The direction derivative of the scalar function f(x, y, z) = x2 + 2y2 + z at the point P = (1, 1, 2) in the direction of the vector
At (1, 1, 2)
So, direction derivative of at P(1, 1, 2) in direction of
The maximum value of the directional derivative of the function φ = 2x2 + 3y2 + 5z2 at a point (1, 1, -1) is
Hence, vector field is irrotational and divergence-free.
Directional derivative of φ is
As x = 1, y = 1, z = - 1
Magnitude of directional derivative is
A velocity vector is given as Then, the divergence of this velocity vector at (1, 1, 1) is
= 5y + 4y + 6yz
If a vector has a constant magnitude, then
is of constant magnitude.
where P is a vector, is equal to
We have from the property of vector triple product
then the value of curl
Find the area of a triangle formed by the tips of vectors
In figure from A, draw AD ⊥ BC. So, in right angled
So, from area of ΔABC
The angle between two unit magnitude coplanar vectors P(0.866,0.500,0) and Q(0.259,0.966,0) will be
The directional derivative of in the direction of
Problem is to find directional derivative of
f at A(1, 1) along the vector
The inner (dot) product o f two non-zero vectors and is zero. The angle (degrees) between the two vectors is
A sphere of unit radius is centred at the origin. The unit normal at point (x, y, z) on the surface of the sphere is the vector
Equation of sphere of unit radius having centre at origin is
f (x, y, z) = x2 + y2 + z2 - 1 = 0
being unit normal at (x,y, z) on the surface.
Let x and y be two vectors in a three-dimensional space and < x, y > denote their dot product. Then the determinant,
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.
For an incompressible flow, the x and y components of the velocity vector are = 3(y | z), where x, y and z are in metre, all velocities are in m/s. Then, the z component of the velocity vector of the flow for the boundary condition at z = 0, is
(where, vz = 0 at z = 0 and fluid is incompressible) For incompressible fluid flow, equation of continuity is
The gradient of the scalar fleld f(x, y) = y2 - 4xy at (1,2) is
The value of the surface integral evaluated over the surface of a cube having sides of length a is is unit normal vector)
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
Value of the integral where C is the square cut from the first quadrant by the lines x = 1 and y = 1 will be (use Green’s theorem to change the line integral into double integral)
From Green’s theorem
Determine the following integral
where, is the position vector field and S is the surface of a sphere of radius R.
S being surface of a sphere of radius R.
(using Gauss’s divergence theorem)
are three points having coordinates (3, -2, -1), (1, 3,4) and (2, -1, -2) in XY and Z-plane, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given by
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).
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).
The area of the loop of Descartes’s Folium x3 + y3 = 3 axy is
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
The divergence of a vector field A is always equal to zero, if the vector field A can be expressed as
Now, we consider
Div curl B
Thus, curl B = A and div A = 0