By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems with confidence and precision.
Calculus is a fundamental mathematical tool used extensively in welding engineering to analyse rates of change and accumulated quantities, such as heat input, material removal, and weld bead geometry. Welding professionals in Kenya apply differentiation and integration to optimise welding parameters, control thermal cycles, and predict mechanical behaviour of welded structures. Mastery of calculus enables precise calculations that improve weld quality, reduce defects, and enhance structural performance in various sectors including manufacturing, construction, and maintenance.
Differentiation is the process of finding the rate at which a quantity changes with respect to another variable. In welding, it helps determine how welding parameters such as temperature, stress, and displacement vary along the weld or over time. Calculating first, second, and third derivatives allows welding engineers to analyse velocity, acceleration, and jerk in thermal or mechanical processes influencing weld integrity.
The first derivative of a function represents the rate of change or slope of the function at any point. It is denoted as dy/dx for a function y = f(x). The first derivative is crucial in welding to understand how temperature or stress changes along the weld length or during a welding cycle.
dy/dx = limit as Δx → 0 of [f(x + Δx) − f(x)] ÷ Δx
Worked examples:
Example 1: A welding rod temperature T(x) along its length x (in cm) is given by T(x) = 5x² + 3x + 20 °C. Find the rate of change of temperature at x = 4 cm.
Given: T(x) = 5x² + 3x + 20, x = 4 cm
dT/dx = d/dx (5x² + 3x + 20)
dT/dx = 10x + 3
At x = 4:
dT/dx = 10(4) + 3
dT/dx = 40 + 3 = 43
Answer: 43 °C/cm
Example 2: The displacement s(t) (in cm) of a weld bead edge during cooling is s(t) = 2t³ − 5t² + 4t, where t is time in seconds. Find the velocity at t = 3 s.
Given: s(t) = 2t³ − 5t² + 4t, t = 3 s
v(t) = ds/dt = d/dt (2t³ − 5t² + 4t)
v(t) = 6t² − 10t + 4
At t = 3:
v(3) = 6(3)² − 10(3) + 4
v(3) = 6(9) − 30 + 4 = 54 − 30 + 4 = 28
Answer: 28 cm/s
Example 3: Stress σ(x) along a weld is given by σ(x) = 4x³ − 7x + 10 MPa. Find the rate of change of stress at x = 2 m.
Given: σ(x) = 4x³ − 7x + 10, x = 2 m
dσ/dx = d/dx (4x³ − 7x + 10)
dσ/dx = 12x² − 7
At x = 2:
dσ/dx = 12(2)² − 7
dσ/dx = 12(4) − 7 = 48 − 7 = 41
Answer: 41 MPa/m
The second derivative d²y/dx² gives the rate of change of the first derivative and represents acceleration or curvature of a function. In welding, it helps analyse how the rate of temperature change or stress variation itself changes, which is important in predicting weld distortions or crack propagation.
d²y/dx² = d/dx (dy/dx)
Worked examples:
Example 1: For the temperature function T(x) = 5x² + 3x + 20, find the second derivative and evaluate it at x = 4 cm.
Given: T(x) = 5x² + 3x + 20, x = 4 cm
First derivative:
dT/dx = 10x + 3
Second derivative:
d²T/dx² = d/dx (10x + 3) = 10
At x = 4:
d²T/dx² = 10
Answer: 10 °C/cm²
Example 2: Displacement s(t) = 2t³ − 5t² + 4t (cm). Find the acceleration at t = 3 s.
Given: s(t) = 2t³ − 5t² + 4t, t = 3 s
First derivative (velocity):
v(t) = 6t² − 10t + 4
Second derivative (acceleration):
a(t) = dv/dt = 12t − 10
At t = 3:
a(3) = 12(3) − 10 = 36 − 10 = 26
Answer: 26 cm/s²
Example 3: Stress σ(x) = 4x³ − 7x + 10 MPa. Find the second derivative and its value at x = 2 m.
Given: σ(x) = 4x³ − 7x + 10, x = 2 m
First derivative:
dσ/dx = 12x² − 7
Second derivative:
d²σ/dx² = d/dx (12x² − 7) = 24x
At x = 2:
d²σ/dx² = 24(2) = 48
Answer: 48 MPa/m²
The third derivative d³y/dx³ measures the rate of change of the second derivative and is used to analyse the jerk or sudden changes in acceleration. In welding, this can be important in dynamic processes such as robotic welding arm movements or thermal shock analysis.
d³y/dx³ = d/dx (d²y/dx²)
Worked examples:
Example 1: Given T(x) = 5x² + 3x + 20, find the third derivative.
