Engineering Mathematics Test 4

1. \({\vec F_1} = \left( {x + y} \right)\hat i + \left( {2y - 3z} \right)\hat j + \left( {3x - 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_1} = 1 + 2 - 4 = - 1 \ne 0\) 

Therefore, \({\vec F_1}\) is not solenoidal.

2. \({\vec F_2} = \left( {x - y} \right)\hat i + \left( {2y - z} \right)\hat j + \left( {3x - 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_2} = 1 + 2 - 4 = - 1 \ne 0\) 

Therefore, \({\vec F_2}\) is not solenoidal.

3. \({\vec F_3} = \left( {2x - y} \right)\hat i + \left( {y - 3z} \right)\hat j + \left( {3x + 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_3} = 2 + 1 + 4 = 8 \ne 0\) 

Therefore, \({\vec F_3}\) is not solenoidal.

4. \({\vec F_4} = \left( {2x + y} \right)\hat i + \left( {2y - 3z} \right)\hat j + \left( {3x - 4z} \right)\hat k\) 

\(\nabla \cdot {\vec F_4} = 2 + 2 - 4 = 0\) 

Therefore, \({\vec F_4}\) is solenoidal.

Concept:

For a vector field, V defined as Vx î + Vy ĵ + Vz k̂, the divergence (∇.V) is given by:

\(∇.V= \left( {\frac{\partial }{{\partial x}}\hat i + \frac{\partial }{{\partial y}}\hat j + \frac{\partial }{{\partial z}} \hat k} \right).\left( {V_x \hat i + V_y \hat j + V_z \hat k} \right)\)

\(∇.V=\frac{\partial V_x }{\partial x}+\frac{\partial V_y }{\partial y}+\frac{\partial V_z }{\partial z}\)

Also, for a Scalar function A, the gradient is defined as:

\(\nabla A = \frac{{\partial {A_x}}}{{\partial x}}{a_x} + \frac{{\partial {A_y}}}{{\partial y}}{a_y} + \frac{{\partial {A_z}}}{{\partial z}}{a_z}\)

Calculation:

For the given vector field \(\vec F\):

\(div\;\vec F = \nabla \cdot \vec F = 2xy - 2 + {x^2}\)

\(grad\left( {div\;\vec F} \right) = \left( {2x + 2y} \right)\hat i + \left( {2x} \right)\hat j\)

At given point (1, 2, 3)

\(grad\left( {div\;\vec F} \right) = 6\hat i + 2\hat j\)

f(x, y, z) = (4x³y – 3y²z² + 4z) lny

\(fy = \frac{{\partial f}}{{\partial y}} = \left( {4{x^3} - 6y{z^2}} \right)\ln y + \left( {4{x^3}y - 3{y^2}{z^2} + 4z} \right)\frac{1}{y}\)

At point x = 2, y = 1, z = 2

f_y = 0 + (4(8)(1) - 3(4)(1) + 4(2))

= 28

Concept:

If z is homogeneous function of x & y of degree n and z = f(u), then by Euler’s theorem:

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = n\frac{{f\left( u \right)}}{{f'\left( u \right)}}\)

Given, \(u = \ln \left( {\frac{{{x^3} + {y^3}}}{{x + y}}} \right)\)

\(z = \frac{{{x^3} + {y^3}}}{{x + y}}\)

z is homogenous function of x & y with degree 2.

Now, z = e^u

Thus, by Euler’s theorem:

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 2\frac{{{e^u}}}{{{e^u}}}\)

\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 2\)

Let \(\vec u = - x\hat i + y\hat j + z\hat k\)

\(\vec v = y{z^2}\hat i + x{z^2}\hat j + 2xyz\hat k\)

\(\vec u \times \vec v = \left| {\begin{array}{*{20}{c}} i&j&k\\ { - x}&y&z\\ {y{z^2}}&{x{z^2}}&{2xyz} \end{array}} \right|\)

\( = \hat i\left( {2x{y^2}z - x{z^3}} \right) - \hat j\left( { - 2{x^2}yz - y{z^3}} \right) + \hat k\left( { - {x^2}{z^2} - {y^2}{z^2}} \right)\)

\(= \left( {2x{y^2}z - x{z^3}} \right)\hat i + \left( {2{x^2}yz + y{z^3}} \right)\hat j - \left( {{x^2}{z^2} + {y^2}{z^2}} \right)\hat k\)

\(\vec \nabla \cdot \left( {\vec u \times \vec v} \right) = 2{y^2}z - {z^3} + 2{x^2}z + {z^3} - 2{x^2}z - 2{y^2}z = 0\)

\(\vec u \cdot \vec v = \left( { - x\hat i + y\hat j + z\hat k} \right) \cdot \left( {y{z^2}\hat i + x{z^2}\hat j + 2xyz\hat k} \right)\)

= -xyz² + xyz² + 2xyz²

= 2xyz²

f(x, y, z) = -2xyz²

F(x, y) = x³ + y³ + 3xy – 1

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.

