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If is the unit outward drawn normal to any dosed surface S, then
Let T (x, y, z) = xy^{2} + 2z – x^{2}z^{2 }be the temperature at the point (x, y, z). The unit vector in the direction in which the temperature decrease most rapidly at (1, 0, – 1) is
Let T(x, y), z) = xy^{2} + 2z – x^{2}z^{2} be the temperature at a point (x, y, z). Temperature increase most rapidly in the direction of gradient i.e., ∇ T
So, temperature decreases most rapidly in the direction of .
Let S be the oriented surface x^{2} + y^{2} + z^{2 }= 1 with the unit normal n pointing outward. For the vector field F(x, y, z) = xi + yj + zk, the value of is
By Guass is divergence theorem,
,
where V is the volume enclosed by S
since the volume V enclosed by a sphere of unit radius is equal to (4/3)π(1^{3}) = (4/3)π .
Let denote the force field on a particle traversing the path L from (0, 0, 0) to (1, 1, 1) along the curve of intersection of the cylinder y = x^{2} and the plane z = x. The work done by is
Work done by ,
A vector point function F is said to be solenoidal, if and only if
The work done in moving a particle in the force field along the line joining (0 , 0, 0) to (2 ,1, 3) is
Correct Answer : B
Explanation : Straight line x = 2t, y = t, z = 3t 0 ≤ t ≤ 1
Work done = ∫F · dr
= ∫(0 to 1) F·dr/dt * dt
= ∫(0 to 1) F·(dr/dt)dt
= ∫(0 to 1)[3(2t)^{2}i + (2.2t.3t − t)j + 3tk] · [2i + j + 3k]dt
= ∫(0 to 1)[24t^{2} + 12t^{2} − t + 9t]dt
= [8t^{3} + 4t^{3}(1/2t^{2}) + 9/2t^{2}](0 to 1)
= 8 + 4 + 4
= 16
Green's theorem gives the relationship between a line integral around a simple closed curve, C, in a plane and a double integral over the plane region R bounded by C. It is a special twodimensional case of the more general Stokes' theorem.
Stokes’ Theorem gives the relationship between a line integral around a simple closed curve, C, in space, and a surface integral over a piecewise, smooth surface.
Green’s theorem in its “curl form”.
where F = P(x,y) i + Q(x,y) j and dr = dx i + dy j
is as follows: (curl form of Green’s Theorem)
c∫ F(x,y) . dr = c∫ F . T ds = r∫ ∫ curl F dA
where curl F is the zcomponent of curl F = curl F . k
for stoke’s theorem
If φ = sin (x + y + 2z) + x^{2} y^{3}z, then curl grad φ is equal to
Correct Answer : a
Explanation :
a * (b * c) + b * (c * a) + c * (a * b) = 0
L.H.S.
a * (b * c) + b * (c * a) + c * (a * b)
= a * ( a) + b * (b) + c * ( c)
[∵ b * c = a ; c * a = b and a * b = c ]
= a * a + b * b + c * c
= 0 + 0 + 0
[∵ a * a= 0; b * b = 0 ; c * c= 0]
= 0 + 0 + 0
= 0
= R.H.S.
a * (b * c) + b * (c * a) + c * (a * b) = 0,
Hence proved
Consider the vector field , where a is a constant. If , then the value of a is
⇒ – (ax + y + a) + 1 + x + y = 0
⇒ (– a + 1)x + (– a + 1) = 0
⇒ (– a + 1) (1 + x) = 0
⇒ a = 1 (because x is not constant).
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