Given: T(x) = 5x² + 3x + 20
First derivative:
dT/dx = 10x + 3
Second derivative:
d²T/dx² = 10
Third derivative:
d³T/dx³ = d/dx (10) = 0
Answer: 0 °C/cm³
Example 2: For displacement s(t) = 2t³ − 5t² + 4t (cm), find the jerk at t = 3 s.
Given: s(t) = 2t³ − 5t² + 4t, t = 3 s
First derivative (velocity):
v(t) = 6t² − 10t + 4
Second derivative (acceleration):
a(t) = 12t − 10
Third derivative (jerk):
j(t) = da/dt = 12
At t = 3:
j(3) = 12
Answer: 12 cm/s³
Example 3: Stress σ(x) = 4x³ − 7x + 10 MPa. Find the third derivative and evaluate it at x = 2 m.
Given: σ(x) = 4x³ − 7x + 10, x = 2 m
First derivative:
dσ/dx = 12x² − 7
Second derivative:
d²σ/dx² = 24x
Third derivative:
d³σ/dx³ = d/dx (24x) = 24
At x = 2:
d³σ/dx³ = 24
Answer: 24 MPa/m³
A function states how one welding quantity depends on another, for example plate temperature against time or bead height against travel speed. Each family of function has its own differentiation formula, and these formulas are the building blocks for every rule that follows. In this section each function type is differentiated up to the third derivative using examples from the welding workshop.
A linear function has the form y = mx + c, and its graph is a straight line of gradient m. The first derivative is the constant m, so the rate of change is the same everywhere, and the second and third derivatives are zero. In a workshop this describes steady processes, such as a preheating torch raising a plate temperature at a constant rate, or the cost of a job rising with hours worked.
d/dx (mx + c) = m
d²y/dx² = 0
d³y/dx³ = 0
Example 1: A plate is preheated so that its temperature is T(t) = 2.5t + 25 °C, where t is time in minutes. Find the heating rate.
Given: T(t) = 2.5t + 25
dT/dt = d/dt (2.5t + 25)
dT/dt = 2.5
Answer: 2.5 °C/min at every instant, so the second and third derivatives are 0
Example 2: The cost of a fabrication job is C(h) = 1200h + 3500 Ksh, where h is the welding time in hours and 3500 Ksh covers set-up and consumables. Find the rate at which cost changes with time.
Given: C(h) = 1200h + 3500
dC/dh = d/dh (1200h + 3500)
dC/dh = 1200
Answer: Ksh 1,200 per hour
Trigonometric functions describe repeating motion, such as the side-to-side weaving of a torch across a joint. The derivatives repeat in a cycle: sine becomes cosine, cosine becomes minus sine, and so on. Angles must always be in radians when differentiating.
d/dx (sin x) = cos x
d/dx (cos x) = −sin x
d/dx (tan x) = sec² x
d/dx (sin ax) = a cos ax
d/dx (cos ax) = −a sin ax
Example: A torch weaves across a joint so that its sideways position is y(t) = 4 sin 2t mm, where t is in seconds. Find the velocity and acceleration at t = 0.5 s and the jerk at t = 0.
Given: y(t) = 4 sin 2t
Velocity:
v(t) = dy/dt = 4 × 2 cos 2t = 8 cos 2t
Acceleration:
a(t) = d²y/dt² = −16 sin 2t
Jerk:
j(t) = d³y/dt³ = −32 cos 2t
At t = 0.5:
v(0.5) = 8 cos(1) = 8 × 0.5403 = 4.32
a(0.5) = −16 sin(1) = −16 × 0.8415 = −13.46
At t = 0:
j(0) = −32 cos(0) = −32
Answer: v = 4.32 mm/s, a = −13.46 mm/s², j = −32 mm/s³
Logarithmic functions describe quantities that grow quickly at first and then slow down. The derivative of ln x is 1/x, so the rate of change falls as x increases. For a function containing ln(ax + b), the chain factor a appears in the numerator.
d/dx (ln x) = 1/x
d/dx (ln(ax + b)) = a/(ax + b)
d/dx (log₁₀ x) = 1/(x ln 10)
Example: Suppose an empirical fit gives the depth of the heat-affected zone as d(t) = 2 + 1.5 ln t mm after t seconds of arc dwell (t ≥ 1). Find the first, second and third derivatives at t = 5 s.
