Engineering Mathematics Test 4

Concept:

The divergence of any vector field  \(\vec A\) is defined as:

\(Div= \vec \nabla .\vec A\)

The nabla operator is defined as:

\(\vec \nabla ={\hat i\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}}}+\hat k\frac{\partial }{\partial z}\)

Calculation:

\(f = {x^3}y\vec i - \left( {{z^2} - 2y} \right)\vec j + 5{y^2}z\vec k\)

\(Div\left( f \right) = \frac{\partial }{{\partial x}}\left( {{x^3}y} \right) + \frac{\partial }{{\partial y}}\left( { - \left( {{z^2} - 2y} \right)} \right) + \frac{\partial }{{\partial z}}\left( {5{y^2}z} \right)\)

= 3x²y + (+2) + (5y²)

= 3x²y + 5y² + 2

At (2, 3, 4)

Div (f) = 3(2)²(3) + 5(3)² + 2

∴ Div (f) = 36 + 45 + 2 = 83

Concept:

\({\nabla ^2}\left[ {f\left( r \right)} \right] = f''\left( r \right) + \frac{2}{r}f'\left( r \right)\)

Calculation:

\({\nabla ^2}\left[ {\log r} \right] = \frac{{ - 1}}{{{r^2}}} + \frac{2}{r}\left( {\frac{1}{r}} \right)\)

\({\nabla ^2}\left[ {\log r} \right] = \frac{{ - 1}}{{{r^2}}} + \frac{2}{{{r^2}}}\)

\(\therefore {\nabla ^2}\left[ {\log r} \right] = \frac{1}{{{r^2}}}\)

Calculation:

For an Incompressible flow, div \(\vec V = 0\)

\(div\;\left( {\vec V} \right) = \vec \nabla \cdot \vec V = \frac{\partial }{{\partial x}}\left( {2x + 3y + 4z} \right) + \frac{{\partial \left( {5x + cy + 6z} \right)}}{{\partial y}} + \frac{\partial }{{\partial z}}\left( {8x + 9y} \right){\rm{\;}} = {\rm{\;}}2{\rm{\;}} + {\rm{\;C}}\)

For incompressible flow, \(\vec \nabla \cdot \vec V = 0\)

⇒ 2 + C = 0 ⇒ C = -2

Concept:

If \(\vec a\) and \(\vec b\) are two vectors given in the component form as \(\vec a = {a_1}\hat i + {b_1}\hat j + {c_1}\hat k,\;and\;\vec b = {a_2}\hat i + {b_2}\hat j + {c_2}\hat k\). Then, their cross or vector product is,

\(\vec a \times {\rm{\;}}\vec b = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\end{array}} \right|\)

If \(\vec a\) and \(\vec b\) are two vectors given in the component form as \(\vec a = {a_1}\hat i + {b_1}\hat j + {c_1}\hat k,\;and\;\vec b = {a_2}\hat i + {b_2}\hat j + {c_2}\hat k\). Then, both are perpendicular if their dot product is zero.

Calculation:

\(\vec a = \lambda \hat i - 9\hat j - \hat k,\;\vec b = 3\hat i + 3\hat j + \hat k\;and\;\vec c = 4\hat i + 2\hat j + \hat k\)

\(\vec b \times {\rm{\;}}\vec c = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\3&3&1\\4&2&1\end{array}} \right|\)

\(= \hat i\left( {3 - 2} \right) - \hat j\left( {3 - 4} \right) + \hat k\left( {6 - 12} \right)\)

\(= \hat i + \hat j - 6\hat k\)

vector \(\vec a\) is perpendicular to \(\vec b \times {\rm{\;}}\vec c\)

\(\Rightarrow \vec a.\left( {\vec b \times {\rm{\;}}\vec c} \right) = 0\)

\(\Rightarrow \left( {\lambda \hat i - 9\hat j - \hat k} \right).\left( {\hat i + \hat j - 6\hat k} \right) = 0\)

⇒ λ – 9 + 6 = 0

Explanation:

Given:

curves x² = 4y and y² = 4x

Substituting y = x²/4 in the second curve to obtain the point of intersection, we get:

x = 4 and y = 4

The given integral is in the form of M dx + N dy

By using Green’s theorem,

\(\mathop \smallint \limits_C Mdx + Ndy = \mathop \int\!\!\!\int \limits_R \left( {\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}}} \right)dx\;dy\)

\(M = 2{x^2} - y \Rightarrow \frac{{\partial M}}{{\partial y}} = - 1\)

\(N = 3x{y^2} \Rightarrow \frac{{\partial N}}{{\partial x}} = 3{y^2}\)

\(= \int\!\!\!\int \left( {3{y^2} + 1} \right)dy\;dx\)

\(= \mathop \smallint \limits_0^4 \mathop \smallint \limits_{y = \frac{{{x^2}}}{4}}^{2\sqrt x } \left( {3{y^2} + 1} \right)dy\;dx\)

\(= \mathop \smallint \limits_0^4 \left[ {{y^3} + y} \right]_{\frac{{{x^2}}}{4}}^{2\sqrt x }dx\)

\(= \mathop \smallint \limits_0^4 \left[ {8x\sqrt x + 2\sqrt x - \frac{{{x^6}}}{{64}} - \frac{{{x^2}}}{4}} \right]dx\)

\(= \left[ {\frac{{8{x^{\frac{3}{2} + 1}}}}{{\frac{3}{2} + 1}} + \frac{{2{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} - \frac{{{x^7}}}{{7\left( {64} \right)}} - \frac{{{x^3}}}{{3\left( 4 \right)}}} \right]_0^4\)

