Application of Partial Derivatives in Engineering Mathematics (original) (raw)
Last Updated : 23 Jul, 2025
Partial Derivatives can be used to find the maximum and minimum value (if they exist) of a two-variable function. We try to locate a stationary point with zero slope and then trace maximum and minimum values near it. The practical application of maxima/minima is to maximize profit for a given curve or minimize losses.
Let f(x, y) be a real-valued function and let (pt, pt') be the interior points in the domain of f(x, y) then,
- pt, pt' is called a point of local maxima if there is an h > 0 such that f(pt, pt') ≥ f(x,y), for all x in (pt – h, pt' + h), x≠a The value f(pt, pt') is called the local maximum value of f(x,y).
- pt, pt' is called a point of local minima if there is an h < 0 such that f(pt, pt') ≥ f(x,y), for all x in (pt – h, pt' + h), x≠a The value f(pt, pt') is called the local minimum value of f(x,y).
Partial Derivatives
A partial derivative of a function of several variables is its derivative with respect to one of those variables, with all other variables held constant. For a function f(x,y), the partial derivative with respect to x, denoted as ∂f/∂x, measures the rate at which f changes as x changes, while y remains fixed.
Partial derivatives are extensively used in engineering to model and solve problems involving multiple variables. These derivatives help in understanding how a system changes with respect to one variable while keeping others constant, providing essential insights into the behavior of physical systems.
**Algorithm to Find Maxima and Minima
- Find the values of x and y using
**f xx =0 and **f yy =0
[**NOTE: fxx and fyy are the partial double derivatives of the function with respect to x and y respectively.]
- The obtained result will be considered as **stationary/turning points for the curve.
- Create 3 new variables r,t, and s.
- Find the values of r,t and s using
**r=f xx, t=f yy , s=f xy
- Perform the below operation :
**If (**rt-s 2 )| (stationary pts) >0 ****(Maxima/Minima) exists**
**If (rt-s 2 )| (stationary pts) <0 (No Maxima/Minima)/(Saddle point)
- and now check for r below :
If r >0 (Minima)
If **r <0 (Maxima)
Applications of Partial Derivatives
Here are the some applications of the partial derivatives in the engineering mathematics :
- **Heat Transfer (**Fourier's Law****)**: In heat conduction, temperature varies with both time and space. Partial derivatives describe how temperature changes across a solid.
- Equation example:
q = −k (∂T / ∂x)
- where q is heat flux, k is thermal conductivity, and ∂T / ∂x is the temperature gradient.
- **Fluid Dynamics (Navier-Stokes Equations): These partial differential equations describe the motion of fluid substances (liquids and gases). They involve velocity components, pressure, density, and viscosity.
Application areas: weather forecasting, aerodynamics, and oceanography. - **Structural Analysis (**Stress and Strain Analysis): Partial derivatives help determine how internal forces (stress) and deformations (strain) vary across a structure. Important for designing buildings, bridges, aircraft, and other load-bearing systems.
- **Electromagnetics (**Maxwell’s Equations): These fundamental equations describe how electric and magnetic fields evolve and interact. They rely on partial derivatives with respect to both space and time.
Essential in: electrical engineering, telecommunications, antenna design, etc. - **Optimization Problems (Maximizing Efficiency / Minimizing Cost): Engineers use partial derivatives to find critical points of multivariable functions representing system performance.
- First-order derivatives locate critical points.
- Second-order partial derivatives help classify them as minima, maxima, or saddle points.
- Applications: manufacturing, energy systems, logistics, etc.
Examples on the above Applications
**Example 1 : The function f(x,y)=x2y−3xy+2y+x has
(a) No local extremum.
(b) One local minimum but no local maximum.
(c) One local maximum but no local minimum.
(d) One local minimum and one local maximum.
**Solution:
r = ∂2f / ∂x2 = 2y
s = ∂2f / ∂x ∂y = 2x−3
t = ∂2f / ∂y2 = 0
Since, rt−s2≤0, (if rt-s2< 0 then we have no maxima or minima, if = 0 then we can't say anything).
Maxima will exist when rt−s2>0 and r<0.
Minima will exist when rt−s2>0 and r>0.
**As rt−s 2 is never greater than 0 so we have no local extremum.
**Answer: A
****Example 2 :**Find the local minima of the function f(x , y) = 2x2 + 2xy + 2y2 - 6x
fx(x,y) = 4x + 2y - 6=0 (1)
fy(x,y) = 2x + 4y=0 (2)
On solving (1) and (2) we get,
x=2,y=-1
r = ∂2f / ∂x2 = 4
s = ∂2f / ∂x∂y = 2
t = ∂2f / ∂y2=4
rt−s2=12
As rt−s2>0 and r>0.
**Answer : (2,-1) is the point of local minima.
****Example 3 :**Find the maxima/minima of f(x , y) = x2+y2 + 6x +12
fx(x,y) = 2x+6=0 (1)
fy(x,y) = 2y=0 (2)
On solving (1) and (2) we get,
**x=-3,y=0
r=∂2f/∂x2=2
s=∂2f/∂x∂y=0
t=∂2f/∂y2=2
As rt−s2>0 and r>0.