Given: d(t) = 2 + 1.5 ln t
First derivative:
dd/dt = 1.5/t
Second derivative:
d²d/dt² = −1.5/t²
Third derivative:
d³d/dt³ = 3/t³
At t = 5:
dd/dt = 1.5/5 = 0.3
d²d/dt² = −1.5/25 = −0.06
d³d/dt³ = 3/125 = 0.024
Answer: 0.3 mm/s, −0.06 mm/s², 0.024 mm/s³. The zone deepens, but ever more slowly
Exponential functions model cooling and decay, such as a weld cooling towards room temperature. The function e^(ax) is special because its derivative is the same function multiplied by the constant a. A negative a gives decay, so the first derivative is negative while the quantity falls.
d/dx (e^x) = e^x
d/dx (e^(ax)) = a e^(ax)
d/dx (a^x) = a^x ln a
Example: A weld cools according to T(t) = 25 + 475 e^(−0.02t) °C, where t is in seconds. Find the cooling rate at t = 50 s and the first three derivatives.
Given: T(t) = 25 + 475 e^(−0.02t)
dT/dt = 475 × (−0.02) e^(−0.02t) = −9.5 e^(−0.02t)
d²T/dt² = 0.19 e^(−0.02t)
d³T/dt³ = −0.0038 e^(−0.02t)
At t = 50, e^(−1) = 0.367879:
dT/dt = −9.5 × 0.367879 = −3.49
d²T/dt² = 0.19 × 0.367879 = 0.0699
Answer: dT/dt = −3.49 °C/s (cooling), d²T/dt² = 0.0699 °C/s² (cooling is slowing down)
Most welding relationships are combinations of simple functions, for example a product of current and voltage or a quotient of force and area. The rules below allow such combined functions to be differentiated without returning to the limit definition. Each rule can be applied again to reach the second and third derivatives.
The power rule differentiates any term of the form x^n: multiply by the power and reduce the power by one. It works for whole, negative and fractional powers, and a constant multiplier is simply carried through.
d/dx (x^n) = n x^(n−1)
d/dx (k x^n) = k n x^(n−1)
Example 1: A hemispherical weld pool of radius r mm has volume V = (2/3)πr³ mm³. Find how fast the volume changes with radius at r = 4 mm, and the second derivative.
Given: V = (2/3)πr³, r = 4 mm
dV/dr = (2/3)π × 3r² = 2πr²
d²V/dr² = 4πr
At r = 4:
dV/dr = 2π(16) = 100.53
d²V/dr² = 4π(4) = 50.27
Answer: dV/dr = 100.53 mm²(the pool surface area), d²V/dr² = 50.27 mm
Example 2: Heat input per unit length is H = 2400 v^(−1) J/mm, where 2400 W is the arc power and v is the travel speed in mm/s. Find the rate of change of H with speed at v = 4 mm/s.
Given: H = 2400 v^(−1), v = 4
dH/dv = 2400 × (−1) v^(−2) = −2400/v²
At v = 4:
dH/dv = −2400/16 = −150
Answer: −150 J/mm per mm/s. Heat input falls as travel speed rises
When a quantity is the product of two functions u and v, the derivative is the first function times the derivative of the second, plus the second function times the derivative of the first.
d/dx (uv) = u dv/dx + v du/dx
Example 1: During a current slope-up, arc voltage is V(t) = 20 + 0.5t volts and current is I(t) = 100 + 4t amperes. Find the rate of change of arc power P = VI at t = 10 s.
Given: u = V = 20 + 0.5t, v = I = 100 + 4t, t = 10
dV/dt = 0.5, dI/dt = 4
dP/dt = V dI/dt + I dV/dt
At t = 10: V = 25 V, I = 140 A
dP/dt = 25(4) + 140(0.5) = 100 + 70 = 170
Answer: 170 W/s
Example 2: After a hammer blow, a welded bracket vibrates with displacement y(t) = t e^(−2t) mm. Find the velocity at t = 1 s.
Given: u = t, v = e^(−2t)
du/dt = 1, dv/dt = −2e^(−2t)
dy/dt = t(−2e^(−2t)) + e^(−2t)(1) = e^(−2t)(1 − 2t)
At t = 1:
dy/dt = e^(−2)(1 − 2) = −0.1353
Answer: −0.135 mm/s (moving back towards its rest position)
The chain rule differentiates a function of a function. If y depends on u, and u depends on x, multiply the two derivatives. It is needed whenever the variable sits inside a bracket, a power, an exponent or a trigonometric angle.
dy/dx = dy/du × du/dx
Example 1: The temperature across a plate follows a Gaussian heat distribution T(x) = 100 + 500 e^(−0.02x²) °C, where x is the distance in cm from the arc. Find dT/dx at x = 5 cm.
