Vector Calculus - 4 - Mathematics MCQ

Since,


At (1, 1, 2)

So, direction derivative of at P(1, 1, 2) in direction of 

Hence, vector field is irrotational and divergence-free.
Given,

Directional derivative of φ is
  (given)
As x = 1, y = 1, z = - 1

Magnitude of directional derivative is

Correct Answer :- d

Explanation : V = 5xy i + 2y^2 j + 3yz^2 k

(div V) = d/dx(5xy) + d/dx(2y^2) + d/dx(3yz^2)

= 5y + 4y + 6yz

Points = (1,1,1)

= 5(1) + 4(1) + 6(1)(1)

= 15

 is of constant magnitude.

We have from the property of vector triple product 

In figure from A, draw AD ⊥ BC. So, in right angled

So, from area of ΔABC

We have,

Given,


Problem is to find directional derivative of
f at A(1, 1) along the vector 
We take,

Equation of sphere of unit radius having centre at origin is
f (x, y, z) = x² + y² + z² - 1 = 0
So, 


 being unit normal at (x,y, z) on the surface.

and let 

So,  

Now, on putting
D = 0 or

Vector   are linearly dependent.
So, linearly dependent ⇒ D = 0 and for linearly independent ⇒ D ≠ 0
or Positive and negative
We can also see that D = (x₂y₁ - x₁y₂)² cannot be negative.
So, linearly independent ⇒ D is positive.

Given

(where, v_z = 0 at z = 0 and fluid is incompressible) For incompressible fluid flow, equation of continuity is

or  
or 
So, 

 ...(i)
S, represents the surface of a cube having sides each of length   being unit normal vector to the surface.
Problem is to evaluate I. We know from Gauss’s divergence theorem

So, using Eq. (ii) for Eq. (i), we get

From Green’s theorem

So, 

S being surface of a sphere of radius R.

(using Gauss’s divergence theorem)

and  Now, we have to find the distance of point P to the plane OQR according to figure.

We draw a perpendicular from point P to plane OQR and let it meet with perpendicular at S(x, y, z).
Then,



Now  is normal to given plane OQR  and it is also normal to 
So, vector normal to 




It also have the same direction ratios as a vector normal to 
Since  is itself normal to 

From Eqs. (iv) and (v), the values of x and z putting in Eq. (Hi) respectively, we get y = 0
Then from Eq. (iv), x = 1
and from Eq. (v), z = -2
So, points of 5 is (1, 0, -2).
So,  

Putting y = tx, we get the parametric equation of the contour of the folium as 
The loop is described as t varies from 0 to
 where θ varies

Let 


Now, we consider
Div curl B

Thus, curl B = A and div A = 0