**Answer : (-3,0) is the point of local minima
**Example 4 :Heat Conduction Problem: The temperature T(x,y) of a thin plate is given by T(x,y) = x2 + 2xy + y2. Find the rate of change of temperature with respect to x at the point (1,2).
**Solution:
∂T/∂x = 2x + 2y
At (1,2): ∂T/∂x = 2(1) + 2(2) = 6
The rate of change of temperature with respect to x at (1,2) is 6 units/x.
**Answer : 6 units/x
**Example 5 : Fluid Dynamics Problem: The velocity field of a fluid is given by v(x,y) = 3x2y i + (x3 + y2) j. Find the acceleration in the x-direction at the point (2,1).
**Solution:
ax = ∂vx/∂t + vx(∂vx/∂x) + vy(∂vx/∂y)
∂vx/∂x = 6xy
∂vx/∂y = 3x2
At (2,1): ax = 0 + (12)(6) + (5)(6) = 102 units/s2
**Answer : 102 units/s2
**Example 6 :Optimization in Structural Engineering Problem : The deflection of a beam is given by w(x,t) = (L3 / 48EI) * (4x3/L3 - 3x/L) * sin(πt/T), where L is the length, E is Young's modulus, I is the moment of inertia, and T is the period. Find the maximum deflection with respect to x.
**Solution:
∂w/∂x = (L3 / 48EI) * (4x3/L3- 3/L) * sin(πt/T)
Set ∂w/∂x = 0:
12x2/L3 - 3/L = 0
x2 = L2/4
x = L/2
The maximum deflection occurs at the midpoint of the beam.
**Answer : x = L/2
**Example 7 : Electromagnetics Problem: The electric potential in a region is given by V(x,y,z) = 3x2y - 2yz2 + 5xz. Find the electric field components at (1,1,1).
**Solution:
Ex = -∂V/∂x = -6xy - 5z
Ey = -∂V/∂y = -3x2 + 2z2
Ez = -∂V/∂z = 4yz - 5x
At (1,1,1):
Ex = -11, Ey = -1, Ez = -1
The electric field at (1,1,1) is E = -11i - j - k.
**Answer : E = -11i - j - k.
**Example 8 : Thermodynamics Problem : The pressure P of an ideal gas is given by P(V,T) = nRT/V, where n is the number of moles, R is the gas constant, V is volume, and T is temperature. Find (∂P/∂V)T and (∂P/∂T)V.
**Solution:
(∂P/∂V)T = -nRT/V2
(∂P/∂T)V = nR/V
**Example 9 : Control Systems Problem: A transfer function G(s) = K / (s2 + 2ζωns + ωn2) represents a second-order system. Find ∂G/∂K and ∂G/∂ζ.
**Solution:
∂G/∂K = 1 / (s2+ 2ζωns + ωn2)
∂G/∂ζ = (-2Kωns) / (s2 + 2ζωns + ωn2)2
**Example 10 : Elasticity Problem: The strain energy density of a material is given by U(ε1,ε2) = (E/2) * (ε12 + ε22 + 2νε1ε2), where E is Young's modulus, ν is Poisson's ratio, and ε1, ε2 are principal strains. Find the stress σ1.
**Solution:
σ1 = ∂U/∂ε1 = E(ε1 + νε2)
**Practice Questions
- The displacement of a particle is given by s(t) = 3t2 - 2t + 1. Find the velocity and acceleration at t = 2.
- The pressure in a fluid is given by P(x,y,z) = 5x2y + 3yz - 2z2. Calculate the pressure gradient at the point (1,2,3).
- A company's profit function is P(x,y) = 100x + 80y - 2x2 - xy - y2, where x and y are the quantities of two products. Find the values of x and y that maximize profit.
- The temperature distribution in a plate is T(x,y) = 100 - 2x2 - 3y2. Find the direction of maximum temperature decrease at the point (2,1).
- The magnetic field in a region is given by B(x,y,z) = 2xyi + (x2 - z2)j + 3yzk. Calculate curl B at (1,1,1).
- A chemical reaction rate is given by r(T,C) = k0 * exp(-E/RT) * Cn, where T is temperature and C is concentration. Find ∂r/∂T and ∂r/∂C.
- The deflection of a membrane is w(x,y) = A * sin(πx/L) * sin(πy/W), where L and W are the membrane dimensions. Find the points of maximum deflection.
- In a heat exchanger, the temperature difference is ΔT(x,t) = ΔT0 * exp(-Ux/mc) * (1 - exp(-t/τ)), where U, m, c, and τ are constants. Find ∂(ΔT)/∂x and ∂(ΔT)/∂t.
- The stress-strain relationship for a material is σ = E * (ε - α * ΔT), where E is Young's modulus, α is the thermal expansion coefficient, and ΔT is the temperature change. Find ∂σ/∂ε and ∂σ/∂T.
- A control system has a transfer function G(s) = K / (Ts + 1)2. Find ∂G/∂K and ∂G/∂T.