Partial differentiation and its applications notes — Unit 3

Ordinary differentiation (dy/dx) works when a function depends on only ONE variable. But in engineering, almost everything depends on multiple variables. For example, the volume of a cylinder V = πr²h depends on both radius 'r' and height 'h'. If we heat the cylinder, both r and h will expand.

⇒The Core Concept

Partial differentiation asks: 'How does the function change if I vary ONE variable while freezing all the others?'

Let's perform a dry run of finding first and second order partial derivatives for a function.

z = x³ + y³ - 3axy

1. 1.1. Find ∂z/∂x. Treat 'y' and 'a' as constants.
2. 2. ∂(x³)/∂x = 3x²
3. 3. ∂(y³)/∂x = 0 (since y is a constant here)
4. 4. ∂(-3axy)/∂x = -3ay * (1) = -3ay
5. 5. Result: ∂z/∂x = 3x² - 3ay
6. 6.
7. 7.2. Find ∂z/∂y. Treat 'x' and 'a' as constants.
8. 8. ∂(x³)/∂y = 0
9. 9. ∂(y³)/∂y = 3y²
10. 10. ∂(-3axy)/∂y = -3ax * (1) = -3ax
11. 11. Result: ∂z/∂y = 3y² - 3ax

Just like ordinary derivatives, we can differentiate again to get second-order partial derivatives.

Let's prove Clairaut's theorem holds for our previous example: z = x³ + y³ - 3axy.

From Page 2, we found:
∂z/∂x = 3x² - 3ay
∂z/∂y = 3y² - 3ax

1. 1.Let's find ∂²z/∂y∂x. This means taking ∂z/∂x and differentiating it w.r.t y.
2. 2.∂(3x² - 3ay)/∂y = 0 - 3a(1) = -3a
3. 3.
4. 4.Now let's find ∂²z/∂x∂y. This means taking ∂z/∂y and differentiating it w.r.t x.
5. 5.∂(3y² - 3ax)/∂x = 0 - 3a(1) = -3a

A function f(x, y) is called a 'Homogeneous Function' of degree 'n' if, when you replace x with tx and y with ty, you can factor out exactly tⁿ, leaving the original function completely untouched.

f(tx, ty) = tⁿ * f(x, y)

1. 1.Example: f(x,y) = x² + xy + y²
2. 2.Replace x with (tx) and y with (ty):
3. 3.f(tx, ty) = (tx)² + (tx)(ty) + (ty)²
4. 4. = t²x² + t²xy + t²y²
5. 5.Factor out t²:
6. 6. = t² (x² + xy + y²)
7. 7. = t² * f(x, y)
8. 8.Therefore, it is homogeneous of degree 2.

If a function is Homogeneous, you don't need to do massive partial differentiations to find complex combinations. Euler discovered a beautiful shortcut.

There is also a second-order deduction of Euler's theorem that is frequently tested:

x²(∂²f/∂x²) + 2xy(∂²f/∂x∂y) + y²(∂²f/∂y²) = n(n-1)f

Most exam questions don't give you a directly homogeneous function. They give you a function wrapped in a trigonometric or inverse function. You must 'unwrap' it first.

Given: u = sin⁻¹( (x² + y²) / (x + y) )
Prove: x(∂u/∂x) + y(∂u/∂y) = tan u

1. 1.Step 1: Check if 'u' is homogeneous. It is NOT, because the sin⁻¹ blocks us from factoring out 't'.
2. 2.Step 2: Unwrap it. Move sin⁻¹ to the other side.
3. 3.sin u = (x² + y²) / (x + y)
4. 4.Let f = sin u = (x² + y²) / (x + y)
5. 5.Step 3: Now check if 'f' is homogeneous.
6. 6.Numerator degree = 2. Denominator degree = 1.
7. 7.Degree of f = 2 - 1 = 1. So, f is homogeneous of degree n = 1.

Now apply Euler's Theorem to the unwrapped function 'f', NOT to 'u'.

By Euler's Theorem on f:
x(∂f/∂x) + y(∂f/∂y) = n * f

We know n = 1, and f = sin u. Substitute these in:
x [∂(sin u)/∂x] + y [∂(sin u)/∂y] = 1 * (sin u)

Apply chain rule to differentiate sin u:
x [cos u * ∂u/∂x] + y [cos u * ∂u/∂y] = sin u

Factor out cos u:
cos u * [ x(∂u/∂x) + y(∂u/∂y) ] = sin u

Divide by cos u:
x(∂u/∂x) + y(∂u/∂y) = sin u / cos u
x(∂u/∂x) + y(∂u/∂y) = tan u

Partial derivatives assume independent variables don't affect each other. But what if x and y are both dependent on a third variable, like time 't'? For example, as a balloon is heated over time 't', its pressure 'P' changes, but its volume 'V' and temperature 'T' also change over time 't'.

