Application of Partial Derivatives in Engineering Mathematics

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: f_xx and f_yy 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-s²)|_{(stationary pts)}>0(Maxima/Minima) exists

If (rt-s²)|_{(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−s² 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

f_x(x,y) = 4x + 2y - 6=0 (1)
f_y(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

f_x(x,y) = 2x+6=0 (1)
f_y(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

1. The displacement of a particle is given by s(t) = 3t² - 2t + 1. Find the velocity and acceleration at t = 2.
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).
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.
4. 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).
5. 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).
6. 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.
7. 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.
8. 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.
9. 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.
10. A control system has a transfer function G(s) = K / (Ts + 1)². Find ∂G/∂K and ∂G/∂T.