\(= 8\left( {\frac{2}{5}} \right){x^{\frac{5}{2}}} + \frac{4}{3}{x^{\frac{3}{2}}} - \frac{{{x^7}}}{{7\left( {64} \right)}} - \frac{{{x^3}}}{{12}}\)

\(= \frac{{16}}{5}\left( {32} \right) + \frac{4}{3}\left( 8 \right) - \frac{{{4^7}}}{{7 \times 64}} - \frac{{{4^3}}}{{12}}\)

Explanation:

\(\oint \bar A \cdot d\bar r = \mathop \smallint \nolimits_C \left( {ydx + zdy + xdz} \right)\)

But, z = 0    ∴ dz = 0

\(\therefore \oint \bar A \cdot d\bar r = \mathop \smallint \nolimits_C \left( {ydx} \right)\)

Along circle, x² + y² = 1

Put, x = cost y = sin t

Now,

dx = - sin t dt dy = cos t dt

And t varies from 0 to 2π

\(\mathop \oint \nolimits_c \bar A \cdot d\bar r = \mathop \smallint \limits_{t = 0}^{t = 2\pi } \sin t\left( { - \sin t} \right)\)

\(\mathop \oint \nolimits_c \bar A \cdot d\bar r = - \mathop \smallint \limits_0^{2\pi } \left[ {\frac{{1 - \cos 2t}}{2}} \right]dt\)

\(\mathop \oint \nolimits_c \bar A \cdot d\bar r = - \left[ {\frac{x}{2}\;\frac{{ - \sin 2t}}{4}} \right]_0^{2\pi }\)

Explanation:

∇ ⋅ (∇f × ∇g) = ∇g ⋅ curl (∇f) - ∇f ⋅ [curl (∇g)]

[∵ div (A̅ × B̅) = B̅ ⋅ curl A̅ - A̅ ⋅ curl B̅]

We have, curl (grad ϕ) = 0̅    

∴ ∇ ⋅ (∇f × ∇g) = ∇g ⋅ 0̅ - ∇f ⋅ 0̅

Concept:

A vector F is said to be solenoidal when ∇. F = 0

A vector F is said to be irrotational when ∇ × F = 0

Calculation:

\(\vec V = \left( {x + y + az} \right)i + \left( {bx + 2y - z} \right)j + + \left( { - x + cy + 2z} \right)k\)

\(\nabla \times \vec V = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {\left( {x + y + az} \right)}&{\left( {bx + 2y - z} \right)}&{\left( { - x + cy + 2z} \right)} \end{array}} \right| = 0\)

⇒ i (c + 1) – j (-1 – a) + k (b – 1) = 0

⇒ a = -1, b = 1, and c = -1

\(\vec V = \left( {x + y - z} \right)i + \left( {x + 2y - z} \right)j + \left( { - x - y + 2z} \right)k\)

Divergence of the given vector is

\(div\;\vec V = \nabla .\vec V = \frac{\partial }{{\partial x}}\left( {x + y - z} \right) + \frac{\partial }{{\partial y}}\left( {x + 2y - z} \right) + \frac{\partial }{{\partial z}}\left( { - x - y + 2z} \right)\)

Explanation:

Gradient of scalar gives the vector

∇ϕ = ∇(x²yz + 4xz²)

= (2xyz + 4z²) î + x²zĵ + (x²y + 8xz) k̂

At, (1, -2, -1)

= 8 î - ĵ - 10 k̂

Unit vector in the direction of 2c – ĵ - 2k̂ is

\(a = \frac{{2\hat i - \hat j - 2\hat k}}{{\sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }} = \frac{2}{3}\hat i - \frac{1}{3}\hat j\frac{{ - 2}}{3}\hat k\)

Required directional derivative.

\(\frac{{\nabla\phi }}{{\left| {\nabla\phi } \right|}}.\hat a\)

\(= \left( {8\;\hat i - \hat j - 10\;\hat k} \right)\left( {\frac{2}{3}\hat i - \frac{1}{3}\hat j\frac{{ - 2}}{3}\hat k} \right)\)

\(= \frac{{16}}{3} + \frac{1}{3} + \frac{{20}}{3} = \frac{{37}}{3} = 12.33\)

Explanation:

By Stokes theorem

\(\begin{array}{l} I = \mathop \int\!\!\!\int \limits_s^\; \left( {\nabla \times F} \right).ds = \mathop \oint \limits_C^\; F.dr\\ I = \mathop \oint \limits_C^\; F.dr = \mathop \oint \limits_C^\; \left( {3y\hat i - xz\hat j + y{z^2}\hat k} \right).\left( {dx\hat i + dy\hat j + dz\hat k} \right)\\ = \mathop \oint \limits_C^\; 3ydx - xzdy + y{z^2}dz \end{array}\)

S ≡ x² + y² = 16, z = 8

Let x = 4 cos θ, y = 4sin θ

C ≡ x² + y² = 16, θ = 0 to 2π

\(\begin{array}{l} I = \mathop \smallint \limits_0^{2\pi } 12\sin \theta \left( { - 4\sin \theta } \right)d\theta - 32\cos \theta \left( {4\cos \theta } \right)d\theta + 256\sin \theta \left( 0 \right)\\ = - 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {48{{\sin }^2}\theta + 128{{\cos }^2}\theta } \right)d\theta \\ = - 4\left( {48 \times \frac{1}{2} \times \frac{\pi }{2} + 128 \times \frac{1}{2} \times \frac{\pi }{2}} \right) \end{array}\)

= -176 π