Engineering Mathematics Test 4

Calculation:

When any vector is differentiated, it rotates by 90°.

Dot product:

\(\vec a \cdot \vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta \)

Where, θ = angle between \(\vec a\;\& \;\vec b\) 

\(\vec R\) rotates by 90° when differentiated, \(\frac{{d\vec R}}{{dt}}\;and\;\vec R\) are perpendicular to each other

\( \Rightarrow \vec R \cdot \frac{{d\vec R}}{{dt}} = 0\)

\(\nabla \times \left( {\nabla \times {\rm{\vec V}}} \right) = \nabla \left( {\nabla .{\rm{\vec V}}} \right) - {\nabla ^2}{\rm{\vec V}}\)

We know that divergence of Curl of a vector is always zero.

\(\nabla .\left( {\nabla \times {\rm{\vec V}}} \right) = 0\)

Area of triangle is, \(A = \frac{1}{2}\left| {AB \times AC} \right|\)

× represents the curl operation.

A(2, -5, 1), B(0, 2, 4) and C(0, 3, 1)

AB = -2i + 7j + 3k

AC = -2i + 8j

\(AB \times AC = \left| {\begin{array}{*{20}{c}}i&j&k\\{ - 2}&7&3\\{ - 2}&8&0\end{array}} \right|\)

= -24i + 6j – 2k

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:

Angular velocity = \(\frac{1}{2} \times \;\)Curl of Linear velocity

i.e. \(\omega = \frac{1}{2} \times \left( {\vec \nabla \times \vec V} \right)\) 

The curl is a vector operator that describes the infinitesimal rotation of a vector field in three-dimensional space.

The curl of a scalar field is undefined. It is defined only for 3D vector fields.

\(Curl = \nabla \times F = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{F_1}}&{{F_2}}&{{F_3}} \end{array}} \right|\)

Calculation:

\(\vec V = 2{x^2}y\hat i + xy{z^2}\hat j + xy\hat K\)

\(\therefore \vec \nabla \times \vec V = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {2{x^2}y}&{xy{z^2}}&{xy} \end{array}} \right|\)

= î (x – 2xyz) – ĵ (y - 0) + k̂ (yz² – 2x²)

\(\therefore {\left( {\vec \nabla \times \vec V} \right)_{1,1,1}} = i\left( {1 - 2} \right) - \hat j\left( {1 - 0} \right) + \hat k\left( {1 - 2} \right)\) = -i – j – k

\( \Rightarrow \omega = \frac{1}{2}\left( { - i - \hat j - k} \right)\)

\(\therefore \left| \omega \right| = \sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = \frac{{\sqrt 3 }}{2}\;units\)

We know that,

Curl grad f = ∇ × ∇f = 0

∇.(∇ × ∇f) + ∇.∇f

= 0 + ∇.∇f = ∇.∇f

We also know that,

\({\nabla ^2}f = \nabla .\nabla {\rm{f}} = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}\)

\(\frac{{\partial f}}{{\partial x}} = 2xy + {z^2} \Rightarrow \frac{{{\partial ^2}f}}{{\partial {x^2}}} = 2y\)

\(\frac{{\partial f}}{{\partial y}} = {x^2} + 2yz \Rightarrow \frac{{{\partial ^2}f}}{{\partial {y^2}}} = 2z\)

\(\frac{{\partial f}}{{\partial z}} = {y^2} + 2zx \Rightarrow \frac{{{\partial ^2}f}}{{\partial {z^2}}} = 2x\)

∇²f = 2y + 2z + 2x

\(\vec F\) is solenoid then \(\bar \nabla .\bar F = 0\)

\(\vec F = \left| {\vec F} \right|\hat r\)

\(\rm \Rightarrow \left| {\bar F} \right|\overrightarrow {\frac{r}{{\left| {\vec r} \right|}}}\ \ \)

\(\Rightarrow \vec F = {r^{n - 1}}\frac{{\vec r}}{r} = {r^{n - 2}}\vec r=r^{n-2}(x\vec i+y\vec j+z\vec k)\)     

Consider \(\rm \frac {\partial}{\partial x}(r^{n-2}.x)=r^{n-2}+x.(n-2)r^{n-3}.\frac {x}{r}=r^{n-2}+x^2.(n-2)r^{n-4}\)

\(\rm \frac {\partial}{\partial y}(r^{n-2}.y)=r^{n-2}+y.(n-2)r^{n-3}.\frac {y}{r}=r^{n-2}+y^2.(n-2)r^{n-4}\)

\(\rm \frac {\partial}{\partial z}(r^{n-2}.z)=r^{n-2}+z.(n-2)r^{n-3}.\frac {z}{r}=r^{n-2}+z^2.(n-2)r^{n-4}\)

\(\rm \therefore {\rm{\Delta }}.\vec F = 3{r^{n - 2}} + \left( {n - 2} \right){r^{n - 2}} = \left( {n + 1} \right){r^{n - 2}}\)

∴ we get \(\rm n = -1\) for solenoid.

Position vector \(\vec r = x\hat i + y\hat j + zk\)

\(\left| {\vec r} \right| = \sqrt {{x^2} + {y^2} + {z^2}}\)

\(\phi = \ln \left| {\vec r} \right|\)

\(Gradient\;\nabla \phi = \left( {i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}} + k\frac{\partial }{{\partial z}}} \right).\left\{ {\ln \sqrt {{x^2} + {y^2} + {z^2}} } \right\}\)

\(= \frac{1}{2}\left\{ {i\frac{\partial }{{\partial x}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + j.\frac{\partial }{{\partial y}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + k\frac{\partial }{{\partial z}}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right\}\)

\(= \frac{1}{2}\left\{ {i.\frac{{2x}}{{{x^2} + {y^2} + {z^2}}} + j\frac{{2y}}{{{x^2} + {y^2} + {z^2}}} + k\frac{{2z}}{{{x^2} + {y^2} + {z^2}}}} \right\}\)

\(\nabla \phi = \left\{ {\frac{{xi + yj + zk}}{{{x^2} + {y^2} + {z^2}}}} \right\}\)

\(Gradient\;\nabla \phi = \frac{{\vec r}}{{\vec r.\vec r}}\)