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Test: Directional Derivatives - Civil Engineering (CE) MCQ


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15 Questions MCQ Test Engineering Mathematics - Test: Directional Derivatives

Test: Directional Derivatives for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Directional Derivatives questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Directional Derivatives MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Directional Derivatives below.
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*Answer can only contain numeric values
Test: Directional Derivatives - Question 1

The directional derivative of ϕ = xy + yz + zx along the tangent to the curve at x = t, y = t2, z = t3 at P(1,1,1) is A/√14, then the value of A is:


Detailed Solution for Test: Directional Derivatives - Question 1

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

Test: Directional Derivatives - Question 2

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:

Detailed Solution for Test: Directional Derivatives - Question 2

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.

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*Answer can only contain numeric values
Test: Directional Derivatives - Question 3

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


Detailed Solution for Test: Directional Derivatives - Question 3

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), 

Test: Directional Derivatives - Question 4

The directional derivative of f(x, y, z) = xyz at point (-1, 1, 3) in the direction of vector î - 2ĵ + 2k̂ is

Detailed Solution for Test: Directional Derivatives - Question 4

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 = 

Test: Directional Derivatives - Question 5

The maximum value of the directional derivative of the function ϕ = 2x2 + 3y2 + 5z2 at point (1, 1, -1) is

Detailed Solution for Test: Directional Derivatives - Question 5

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,

∴ 

Test: Directional Derivatives - Question 6

The value of the directional derivative of the function θ (x, y, z) = xy2 + yz2 + zx2 at the point (2, -1, 1) in the direction of the vector p = i + 2j + 2k is

Detailed Solution for Test: Directional Derivatives - Question 6

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

Test: Directional Derivatives - Question 7

The directional derivative of the function f(x, y) = x2 + y2 along a line directed from (0,0) to (1,1), evaluated at the point x = 1, y = 1 is

Detailed Solution for Test: Directional Derivatives - Question 7

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

*Answer can only contain numeric values
Test: Directional Derivatives - Question 8

The smaller angle (in degrees) between the planes x + y + z = 1 and 2x - y + 2z = 0 is __________.


Detailed Solution for Test: Directional Derivatives - Question 8

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°

Test: Directional Derivatives - Question 9

If |p̅ (s)| is a non-zero constant, then direction of  is:

Detailed Solution for Test: Directional Derivatives - Question 9

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.

*Answer can only contain numeric values
Test: Directional Derivatives - Question 10

Find the greatest value of the directional derivatives of the function f = x2yz3 at (2, 1, -1) is


Detailed Solution for Test: Directional Derivatives - Question 10

Concept:

The greatest value of directional derivative = |∇ ϕ|

Analysis:

f = x2yz3  (2, 1, -1)

∇f = -4i - 4j + 12k

|Δ f| = 13.26

Test: Directional Derivatives - Question 11

The directional derivative of 1/r in the direction of is

Detailed Solution for Test: Directional Derivatives - Question 11

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:

Test: Directional Derivatives - Question 12

At point (1, 0, 3) on the surface 2x2 + 3y2 + z2 – 11 = 0, the directional derivative in the direction 
 is

Detailed Solution for Test: Directional Derivatives - Question 12

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 = 

Test: Directional Derivatives - Question 13

The magnitude of the directional derivative of the function f(x, y) = x2 + 3y2 in a direction normal to the circle x2 + y2 = 2, at the point (1, 1), is

Detailed Solution for Test: Directional Derivatives - Question 13

F(x,y) = x2 + 3y2

ϕ = x+ 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

*Answer can only contain numeric values
Test: Directional Derivatives - Question 14

A function is defined in Cartesian coordinate system as f(x, y) = xey. The value of the directional derivative of the function (in integer) at the point (2, 0) along the direction of the straight line segment from point (2, 0) to point (1/2, 2) is ______


Detailed Solution for Test: Directional Derivatives - Question 14

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

Test: Directional Derivatives - Question 15

The magnitude of the directional derivative of the function f(x, y) = x2 + 3y2 in a direction normal to the circle x2 + y2 = 2, at the point (1, 1), is

Detailed Solution for Test: Directional Derivatives - Question 15

F(x,y) = x2 + 3y2

ϕ = x+ y2 – 2

point P (1, 1)

Normal to circle x2 + y2 - 2 is

F(x,y) = x2 + 3y2

ϕ = x+ 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

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