2. • Directional derivative: Component of ∇∅ in the direction of a unit vector ā is
given by ∇∅.ā and is called the directional derivative of ∅ in the direction of ā.
• In particular ∇∅.ī = ∅ is the directional derivative in the direction of ī.
Physically ∇∅.ā is the rate of changes of ∅ in the direction of ā at any point
p(x,y,z).
Note:
1. The directional derivative of scalar point function ∅ at a point p(x,y,z) in
the direction of a unit vector ē is ē.∇∅
DIRECTIONAL DERIVATIVES
2. ∇∅ is a normal vector to the level surface ∅(x,y,z)=c at any point.
3. Maximum Value of directional derivative of ∅ at any point p is |∇∅|
and -|∇∅|is the minimum value of directional derivative of ∅ at point P.
3. 4. If [ā,b̄,c̄]=0 then ā,b̄,c̄ are co-planar.
5. grad ∅=∇∅ is a vector whose three components are , , .
6. ∇∅ may be considered as the result produced when the operator
acts on ∅ (or) ∇∅ may be assumed to be the
multiplication of vector by a scalar ∅.
❑The symbol ‘∇‘ is knownas Vector Differential Operator due
to it’s dual character(∇ is a operator and a vector).
4. Problems based on directional derivatives
1. Find the directional derivative of f=xy+yz+zx in the direction of Vector
ī+2j̄+2k̄ at the point (1,2,0)
Sol: given f=xy+yz+zx
grad f = ∇f = ∑ ī
grad f = ī(y+z)+ j̄(x+z)+k̄(y+x)
if ē is the unit vector in the direction of Vector ī+2j̄+2k̄, then
ē= ī+2j̄+2k̄/✓1^2+2^2+2^2
= 1/3 (ī+2j̄+2k̄)
5. • Directional derivative of f along the given direction= ē.∇f
= 1/3 (ī+2j̄+2k̄) . (Y+z)ī + (x+z)j̄ + (x+y)k̄ at (1,2,0)
= 1/3 [ (Y+z) + 2(x+z) + 2(x+y)] at (1,2,0)
= 1/3 [(2+0) + 2(1+0) + 2(1+2)
= 1/3 (2+2+6)
= 1/3 (10)
= 10/3
6. 2. Find a unit normal vector to the surface x^2+y^2+2z^2=26 at the point (2,2,3).
Sol: let f= x^2+y^2+2z^2-26
∇f = 2xī + 2yj̄ + 4zk̄
∇f = 4ī + 4j̄ + 12k̄
• Unit normal vector = f/| f|
= 4ī+ 4j̄+ 12k̄/✓16+16+144
= ī + j̄ + 3k̄/✓11
(2,2,3)