⇒The Total Derivative Formula

If z = f(x, y), and x = g(t), y = h(t), then the total change in z with respect to t is the sum of the changes coming through x and through y.

dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt)

Sometimes, x and y are locked together in a tangled equation like x³ + y³ - 3axy = 0, and you can't isolate y = f(x). You can find dy/dx using the total derivative concept.

1. 1.Why does this work?
2. 2.1. Take the total derivative of f(x,y)=c with respect to x.
3. 3.2. (∂f/∂x)(dx/dx) + (∂f/∂y)(dy/dx) = 0
4. 4.3. Since dx/dx = 1, we get: ∂f/∂x + (∂f/∂y)(dy/dx) = 0
5. 5.4. Rearrange to solve for dy/dx.

This formula is a lifesaver. Instead of doing messy implicit differentiation with the product rule, just do two simple partial derivatives and divide them.

Find dy/dx if xʸ = yˣ

Step 1: Move everything to one side to form f(x,y) = 0.
f(x,y) = xʸ - yˣ = 0

1. 1.Step 2: Find ∂f/∂x (treat y as constant).
2. 2.Derivative of xʸ is y*xʸ⁻¹ (like derivative of x² is 2x).
3. 3.Derivative of yˣ is yˣ log(y) (like derivative of 2ˣ is 2ˣlog(2)).
4. 4.∂f/∂x = yxʸ⁻¹ - yˣlog(y)
5. 5.
6. 6.Step 3: Find ∂f/∂y (treat x as constant).
7. 7.Derivative of xʸ is xʸ * log(x).
8. 8.Derivative of yˣ is x*yˣ⁻¹.
9. 9.∂f/∂y = xʸlog(x) - xyˣ⁻¹

Step 4: Plug into the magic formula.
dy/dx = - (∂f/∂x) / (∂f/∂y)
dy/dx = - [ y*xʸ⁻¹ - yˣ*log(y) ] / [ xʸ*log(x) - x*yˣ⁻¹ ]
dy/dx = [ yˣ*log(y) - y*xʸ⁻¹ ] / [ xʸ*log(x) - x*yˣ⁻¹ ]

Done in 4 lines. If you tried to take logs of both sides and differentiate implicitly with product rules, it would take a full page.

In calculus, if we change variables from (x) to (u), we multiply by dx/du. But what if we change from 2D coordinates (x,y) to 2D coordinates (u,v)? We can't just divide vectors. We use a matrix determinant called the Jacobian.

The Jacobian is the ultimate test to see if two functions are secretly the same thing just disguised algebraically. If two functions u and v are functionally dependent (meaning u = f(v)), their Jacobian will be exactly ZERO.

If J(u,v / x,y) = 0 everywhere, then u and v are functionally dependent.
If J(u,v / x,y) ≠ 0, they are independent.

1. 1.Exam Question Pattern:
2. 2.1. You are given u = (x+y)/(1-xy) and v = tan⁻¹x + tan⁻¹y.
3. 3.2. Question asks: 'Are u and v functionally dependent? If so, find the relation.'
4. 4.3. Calculate the 2x2 Jacobian determinant.
5. 5.4. Prove it equals 0.
6. 6.5. Look at the formulas and realize v = tan⁻¹(u). That's the relation!

Find J(x,y,z / r,θ,φ) for Spherical Coordinates:

x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ

1. 1.Step 1: Set up the 3x3 determinant.
2. 2.Row 1: ∂x/∂r, ∂x/∂θ, ∂x/∂φ
3. 3.Row 2: ∂y/∂r, ∂y/∂θ, ∂y/∂φ
4. 4.Row 3: ∂z/∂r, ∂z/∂θ, ∂z/∂φ
5. 5.
6. 6.Step 2: Differentiate.
7. 7.Row 1: sinθ cosφ, r cosθ cosφ, -r sinθ sinφ
8. 8.Row 2: sinθ sinφ, r cosθ sinφ, r sinθ cosφ
9. 9.Row 3: cosθ, -r sinθ, 0

Step 3: Expand the determinant along Row 3 (since it has a zero).
= cosθ [ r²cosθsinθcos²φ - (-r²cosθsinθsin²φ) ]
  - (-r sinθ) [ r sin²θcos²φ - (-r sin²θsin²φ) ]

Factor out the common terms and use sin²φ + cos²φ = 1:
= cosθ [ r²cosθsinθ (1) ] + r sinθ [ r sin²θ (1) ]
= r² cos²θ sinθ + r² sin³θ
= r² sinθ (cos²θ + sin²θ)
= r² sinθ

For a 1D function y = f(x), we found max/min by setting f'(x) = 0 and checking if f''(x) is positive or negative. For a 3D surface z = f(x,y), we have to check multiple directions.

