Diploma Continuous Assessment Test -2/ April – 2021

Course: First year Diploma course in Engineering & Technology

Subject & Code : Engineering Mathematics – II (40022)

Time : 2 Hours                   Date:  27-04-2021                                              Max. Marks: 50

PART – A

 Answer all questions                                          ( 6×1=6 marks)

\[1.\ Evaluate:\ [\overrightarrow{i} + \overrightarrow{j}\ \overrightarrow{j} + \overrightarrow{k}\ \overrightarrow{k} + \overrightarrow{i}]\ \hspace{20cm}\]

Soln:

\[\overrightarrow{a}= \overrightarrow{i}\ + \overrightarrow{j}\]
\[\overrightarrow{b}= \overrightarrow{j}\ + \overrightarrow{k}\]
\[\overrightarrow{c}= \overrightarrow{k} + \overrightarrow{i}\]
\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1 \\ \end{vmatrix}\]

=    1  ( 1 - 0 )  - 1 ( 0 - 1 )  + 0 (  0 - 1 )

=   1 ( 1 )  -1 (- 1 ) +  0

=   1 + 1

=   2

2.            Define vector Differential operator.

Soln:

\[The\ vector\ differential\ operator\ 'DEL'\ denoted\ as\ '\nabla'\ is\ defined\ by\]
\[\nabla = \frac{\partial}{\partial\ x}\overrightarrow{i} + \frac{\partial}{\partial\ y}\overrightarrow{j} + \frac{\partial}{\partial\ z}\overrightarrow{k}\]
\[\overrightarrow{i},\ \overrightarrow{j},\ \overrightarrow{k}\ are\ unit\ vectors\ in\ x,\ y,\ z\ directions\]
\[3.\ If\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 - zy^3)\overrightarrow{k}, then\ find\ div\ \overrightarrow{F}\ \hspace{20cm}\]

Soln:

\[Let\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 - zy^3)\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(xyz)}+ \frac{\partial}{\partial\ y} {(3x^2y)}+ \frac{\partial}{\partial\ z} {(xy^2-xy^3)}\]

=   yz  + 3x2 ( 1 )  + ( 0 - y3 )

=  yz  + 3x2 - y3

\[ 4.\ Evaluate: \int(x^2 -x -1)\ dx\ \hspace{20cm}\]

Soln:

\[\int(x^2 -x -1)\ dx = \frac{x^3}{3} - \frac{x^2}{2} - x + c\]
\[5.\ Evaluate\ :\ \int\ \frac{2x}{1 + x^2}\ dx\ \hspace{20cm}\]

Soln:

\[put\ u\ =\ 1 + x^2\]
\[\frac{du}{dx}= \ 2x\]

du  = 2 dx

\[\int\ \frac{2x}{1 + x^2}\ dx= \int\ \frac{du}{u}\]

=  log u + c

\[= log(1 + x^2) + c\]
\[\int\ \frac{2x}{1 + x^2}\ dx= log(1 + x^2) + c\]
\[6.\ Evaluate:\ \int \frac{dx}{x^2 + 16}\ \hspace{20cm}\]

Soln:

\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{x^2 + 16} = \int \frac{dx}{x^2 + 4^2}\]
\[ = \frac{1}{4}\ {tan}^{-1} (\frac{x}{4}) +c\]

PART – A

Answer all questions                                          ( 7×2=14 marks)

\[7.\ Prove\ that\ the\ vectors\ 2\overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k},\ 3\overrightarrow{i} + 4\overrightarrow{j}+ \overrightarrow{k}\ and\ \overrightarrow{i} - 2\overrightarrow{j}+ \overrightarrow{k}\ are\ coplanar. \ \hspace{20cm}\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{b}= 3\overrightarrow{i}+ 4\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{c}= \overrightarrow{i}- 2\overrightarrow{j}+ \overrightarrow{k}\]
\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} 2 & 1 & 1\\ 3 & 4 & 1\\ 1 & -2 & 1\\ \end{vmatrix}\]

=    2  ( 4 + 2 )  - 1 (3 - 1 )  + 1 ( -6 - 4 )

=   2 ( 6 )  - 1 (2 ) +  1 ( - 10 )

=   12 - 2  -  10   =    10  -  10

\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] = 0\]

The given points are coplanar.

8. Find the unit normal to the surface xy +yz + zx= 3 at the point (1, 1, 1).

Soln:

\[\phi(x,y,z)\ = c\ (being\ constant)\ represents\ a\ surface\]
\[then\ unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (x,y,z)\ is\ \frac{\nabla\phi}{|\nabla\phi|}\]
\[Here\ \phi = xy +yz + zx\]
\[\nabla\ \phi=(\frac{\partial\ \phi}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ \phi}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ \phi}{\partial\ z})\overrightarrow{k}------------(1)\]
\[\frac{\partial\ \phi}{\partial\ x} = y \frac{\partial}{\partial\ x}(x) + 0 + z \frac{\partial}{\partial\ x}(x)\]

= y ( 1 )  + z ( 1 )

\[\frac{\partial\ \phi}{\partial\ x} = y + z\]
\[\frac{\partial\ \phi}{\partial\ y} = x \frac{\partial}{\partial\ y}(y) + z \frac{\partial}{\partial\ y}(y) + 0\]

= x ( 1 )  + z ( 1 )

\[\frac{\partial\ \phi}{\partial\ y} = x + z\]
\[\frac{\partial\ \phi}{\partial\ z} = 0 + y \frac{\partial}{\partial\ z}(z) + x \frac{\partial}{\partial\ z}(z)\]

= y ( 1 )  + x ( 1 )

Equation (1) becomes

\[\nabla\ \phi=(y + z)\overrightarrow{i} + (x + z)\overrightarrow{j} + (y + x)\overrightarrow{k}\]

At the point (1,1, 1),

\[\nabla\ \phi=(1 + 1)\overrightarrow{i} + (1 + 1)\overrightarrow{j} + (1 + 1)\overrightarrow{k}\]
\[\nabla\ \phi= 2\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k}\]
\[|\nabla\phi| = \sqrt{(2)^2 + (2)^2 + (2)^2 }\]
\[= \sqrt{(4 + 4 + 4) }\]
\[= 2\sqrt{3}\]
\[unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (1,1,1)\ =\ \frac{\nabla\phi}{|\nabla\phi|}\]
\[ = \frac{2\overrightarrow{i}+ 2\overrightarrow{j}+ 2\overrightarrow{k}}{2\sqrt{3}}\]