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If Φ is a differentiable scalar function, then div grad Φ is equal to
Correct Answer :- a
Explanation : curl F = ∇ × F
= (∂/∂x, ∂/∂y, ∂/∂z) × (F1, F2, F3)
(∂F3/∂y − ∂F2/∂z , ∂F1/∂z − ∂F3/∂x , ∂F2/∂x − ∂F1/∂y)
= (∂z/∂y − ∂y/∂z , ∂x/∂z − ∂z/∂x, ∂y/∂x − ∂x/∂y)
= (0, 0, 0)
Note that since this curl is 0, the radial vector field
F(x, y, z) = (x, y, z) is irrotational.
Directional derivative of ψ(x,y,z) = xy2 + 4xyz + z2 at the point (1, 2, 3) in the direction of is
Let W be the region bounded by the planes x = 0, y = 0, y = 3, z = 0 and x + 2z = 6. Let S be the boundary of this region. Using gauss’ divergence theorem, evaluate, where
and ň is the outward unit normal vector to S.
By gauss divergence theorem
If is a differentiable vector point function, then the value of div curl
is
Let and
Q. The unit vector perpendicular to the plane containing and
is
Given that
and
Therefore, The unit vector perpendicular to the plane containing vector and
is
Let us consider the scalar point function f(x y, z) =x2 + y2 + z2
Q. The grad of f(x, y, z) is
Grad f=
Let us consider the scalar point function f(x y, z) =x2 + y2 + z2
Q. The directional derivative of f(x, y, z) at the point P(1, 1, 1) along is
(Grad f)(1,1,1)=
Now, the directional derivative o f f at P(1,1,1) along is
Let where a, b and c are constants and S is the surface of unit sphere.
Q. The value of is
By Gauss divergence theorem,
Let where a, b and c are constants and S is the surface of unit sphere.
Q. The value of is
By Gauss divergence theorem,
Since, V is enclosed by a sphere of unit radius. Thereofore
a + b + c = 1
and
We take,
= 1 - 2 + 1 = 0
So, B is normal to A.
If and curve C is the arc of the curve y = x3 from (0,0) to (2,8), then the value of
Since, C is the curve y = x3 from (0,0) to (2,8)
So, let x = t ⇒ y = t3
If is the position vector of any point on C, then
or
or
At (0, 0) ⇒ t = x = 0 and at (2, 8) ⇒ t = 2
So,
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