Given: u = −0.02x², so du/dx = −0.04x
dT/dx = 500 e^u × (−0.04x)
dT/dx = −20x e^(−0.02x²)
At x = 5, u = −0.5 and e^(−0.5) = 0.606531:
dT/dx = −20(5)(0.606531) = −60.65
Answer: −60.65 °C/cm (temperature falls away from the arc)
Example 2: A spot weld nugget has area A = πr² mm², and its radius grows as r = 2 + 0.5t mm, with t in seconds. Find dA/dt at t = 4 s.
Given: A = πr², r = 2 + 0.5t
dA/dr = 2πr, dr/dt = 0.5
dA/dt = 2πr × 0.5 = πr
At t = 4: r = 2 + 0.5(4) = 4 mm
dA/dt = π(4) = 12.57
Answer: 12.57 mm²/s
When a quantity is one function divided by another, use the quotient rule: bottom times the derivative of the top, minus top times the derivative of the bottom, all divided by the bottom squared.
d/dx (u/v) = (v du/dx − u dv/dx) ÷ v²
Example 1: The temperature rise of an electrode holder above ambient is ΔT(t) = 60t/(t + 3) °C, with t in minutes. Find the heating rate at t = 2 min.
Given: u = 60t, v = t + 3, du/dt = 60, dv/dt = 1
dΔT/dt = [(t + 3)(60) − 60t(1)] ÷ (t + 3)²
dΔT/dt = 180/(t + 3)²
At t = 2:
dΔT/dt = 180/25 = 7.2
Answer: 7.2 °C/min
Example 2: Stress along a weld is σ(x) = (2x² + 3)/(x + 1) MPa, with x in metres. Find dσ/dx at x = 2 m.
Given: u = 2x² + 3, v = x + 1, du/dx = 4x, dv/dx = 1
dσ/dx = [(x + 1)(4x) − (2x² + 3)(1)] ÷ (x + 1)²
dσ/dx = (2x² + 4x − 3)/(x + 1)²
At x = 2:
dσ/dx = (8 + 8 − 3)/9 = 13/9 = 1.44
Answer: 1.44 MPa/m
Differentiation is used in the workshop to locate the best welding settings and to describe how fast a process variable is changing. Setting the first derivative to zero finds the settings at which a quantity peaks or bottoms out, while the derivative itself gives the speed of change at any instant.
Differentiation helps welding engineers find optimum values of welding parameters by locating maxima, minima, or points of inflection in performance functions such as bead width, heat input, or residual stress. At a stationary point the gradient is zero, so setting the first derivative to zero identifies critical points that can improve weld quality.
dy/dx = 0
Procedure for finding and classifying stationary points:
| Sign of d²y/dx² at the point | Type of point | Meaning |
|---|---|---|
| Negative (< 0) | Maximum | Curve is concave down, the quantity peaks here |
| Positive (> 0) | Minimum | Curve is concave up, the quantity is lowest here |
| Zero, with a change of sign | Point of inflection | Curvature reverses, so the rate of change is at its extreme |
Worked examples:
Example 1: The weld bead width w(x) = −2x² + 8x + 10 mm varies with welding speed x (cm/s). Find the welding speed that maximises bead width.
Given: w(x) = −2x² + 8x + 10
dw/dx = −4x + 8
Set dw/dx = 0:
−4x + 8 = 0
4x = 8
x = 2 cm/s
Second derivative:
d²w/dx² = −4 < 0 (maximum)
Maximum width:
w(2) = −2(4) + 8(2) + 10 = −8 + 16 + 10 = 18
Answer: A welding speed of 2 cm/s maximises bead width at 18 mm
Example 2: Heat input H(t) = 3t³ − 12t² + 9t + 20 kJ varies with time t (s). Find the time when heat input is a minimum.
Given: H(t) = 3t³ − 12t² + 9t + 20
First derivative:
dH/dt = 9t² − 24t + 9
Set to zero and divide by 3:
3t² − 8t + 3 = 0
Quadratic formula:
t = [8 ± √(64 − 36)] ÷ 6 = (8 ± √28) ÷ 6
t₁ = (8 − 5.2915) ÷ 6 = 0.45 s
t₂ = (8 + 5.2915) ÷ 6 = 2.22 s
Second derivative:
d²H/dt² = 18t − 24
At t₁ = 0.4514:
18(0.4514) − 24 = −15.87 < 0 (maximum)
At t₂ = 2.2153:
18(2.2153) − 24 = 15.87 > 0 (minimum)
Minimum value:
H(2.2153) = 3(2.2153)³ − 12(2.2153)² + 9(2.2153) + 20 = 13.66
Answer: Heat input is a local minimum of 13.7 kJ at t = 2.22 s
Example 3: Residual stress R(x) = x³ − 6x² + 9x + 4 MPa along weld length x (m). Find the points of inflection.
