 5) \), where \( x \) is the number of units. Find \( \frac{dC}{dx} \).
Given: \( u = x^2 \), \( v = 3x + 5 \)
$$ u' = 2x, \quad v' = 3 $$
$$ \frac{dC}{dx} = u'v + uv' = 2x(3x + 5) + x^{2} \times 3 $$
$$ = 6x^{2} + 10x + 3x^{2} = 9x^{2} + 10x $$
Answer: \(\frac{dC}{dx} = 9x^{2} + 10x\)
Example 2: A quantity surveyor calculates the area \( A(t) = (4t + 1)(t^2 + 3) \) of a site where \( t \) is time in days. Find the rate of change of area with respect to time.
Given: \( u = 4t + 1 \), \( v = t^{2} + 3 \)
$$ u' = 4, \quad v' = 2t $$
$$ \frac{dA}{dt} = u'v + uv' = 4(t^{2} + 3) + (4t + 1)(2t) $$
$$ = 4t^{2} + 12 + 8t^{2} + 2t = 12t^{2} + 2t + 12 $$
Answer: \(\frac{dA}{dt} = 12t^{2} + 2t + 12\)
Example 3: The cost \( C(x) = (2x^{3})(5x^{2} + 1) \). Find \( \frac{dC}{dx} \).
Given: \( u = 2x^{3} \), \( v = 5x^{2} + 1 \)
$$ u' = 6x^{2}, \quad v' = 10x $$
$$ \frac{dC}{dx} = u'v + uv' = 6x^{2}(5x^{2} + 1) + 2x^{3}(10x) $$
$$ = 30x^{4} + 6x^{2} + 20x^{4} = 50x^{4} + 6x^{2} $$
Answer: \(\frac{dC}{dx} = 50x^{4} + 6x^{2}\)
When a function is the division of two functions \( u(x) \) and \( v(x) \), the quotient rule states:
$$ \frac{d}{dx} (\frac{u}{v}) = \frac{v u' - u v'}{v^{2}} $$
This is applicable in quantity surveying when rates or ratios are involved.
Example 1: The cost efficiency function is \( E(x) = \frac{5x^{2}}{x + 1} \). Find \( \frac{dE}{dx} \).
Given: \( u = 5x^{2} \), \( v = x + 1 \)
$$ u' = 10x, \quad v' = 1 $$
$$ \frac{dE}{dx} = \frac{(x + 1)(10x) - 5x^{2}(1)}{(x + 1)^{2}} = \frac{10x^{2} + 10x - 5x^{2}}{(x + 1)^{2}} = \frac{5x^{2} + 10x}{(x + 1)^{2}} $$
Answer: \(\frac{dE}{dx} = \frac{5x^{2} + 10x}{(x + 1)^{2}}\)
Example 2: The function \( R(t) = \frac{t^{3} + 2}{t^{2} - 1} \) models resource usage. Find \( \frac{dR}{dt} \).
Given: \( u = t^{3} + 2 \), \( v = t^{2} - 1 \)
$$ u' = 3t^{2}, \quad v' = 2t $$
$$ \frac{dR}{dt} = \frac{(t^{2} - 1)(3t^{2}) - (t^{3} + 2)(2t)}{(t^{2} - 1)^{2}} $$
$$ = \frac{3t^{4} - 3t^{2} - 2t^{4} - 4t}{(t^{2} - 1)^{2}} = \frac{t^{4} - 3t^{2} - 4t}{(t^{2} - 1)^{2}} $$
Answer: \(\frac{dR}{dt} = \frac{t^{4} - 3t^{2} - 4t}{(t^{2} - 1)^{2}}\)
Example 3: The function \( F(x) = \frac{3x^{2} + 1}{2x^{3} + 5} \). Find \( \frac{dF}{dx} \).
Given: \( u = 3x^{2} + 1 \), \( v = 2x^{3} + 5 \)
$$ u' = 6x, \quad v' = 6x^{2} $$
$$ \frac{dF}{dx} = \frac{(2x^{3} + 5)(6x) - (3x^{2} + 1)(6x^{2})}{(2x^{3} + 5)^{2}} $$
$$ = \frac{12x^{4} + 30x - 18x^{4} - 6x^{2}}{(2x^{3} + 5)^{2}} = \frac{-6x^{4} + 30x - 6x^{2}}{(2x^{3} + 5)^{2}} $$
Answer: \(\frac{dF}{dx} = \frac{-6x^{4} + 30x - 6x^{2}}{(2x^{3} + 5)^{2}}\)
The chain rule differentiates composite functions \( y = f(g(x)) \). It states:
$$ \frac{dy}{dx} = \frac{dy}{dg} \times \frac{dg}{dx} $$
This is essential when cost or volume depends on a function of another variable.
Example 1: The cost function is \( C(x) = (3x + 2)^{4} \). Find \( \frac{dC}{dx} \).
Given: \( y = (3x + 2)^{4} \)
Let \( u = 3x + 2 \), then \( y = u^{4} \)
$$ \frac{dy}{du} = 4u^{3}, \quad \frac{du}{dx} = 3 $$
$$ \frac{dy}{dx} = 4u^{3} \times 3 = 12(3x + 2)^{3} $$
Answer: \(\frac{dC}{dx} = 12(3x + 2)^{3}\)
Example 2: The volume \( V(t) = \sqrt{5t^{2} + 1} \). Find \( \frac{dV}{dt} \).
Given: \( V(t) = (5t^{2} + 1)^{1/2} \)
Let \( u = 5t^{2} + 1 \), \( V = u^{1/2} \)
$$ \frac{dV}{du} = \frac{1}{2} u^{-1/2}, \quad \frac{du}{dt} = 10t $$
$$ \frac{dV}{dt} = \frac{1}{2} u^{-1/2} \times 10t = 5t (5t^{2} + 1)^{-1/2} $$
Answer: \(\frac{dV}{dt} = \frac{5t}{\sqrt{5t^{2} + 1}}\)
Example 3: The cost function \( C(x) = e^{3x^{2} + 1} \). Find \( \frac{dC}{dx} \).
Given: \( C(x) = e^{3x^{2} + 1} \)
Let \( u = 3x^{2} + 1 \)
$$ \frac{dC}{du} = e^{u}, \quad \frac{du}{dx} = 6x $$
$$ \frac{dC}{dx} = e^{3x^{2} + 1} \times 6x = 6x e^{3x^{2} + 1} $$
Answer: \(\frac{dC}{dx} = 6x e^{3x^{2} + 1}\)
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Create a free accountThis chapter covered the fundamental calculus operations essential for engineering mathematics. It began with differentiation of algebraic functions, explaining key rules such as the power, product, quotient, and chain rules. The focus then shifted to differentiating trigonometric and hyperbolic functions, including sine, cosine, tangent, sinh, and cosh. The chapter also addressed the derivatives of inverse trigonometric functions like arcsin, arccos, and arctan. The concept of rate of change was explored through first and second derivatives, highlighting applications in velocity and acceleration. Stationary points were examined to identify maxima, minima, and points of inflection, with emphasis on their role in optimization problems. Integration techniques for algebraic functions were introduced, covering both definite and indefinite integrals, along with their use in calculating areas. The integration of trigonometric, hyperbolic, and logarithmic functions was discussed, including substitution methods. Finally, practical applications of calculus were demonstrated in calculating areas under curves and volumes of solids of revolution.
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