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

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

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

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

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

A) \(u\; = \;\frac{{{x^2}y}}{{x + y}}\)

It is a homogeneous function of degree 2.

By Euler’s theorem, \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}}\; = \;nu\)

Differentiating partially w.r.t. x, we get

\(\begin{array}{l} x\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{\partial u}}{{\partial x}} + y\frac{{{\partial ^2}u}}{{\partial y\partial x}}\; = \;n\frac{{\partial u}}{{\partial x}}\;\\ \Rightarrow \;x\frac{{{\partial ^2}u}}{{\partial {x^2}}} + y\frac{{{\partial ^2}u}}{{\partial y\partial x}}\; = \;\left( {n - 1} \right)\frac{{\partial u}}{{\partial x}}\\ x\frac{{{d^2}u}}{{\partial {x^2}}} + y\frac{{{\partial ^2}u}}{{\partial x\partial y}}\; = \;\left( {n - 1} \right)\frac{{\partial u}}{{\partial x}}\; = \;\left( {2 - 1} \right)\frac{{\partial u}}{{\partial x}}\; = \;\frac{{\partial u}}{{\partial x}} \end{array}\)

B) \(u\; = \;\frac{{\sqrt x - \sqrt y }}{{{x^{\frac{1}{4}}} + {y^{\frac{1}{4}}}}}\)

It is a homogenous function of degree \(\left( {\frac{1}{2} - \frac{1}{4}} \right)\; = \;\frac{1}{4}\)

\(\begin{array}{l} {x^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\;\frac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\frac{{{\partial ^2}u}}{{\partial {y^2}}}\; = \;n\left( {n - 1} \right)u\\ = \;\frac{1}{4}\left( {\frac{1}{4} - 1} \right)\; = \;\frac{{ - 3}}{{16}}u \end{array}\)

C) \(u\; = \;{x^{\frac{1}{2}}} + {y^{\frac{1}{2}}}\)

It is a homogeneous function of degree ½

\(\begin{array}{l} {x^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}} + 2xy\frac{{{\partial ^2}u}}{{\partial x\partial y}} + {y^2}\frac{{{\partial ^2}u}}{{\partial {y^2}}}\; = \;n\left( {n - 1} \right)u\\ = \;\frac{1}{2}\left( {\frac{1}{2} - 1} \right)u\; = \; - \frac{1}{4}u \end{array}\)

D) u = f (y/x)

It is a homogeneous function of degree zero.

\(\Rightarrow \;x\frac{{\partial u}}{{\partial x\;}} + y\frac{{\partial y}}{{\partial y}}\; = \;nu\; = \;0\)

If ϕ(x, y, z) is a potential function, then the vector field is given by

\(\vec F = grad\;\phi = \frac{{\partial \phi }}{{\partial x}}\hat i + \frac{{\partial \phi }}{{\partial y}}\hat j + \frac{{\partial \phi }}{{\partial z}}\hat k\) 

Given vector field is,

\(\vec F = \left( {4xy + {y^2}z} \right)\hat i + \left( {2{x^2} + 2xyz + {z^2}} \right)\hat j + \left( {x{y^2} + 2yz} \right)\hat k\) 

\(\frac{{\partial \phi }}{{\partial x}} = 4xy + {y^2}z,\frac{{\partial \phi }}{{\partial y}} = 2{x^2} + 2xyz + {z^2},\frac{\partial }{{\partial z}} = x{y^2} + 2yz\) 

Consider \(\frac{{\partial \phi }}{{\partial x}} = 4xy + {y^2}z\) 

By integrating w.r.t ‘x’

\(\Rightarrow \phi = \frac{{4{x^2}y}}{2} + x{y^2}z + u\left( {y,z} \right)\) 

⇒ ϕ (x, y, z) = 2x²y + xy²z + u(y, z)

By differentiating w.r.t ‘y’

\(\frac{{\partial \phi }}{{\partial y}} = 2{x^2} + 2xyz + \frac{{\partial u}}{{\partial y}}\) 

We know that,

\(\frac{{\partial \phi }}{{\partial y}} = 2{x^2} + 2xyz + {z^2}\) 

\(\Rightarrow \frac{{\partial u}}{{\partial y}}\left( {y,z} \right) = {z^2}\) 

By integrating w.r.t ‘y’

⇒ u = z²y + v(z)

Now, ϕ (x, y, z) = 2x²y + xy²z + z²y + v(x)

By differentiating w.r.t ‘z’

\(\frac{{\partial \phi }}{{\partial z}} = x{y^2} + 2yz + \frac{{\partial v}}{{\partial z}}\) 

We know that,

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

\( \Rightarrow \frac{{\partial v}}{{\partial z}} = 0\) 

⇒ v = constant

Now, the potential function becomes

ϕ (x, y, z) = 2x²y + xy²z + z²y + c