Given: R(x) = x³ − 6x² + 9x + 4
First derivative:
dR/dx = 3x² − 12x + 9
Second derivative:
d²R/dx² = 6x − 12
Set the second derivative to zero:
6x − 12 = 0
6x = 12
x = 2 m
Check the change of sign: d²R/dx² = −6 at x = 1 and +6 at x = 3, so the curvature reverses at x = 2. The third derivative d³R/dx³ = 6 is not zero, which confirms an inflection.
R(2) = 8 − 24 + 18 + 4 = 6
Answer: Point of inflection at x = 2 m, where the residual stress is 6 MPa
The derivative dy/dx is the instantaneous rate of change of y with respect to x, and its unit is the unit of y divided by the unit of x. A positive value means the quantity is increasing and a negative value means it is decreasing. The average rate of change between two points is [y(x₂) − y(x₁)] ÷ (x₂ − x₁), which approaches the derivative as the interval shrinks.
Example 1: The volume of shielding gas consumed from a cylinder is V(t) = 15t + 0.02t² litres, where t is time in minutes. Find the rate of gas consumption at t = 100 min.
Given: V(t) = 15t + 0.02t²
dV/dt = 15 + 0.04t
At t = 100:
dV/dt = 15 + 0.04(100) = 19
Answer: 19 L/min
Example 2: During preheating, plate temperature is T(t) = 20 + 40t − 2t² °C, where t is in minutes. Find the average rate between t = 2 and t = 6 min, the instantaneous rate at t = 6 min, and the time when heating stops.
Given: T(t) = 20 + 40t − 2t²
T(2) = 20 + 80 − 8 = 92
T(6) = 20 + 240 − 72 = 188
Average rate = (188 − 92) ÷ (6 − 2) = 96 ÷ 4 = 24 °C/min
Instantaneous rate:
dT/dt = 40 − 4t
At t = 6:
dT/dt = 40 − 24 = 16 °C/min
Heating stops when dT/dt = 0:
40 − 4t = 0
t = 10 min
T(10) = 20 + 400 − 200 = 220
Answer: Average rate 24 °C/min, instantaneous rate at 6 min 16 °C/min, and the plate levels off at 220 °C after 10 min
For the temperature profile \( T(x) = 7x^2 - 4x + 15 \) °C along a weld, find the rate of temperature change at \( x = 3 \) cm. (3 marks)
A weld bead displacement is given by \( s(t) = 4t^3 - 9t^2 + 6t \) cm. Calculate the acceleration at \( t = 2 \) s. (4 marks)
The stress distribution along a weld line is \( \sigma(x) = 5x^3 - 3x^2 + 2x + 1 \) MPa. Find the third derivative at \( x = 1 \) m. (4 marks)
The weld bead width \( w(x) = -3x^2 + 12x + 5 \) mm depends on welding speed \( x \) (cm/s). Find the speed that maximises the bead width. (5 marks)
Heat input varies with time as \( H(t) = 2t^3 - 15t^2 + 24t + 10 \) kJ. Determine when heat input is minimum. (6 marks)
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Create a free accountThis chapter focused on the application of calculus in engineering mathematics, beginning with differentiation up to the third order. It explored various types of functions including linear, trigonometric, logarithmic, and exponential, followed by differentiation rules such as the power, product, chain, and quotient rules. Practical applications of differentiation were demonstrated through finding stationary points and calculating rates of change. The chapter then introduced integration, covering standard and definite integrals, and advanced techniques including integration by parts, substitution, and partial fractions. Applications of integration were examined in computing areas between and under curves as well as volumes. Further topics included solving differential equations, evaluating double and triple integrals, and the use of Laplace transforms and Fourier series to analyze complex engineering problems. Together, these concepts provide essential tools for solving real-world engineering challenges involving change and accumulation.
Differentiate the function \(f(x) = 5x^3 - 2x + 7\) and find \(f'(x)\). (2 marks)
Find the first derivative of \(f(\theta) = 4 \sin \theta + 3 \cos \theta\) where \(\theta\) is in radians. (2 marks)
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