Test: Directional Derivatives - Civil Engineering (CE) MCQ

Concept:

The directional derivative is given by:

D.D. = grad(ϕ). 

Calculation:

grad(ϕ) at P(1,1,1) is:

The equation of the curve is represented by r(t).

The tangent of this curve is given by:

The value of tangent at (1,1,1) is:

D.D. = 12√14

Concept:

Directional Derivative = Gradient of function × Unit direction Vector

If F = f(x, y, z) then,

Grad 

For the given direction vector

Unit direction vector = 

and 

Calculation:

We have, f(x, y, z) = x(x2 - y2) - z = x³ - xy² - z

⇒ 

⇒ 

⇒ 

⇒ Unit direction vector 

⇒ Unit direction vector = 

⇒ Directional derivative = 

⇒ Directional derivative = 

∴ The directional derivative of f(x, y, z) = x(x² - y²) - z at A(1, -1, 0) in the direction of p̅ = (2î - 3ĵ + 6k̂) is -8/7.

The directional derivative of f(r) = r7 in the direction of r→ will be,

⇒ 

Where,

Also, 

Calculation:

Using equations (1),

Using equations (4),

Using equations (3),

⇒ 

The directional derivative of r7 in the direction of  at (1, -1, 1), 

Concept:

Directional Derivative = Gradient of function × Unit direction Vector

If F = f(x, y, z) then,

For the given direction vector

Unit direction vector = 

and 

Calculation:

Given, f(x,y,z) = xyz

∴ (Grad f)_P = 

Direction vector 

Unit direction vector

= 

∴ Directional Derivative = 

D. D = 

Concept:

Gradient:

For a scalar function ϕ(x, y, z), the gradient is the maximum rate of change which is given by-

Directional derivative:

It gives the rate of change of scalar point function in a particular direction. The maximum magnitude of the directional derivative is the magnitude of the gradient.

Calculation:

Given:

ϕ = 2x² + 3y² + 5z²

∇ϕ = 4xî + 6yĵ + 10zk̂

∇ϕ_{(1, 1, -1)} = 4î + 6ŷ - 10k̂

Maximum value is given by the magnitude,

∴ 

Given that,

ϕ = xy² + yz² + zx²

directional vector (p) = I + 2j + 2K

Directional derivative = 

∇ϕ at the point (2, -1, 1) is

∇ϕ = ((-1)² + 2(2)(1)) î + (2(2)(-1) + (1)²) ĵ + (2(-1)(1) + (2)²)k̂

= 5î - 3ĵ + 2k̂

Directional derivative = 

= 5 - 6 + 4 / 3

= 1

Concept:

Directional derivative of a function f along the vector  is given by

grad f or ∇ f is defined by the equation,

Calculation:

f (x, y) = x² + y²

Line vector from (0,0) to (1,1) is

(1 – 0) î + (1 – 0) ĵ = î + ĵ

∴ Directional derivative of f (x, y) at (1, 1) in the direction of î + ĵ is

= 2√2

Concept:

Equation of a plane is given as:

ax + by + cz = d

The unit vector normal to the plane is given by:

Calculation:

Given two planes with equation:

x + y + z = 1     … (1)

2x - y + 2z = 0     … (2)

Unit vectors in the direction ⊥ to the respective planes are:

 and  respectively.

[∵ a̅.b̅ = |a̅| |b̅| cos (θ)]

θ = 54.73°

Concept Used:

For any vector , there are two parts one is the direction and the other is magnitude.

For a constant vector the direction as well as magnitude both remains constant.

And the direction of the zero vector is arbitrary.  

Calculation:

| p̅ (s)| is a non-zero constant,

Hence, Its derivation must be zero.

, i.e., zero vector.

Hence, it must have an arbitrary direction. i.e., it can have any direction.

Concept:

The greatest value of directional derivative = |∇ ϕ|

Analysis:

f = x²yz³  (2, 1, -1)

∇f = -4i - 4j + 12k

|Δ f| = 13.26

Concept:

Let f(r) be a function then directional derivative of the function f(r) is given by: 

Calculation:

Given:

f(r) = 1/r

As we know that, if f(r) is a function then directional derivative of the function f(r)is given by: 

f(r) = 1/r

∵ 

Here, we have to find the directional derivative of f(r) in the direction of . It will be given by:

Concept:

If any surface is given by f = f(x, y, z), To find the directional derivative at any point P(x₁, y₁, z₁) on the surface in the direction of the vector. Follow the following steps.

Find, 

Then, find the value of ∇f at point P(x₁, y₁, z₁).

Directional Derivative = 

Calculation:

Given:

f(x, y, z) = 2x² + 3y2 + z² - 11

⇒ î(4x) + ĵ(6y) + k̂(2z)

At point P(1, 0, 3), ∇f = î (4 × 1) + ĵ (6 × 0) + k̂ (2 × 3) = 4î + 6k̂ 

Direction vector (b→) = î + 2ĵ + k̂ 

Directional Derivative = 

F(x,y) = x² + 3y²

ϕ = x² + y² – 2

point P (1, 1)

Normal to circle x² + y² - 2 is

∇ϕ = 2xi + 2yj

At P (1, 1)

Normal Vector, a = 2i + 2j

Magnitude of directional derivative of f along normal vector is ∇f. a

= 2xi + 6yj

At P(1,1)

∇ f = 2i + 6j

Directional derivative, = ∇f . a

= 

= 4√2

Concept

Note

i) Gradient (∇) converts the scalar function into vector function

ii) Divergence (∇.) Converts vector function to scalar (Dot product)

iii) Curl (∇ × ) converts vector function to vector function (Cross product)

Calculation

Given, f(x, y) = xey

Let Point P is ( 2, 0), point Q is (1/2,2)

Unit vector along PQ is 

The gradient of f (x, y) is given by ∇f

Directional derivative at point P (2,0) is

 = i + 2 j 

 The directional derivative of f (x, y) at P (2, 0) along the line   is

F(x,y) = x² + 3y²

ϕ = x² + y² – 2

point P (1, 1)

Normal to circle x² + y² - 2 is

F(x,y) = x² + 3y²

ϕ = x² + y² – 2

point P (1, 1)

Normal to circle x² + y² - 2 is

Normal Vector, a = 2i + 2j

Magnitude of directional derivative of f along normal vector is ∇f. a

= 2xi + 6yj
At P(1,1)

∇ f = 2i + 6j

Directional derivative, = ∇f . a