1. 1.Step 1: Find the first partial derivatives: p = ∂f/∂x, and q = ∂f/∂y.
2. 2.Step 2: Set p = 0 and q = 0. Solve these simultaneous equations to find the 'Stationary Points' (a,b).
3. 3.Step 3: Calculate the second order derivatives at (a,b):
4. 4. r = ∂²f/∂x²
5. 5. s = ∂²f/∂x∂y
6. 6. t = ∂²f/∂y²

Now, we use the (rt - s²) test to classify each stationary point.

Evaluate the expression Δ = rt - s² at the stationary point (a,b).

Find the maxima and minima of f(x,y) = x³ + y³ - 3axy (assume a > 0).

Step 1: First derivatives
p = 3x² - 3ay = 0  => x² = ay
q = 3y² - 3ax = 0  => y² = ax

Step 2: Solve for stationary points
Substitute y = x²/a into the second equation:
(x²/a)² = ax  =>  x⁴/a² = ax  =>  x⁴ - a³x = 0
x(x³ - a³) = 0  =>  x = 0, x = a.
If x = 0, y = 0. Point 1: (0,0)
If x = a, y = a. Point 2: (a,a)

1. 1.Step 3: Second derivatives
2. 2.r = ∂²f/∂x² = 6x
3. 3.s = ∂²f/∂x∂y = -3a
4. 4.t = ∂²f/∂y² = 6y

Now apply the (rt - s²) test to both points.

Test Point 1: (0,0)
r = 6(0) = 0
s = -3a
t = 6(0) = 0
rt - s² = (0)(0) - (-3a)² = -9a²
Since a > 0, -9a² is NEGATIVE. Therefore, (0,0) is a Saddle Point.

Test Point 2: (a,a)
r = 6a
s = -3a
t = 6a
rt - s² = (6a)(6a) - (-3a)² = 36a² - 9a² = 27a²
Since a > 0, 27a² is POSITIVE. The test succeeds.
Now check 'r': r = 6a, which is POSITIVE.
Therefore, (a,a) is a MINIMUM point.

What if you need to find the max/min of f(x,y,z), but the variables are restricted by some constraint, like they must lie on the surface of a sphere φ(x,y,z) = 0? You can't just set derivatives to zero because the variables aren't independent anymore. Enter Lagrange.

1. 1.Step 1: Write the target function f(x,y,z) and the constraint function φ(x,y,z) = 0.
2. 2.Step 2: Construct the Lagrangian Function: F = f + λφ
3. 3.Step 3: Find partial derivatives of F w.r.t x, y, and z, and set them to zero.
4. 4. ∂F/∂x = ∂f/∂x + λ(∂φ/∂x) = 0
5. 5. ∂F/∂y = ∂f/∂y + λ(∂φ/∂y) = 0
6. 6. ∂F/∂z = ∂f/∂z + λ(∂φ/∂z) = 0
7. 7.Step 4: You now have 4 equations (the 3 above + the constraint φ=0) and 4 unknowns (x,y,z,λ).
8. 8.Step 5: Solve algebraically to eliminate λ and find the points x, y, z.

⇒Unit 3 Mastery

• Do you remember to treat the other variables as strict constants in partial diff?
• Can you cross-check your math using Clairaut's theorem (∂²z/∂x∂y = ∂²z/∂y∂x)?
• Can you accurately determine the degree 'n' for a homogeneous function?
• Do you know the 3-step 'unwrap' algorithm for Euler's theorem proofs?
• Can you apply the implicit differentiation magic formula: dy/dx = -(∂f/∂x)/(∂f/∂y)?
• Can you calculate a 3x3 Jacobian without dropping signs?
• Do you know the (rt - s²) test conditions for Max, Min, and Saddle point?
• Can you set up the F = f + λφ equation for Lagrange multipliers?

Unit 3 is heavily algebraic. If you checked all these boxes, and you practice your cross-multiplication and factoring, you will ace this section.