Vector Calculus MCQ Quiz in मल्याळम - Objective Question with Answer for Vector Calculus - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

∫_C F(r) dr can be solved by putting:

x = r cos θ

y = r sin θ

dx = -r sin θ dθ

dy = r cos θ dθ

Application:

Given r = 1

θ = 0 to 45°

The required integral can be written as:

${\int }_C\overrightarrow{F}\cdot dr={\int }_0^{{45}^{\circ }}\left[\left(-r\cos \theta \right)\left(-r\sin \theta \right)d\theta +\left(r\sin \theta \right)\left(r\cos \theta \right)d\theta \right]$

${\int }_0^{{45}^{\circ }}\left(r^2\cos \theta \sin \theta +r^2\sin \theta \cos \theta \right)d\theta$

With r = 1, the above integral becomes:

$=\frac{1}{2}{\int }_0^{{45}^{\circ }}\left(\sin 2\theta +\sin 2\theta \right)d\theta$

r = 1

$={\int }_0^{{45}^{\circ }}\sin 2\theta d\theta$

$=\frac{{\left(-\cos 2\theta \right)}_0^{{45}^{\circ }}}{2}$

$=\frac{-0-\left(-1\right)}{2}=\frac{1}{2}$

By stokes theorem,

$\underset{c}{\overset{}{\int }}F.dR=\underset{s}{\overset{}{\int }}curlF.Nds$

Here F = (x + y) I + (2x - z) J + (y + z) K

$curlF=\left|\begin{array}{ccc}I& J& K\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x+y& 2x-z& y+z\end{array}\right|=2I+k$

Equation of the plane through A, B, C is

$\begin{array}{l}\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1\\ \Rightarrow 3x+2y+z=6\end{array}$

Vector N normal to this plane is

∇(3x + 2y + z - 6) = 3I + 2J + K

$\begin{array}{l}\hat{N}=\frac{3I+2J+K}{\sqrt{9+4+1}}=\frac{1}{\sqrt{14}}\left(3I+2J+k\right)\\ \underset{c}{\overset{}{\int }}\left[\left(x+y\right)dx+\left(2x-z\right)dy+\left(y+z\right)dz\right]=\underset{c}{\overset{}{\int }}F.dR\end{array}$

$=\underset{z}{\overset{}{\int }}curlF.\hat{N}ds$ where S is the triangle ABC

$\begin{array}{l}=\underset{s}{\overset{}{\int }}\left(2I+k\right).\left(\frac{3I+2J+K}{\sqrt{14}}\right)ds\\ =\frac{1}{\sqrt{14}}\left(6+1\right)\underset{s}{\overset{}{\int }}ds\\ =\frac{7}{\sqrt{14}}\left(Areaof\mathrm{\Delta }ABC\right)=\frac{7}{\sqrt{14}}.3\sqrt{14}=21\end{array}$

Green's theorem:

$\oint \left(\mathrm{M}\mathrm{d}\mathrm{x}+\mathrm{N}\mathrm{d}\mathrm{y}\right)=\int \int \left(\frac{\partial \mathrm{N}}{\partial \mathrm{x}}-\frac{\partial \mathrm{M}}{\partial \mathrm{y}}\right)\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}$

Stokes theorem:

$\oint \overrightarrow{A}\cdot d\overrightarrow{l}$ = $\iint \left(\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{A}\right).d\overrightarrow{s}$

Gauss Divergence theorem:

$\oint A.ds=\iiint (\mathrm{\nabla }.A)dv$

$\oint \overrightarrow{F.}\hat{n}ds=\iiint (\mathrm{\nabla }.F)dv$

Concept:

The divergence of the given vector field,

$div\overrightarrow{F}=\mathrm{\nabla }\cdot \overrightarrow{F}\dots \left(1\right)$

Calculation:

Given:

$\overrightarrow{F}=x^{2}z\hat{i}+xy\hat{j}-y{z}^2\hat{k}$

Using equation (1),

$\Rightarrow \mathrm{\nabla }\cdot \overrightarrow{F}=\left(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}\right)\cdot \overrightarrow{F}$

$\Rightarrow \mathrm{\nabla }\cdot \overrightarrow{F}=2xz+x–2yz$

$\therefore div\overrightarrow{F}at\left(1,-1,1\right)=2+1+2=5$

Explanation:

If $f$ is twice continuously differentiable, then its second derivatives are independent of the order in which the derivatives are applied. All the terms cancel in the expression for curl $▽f$, and we conclude that $▽f=0$curl⁡∇f=0." id="MathJax-Element-12-Frame" role="presentation" style=" word-spacing: 0px; position: relative;" tabindex="0">

• curl⁡∇f=0." role="presentation" style=" word-spacing: 0px; position: relative;" tabindex="0">Divergence operates on a vector field but results in a scalar.
• Curl operates on a vector field and results in a vector field.
• Gradient operates on a scalar but results in a vector field.
• Divergence of curl, Curl of the gradient is always zero.
• Thus, the gradient of curl gives the result of curl (which is a vector field) to the gradient to operate upon, which is a mathematically invalid expression.curl⁡∇f=0." role="presentation" style=" word-spacing: 0px; position: relative;" tabindex="0">..curl∇f=0.

