Vector Calculus - 2 - Mathematics MCQ

Correct Answer :- c

Explanation : y = 2x²

dy/dx = 4x 

So F. dr

∫(x = 0 to 1) [3x(2x²)dx . (2x²)² d(2x²)]

∫(x = 0 to 1) (6x² - 16x⁵)dx

= -7/6

We know that in Green’s theorem,

On comparing, we get M = -y² and N = xy

We have,



Thus, directional derivative of f in the direction of   the point P(1, 1, 1),

Problem is to find unit normal vector to the surface of the sphere
f = x² + y² + z² - 1 = 0
We know unit normal vector  is given by



 

Since, From Green’s theorem, we have 

(A is the region of the figure)

On changing in polar coordinates, we get
 

or u = - y²x, v = 2yx²



In two-dim ensional flow, equation of continuity

Fluid is incompressible at this point.


Fluid flow is rotational.
Thus, fluid flow at  is rotational and incompressible.

Given, F(x, y) = In(x² + z²) = log (y + z²)

Coordinate of point p is (4, 0).




or  

Given,

Vectors  are coplanar, if

Since, On comparing  with  we get M = x² + xy and 
N = x² + y²
So, from Green’s theorem

Since, We have 
On comparing Eq. (r) with 

The given problem involves the reciprocal system of vectors a', b', and c', which are the reciprocal vectors to a, b, and c. We need to evaluate the expression:

a × a' + b × b' + c × c'

Step 1: Understanding the Reciprocal Vectors
The reciprocal vectors a', b', and c' satisfy the following relationships with the vectors a, b, and c:

a' · a = 1, a' · b = 0, a' · c = 0
b' · a = 0, b' · b = 1, b' · c = 0
c' · a = 0, c' · b = 0, c' · c = 1
These relationships mean that a', b', and c' are orthogonal to the original vectors a, b, and c, and they satisfy the orthonormal conditions.

Step 2: Cross Products Involving Reciprocal Vectors
The expression involves cross products, and we can use the properties of the reciprocal system to analyze them:

a × a' = 0, because a and a' are parallel by definition.
Similarly, b × b' = 0 and c × c' = 0.
Thus, all three cross products in the expression are zero:

a × a' + b × b' + c × c' = 0 + 0 + 0 = 0

Final Answer:
The value of the expression is 0.

Since div(curl⇀v)=0, the net rate of flow in vector field curl⇀v\) at any point is zero. Taking the curl of vector field ⇀F eliminates whatever divergence was present in ⇀F

If P (x,y,z) is a variable point in a three- dimensional space, O the origin, i, j and k the unit vectors along the x-axis, y-axis and z-axis respectively, then the vector OP given by x i + y j +z k is called the position vector of the point P and is denoted by r.

Divergence of any vector f = f1i+ f2j +f3k denoted by div f is defined as the scalar ( delta f1)/(delta x) + (delta f2)/(delta y) + (delta f3)/(delta z)

where the delta s denote partial derivatives.

Using this definition we find that div r = delta (x)/ delta x +delta (y)/delta y + delta (z)/ delta z = 1 + 1 +1 = 3.

Hence, divergence of a position vector = div r = 3.