Vector Calculus (8hrs) · Engineering Mathematics - I

directional derivative

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directional derivative

1. Definition

For scalar field φ(x,y,z)φ(x,y,z), unit vector u=(l,m,n)u = (l,m,n) with l2+m2+n2=1l²+m²+n²=1 giving direction at P0P_0, the directional derivative is:

Duφ(P0)=limh→0[φ(P0+hu)−φ(P0)]/hD_u φ(P_0) = lim h→0 [φ(P_0 + h u) - φ(P_0)]/h

= rate of change of φφ when moving from P0P_0 a small distance hh along direction uu.

2. Formula via Gradient

If φφ differentiable at P0P_0:

Duφ(P0)=∇φ(P0)⋅u=∣∇φ∣cosθD_u φ(P_0) = ∇φ(P_0) · u = |∇φ| cosθ

where θθ = angle between ∇φ∇φ and uu. Derivation uses chain rule on φ(r0+tu)φ(r_0+ t u).

Consequences:

  • Maximum Duφ=∣∇φ∣D_u φ = |∇φ| when uu is parallel to ∇φ∇φ (θ=0θ=0) — direction of steepest increase.
  • Minimum −∣∇φ∣-|∇φ| when uu opposite ∇φ∇φ — steepest decrease.
  • Zero when u⊥∇φu ⊥ ∇φ — moving along level surface φ=cφ=c, no change (tangent direction).
  • If uu not unit, divide by ∣u∣|u|: Duφ=∇φ⋅u/∣u∣D_u φ = ∇φ·u/|u|.

3. Working Steps (recipe)

  1. Compute ∇φ=(φx,φy,φz)∇φ = (φ_x, φ_y, φ_z) at P0P_0.
  2. Make direction vector a=PQa = PQ (from P0P_0 to QQ) and unitise u=a/∣a∣u = a/|a|.
  3. Dot: Duφ=∇φ⋅uD_u φ = ∇φ·u.

For 2D φ(x,y)φ(x,y), same with u=(cosα,sinα)u=(cosα, sinα).

4. Worked Examples

(a) φ= x² + y² + z² at P(1,2,2) towards Q(2,3,3):
    ∇φ=(2x,2y,2z)=(2,4,4); vector a= Q-P=(1,1,1), |a|=√3 ⇒ u=(1,1,1)/√3
    D_u φ = (2+4+4)/√3 =10/√3 ≈5.77 ; max possible |∇φ|=6 along r direction ✓

(b) φ= x² y z at (1,2,3) along direction i+2j+2k: u=(1,2,2)/3, ∇φ=(2x y z, x² z, x² y)=(12,3,2)
    D= (12·1+3·2+2·2)/3=22/3 ✓

(c) Find direction where φ= x²+ y² decreases fastest at (1,1): ∇φ=(2,2) ⇒ opposite direction (-1,-1)/√2 gives D=-2√2 ✓

(d) Need directional derivative along tangent to curve r(t)? Use u = r'(t)/|r'|.

(e) Temperature T= x²+y²+z²: isotherms spheres; directional derivative along sphere (perpendicular to ∇T) zero ⇒ no temperature change walking on sphere ✓

5. Level Surface Connection

If uu lies in tangent plane to φ=cφ=c at P0P_0, then Duφ=0D_u φ=0 because u⋅∇φ=0u·∇φ=0 — moving on level surface keeps φφ constant to first order.

6. Exam Points

  • State Duφ=∇φ⋅uD_u φ = ∇φ·u — 3 marks, very frequent.
  • Compute DuφD_u φ at PP towards QQ — 5-7 marks, very frequent.
  • Find maximum Duφ=∣∇φ∣D_u φ = |∇φ| and its direction u=∇φ/∣∇φ∣u=∇φ/|∇φ| — 5 marks, very frequent.
  • Show Duφ=0D_u φ=0 for uu tangent to φ=cφ=c — 3 marks.
  • Second-order? Sometimes ask for rate of change of FF along curve r(t)r(t): dF/dt=∇F⋅r′(t)dF/dt = ∇F·r'(t) — 3 marks.
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