Straight line from (0, 0) to (2, 4)

$\frac{x-0}{2-0}=\frac{y-0}{4-0}=t$

x = 2t, y = 4t

dx = 2dt

dy = 4dt

Since x (2t) is from 0 to 2 and y(4t) is from 0 to 4, the limits of t are: 0 to 1

Now, the line integral becomes:

$\underset{0}{\overset{1}{\int }}\left[2\left(4t\right)\hat{i}+\left(2t\right)\hat{j}\right].\left(\hat{i}dx+\hat{j}dy\right)$

$=\underset{0}{\overset{1}{\int }}8tdx+2tdy$

$=\underset{0}{\overset{1}{\int }}16tdt+8tdt$

$=\underset{0}{\overset{1}{\int }}24tdt=12$

Concept:

In 3D, rotations can also be defined as linear transformations, a rotation in 3D can be represented by a matrix equation P'=RP where R is a rotation matrix.

R =  $\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array}\right]$

where θ is the angle of rotation with the positive axis in the clockwise direction.

Given Data and Calculation:

Given that, Coordinate of point P = (1,1,1)

The angle of rotation = 45°

New Coordinate will be,

$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\\ z^{\prime }\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

= $\left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ 0& \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ \sqrt{2}\end{array}\right]$

So the solution will be 1.3×3" id="MathJax-Element-163-Frame" role="presentation" style=" position: relative;" tabindex="0">rotation matrix.(16)p′=Rp" id="MathJax-Element-161-Frame" role="presentation" style=" position: relative;" tabindex="0">p′=Rp

Concept:

If aî + bĵ + ck̂ and pî + qĵ + rk̂ are 2 different sides of the rhombus.

Suppose $\overrightarrow{A}=aî+bĵ+ck̂$ and B = $\overrightarrow{B}=pî+qĵ+rk̂$

Then anyone diagonal of the rhombus is given by $D_1=\overrightarrow{A}+\overrightarrow{B}$

The other diagonal is given by $D_2=\overrightarrow{B}-\overrightarrow{A}$

The magnitude of the vector  $A=\sqrt{a^2+b^2+c^2}$

Calculation:

Given:

2î + 4ĵ - 5k̂ and î + 2ĵ + 3k̂ are 2 different sides of rhombus.

Suppose, $\overrightarrow{A}=2î+4ĵ-5k̂$ and B = $\overrightarrow{B}=1î+2ĵ+3k̂$

Then anyone diagonal of the rhombus is given by $D_1=\overrightarrow{A}+\overrightarrow{B}$

$D_1=\overrightarrow{A}+\overrightarrow{B}$

D₁ = (2î + 4ĵ - 5k̂) + (1î + 2ĵ + 3k̂)

D₁ = 3î + 6ĵ - 2k̂

The magnitude of the vector diagonal D₁

_{$D_1=\sqrt{3^2+6^2+(-2{)}^2}$}

D₁ = 7 

The other diagonal is given by $D_2=\overrightarrow{B}-\overrightarrow{A}$

$D_2=\overrightarrow{B}-\overrightarrow{A}$

D2 = (1î + 2ĵ + 3k̂) - (2î + 4ĵ - 5k̂)

D2 = - 1î - 2ĵ + 8k̂

The magnitude of the vector diagonal D₂

$D_1=\sqrt{(-1{)}^2+(-2{)}^2+8^2}$

$D_1=\sqrt{69}$

∴ The length of diagonals of a rhombus is 7 and $\sqrt{69}$.

Concept:

Directional derivative of a function f along the vector $\hat{u}$ is given by:

$DD=\mathrm{\nabla }f.\frac{\overrightarrow{u}}{\left|u\right|}$

where $\frac{\overrightarrow{u}}{\left|u\right|}=\hat{u}$

grad f or ∇ f is defined by the equation,

$gradf=\mathrm{\nabla }f=i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}$

Calculation:

Given:

z = y²e^2x

$gradz=\mathrm{\nabla }z=2y^2{e}^{2x}\hat{i}+2y{e}^{2x}\hat{j}$

At (2, -1), $\mathrm{\nabla }z=2e^4\hat{i}-2e^4\hat{j}$

Unit vector $\overrightarrow{b}=\alpha \hat{i}+\beta \hat{j}$

Directional derivative $=\mathrm{\nabla }z.\overrightarrow{b}=\left(2e^4\hat{i}-2e^4\hat{j}\right).(\alpha \hat{i}+\beta \hat{j})$ = 2e4α – 2e4β

Given that, the directional derivative is zero.

⇒ 2e⁴α – 2e⁴β = 0 ⇒ α = β

As b is a unit vector, $\sqrt{{\alpha }^2+{\beta }^2}=1$

$\Rightarrow \alpha =\beta =\frac{1}{\sqrt{2}}$

$\left|\alpha +\beta \right|=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$

Given vector  $\overrightarrow{v}=\frac{1}{3}\left(-y^3\hat{i}+x^3\hat{j}+z^3\hat{k}\right)$

Bounded by open surface of hemisphere x² + y² + z² = 1

Such that z ≥ 0, and closed curve (x² + y² = 1)

We have to find

$I=\int \left(\mathrm{\nabla }\times u\right).\hat{n}ds$

⇒ from stokes theorem, we have

$\int {\int }_s\left(\mathrm{\nabla }\times u\right).\hat{n}ds={\oint }_{c}u.dr$

$\Rightarrow I=\frac{1}{3}\left\{-\oint y^{3}dx+\oint x^{3}dy\right\}$

Now, changing the integral with substitution

x = cos θ ⇒ dx = -sin θ dθ

y = sin θ ⇒ dy = cos θ dθ

$I=\frac{1}{3}{\int }_0^{2\pi }{\sin }^4\theta d\theta +{\int }_0^{2\pi }\frac{1}{3}{\cos }^4\theta d\theta$