Vector Calculus MCQ Quiz in मराठी - Objective Question with Answer for Vector Calculus - मोफत PDF डाउनलोड करा

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:

\(\mathop \smallint \nolimits_C \vec F \cdot dr = \mathop \smallint \nolimits_0^{45^\circ } \left[ {\left( { - r\cos \theta } \right)\left( { - r\sin \theta } \right)d\theta + \left( {r\sin \theta } \right)\left( {r\cos \theta } \right)d\theta } \right]\)

\(\mathop \smallint \nolimits_0^{45^\circ } \left( {{r^2}\cos \theta \sin \theta + {r^2}\sin \theta \cos \theta } \right)d\theta \)

With r = 1, the above integral becomes:

\( = \frac{1}{2}\mathop \smallint \nolimits_0^{45^\circ } \left( {\sin 2\theta + \sin 2\theta } \right)d\theta \)

r = 1

\( = \mathop \smallint \nolimits_0^{45^\circ } \sin 2\theta \;d\theta \)

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

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

By stokes theorem,

\(\mathop \smallint \limits_c^\; F.dR = \mathop \smallint \limits_s^\; curl\;F.N\;ds\)

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

\(curl\;F = \left| {\begin{array}{*{20}{c}} I&J&K\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {x + y}&{2x - z}&{y + z} \end{array}} \right| = 2I + k\)

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

\(\begin{array}{l} \hat N = \frac{{3I + 2J + K}}{{\sqrt {9 + 4 + 1} }} = \frac{1}{{\sqrt {14} }}\left( {3I + 2J + k} \right)\\ \mathop \smallint \limits_c^\; \left[ {\left( {x + y} \right)dx + \left( {2x - z} \right)dy + \left( {y + z} \right)dz} \right] = \mathop \smallint \limits_c^\; F.dR \end{array}\)

\(= \mathop \smallint \limits_z^\; curl\;F.\hat Nds\) where S is the triangle ABC

\(\begin{array}{l} = \mathop \smallint \limits_s^\; \left( {2I + k} \right).\left( {\frac{{3I + 2J + K}}{{\sqrt {14} }}} \right)ds\\ = \frac{1}{{\sqrt {14} }}\left( {6 + 1} \right)\mathop \smallint \limits_s^\; ds\\ = \frac{7}{{\sqrt {14} }}\left( {Area\;of\;{\rm{\Delta }}ABC} \right) = \frac{7}{{\sqrt {14} }}.3\sqrt {14} = 21 \end{array}\)

Green's theorem:

\(\oint \left( {{\rm{Mdx}} + {\rm{Ndy}}} \right) = \int\!\!\!\int \left( {\frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}} - \frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}}} \right){\rm{dxdy}}\)

Stokes theorem:

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

Gauss Divergence theorem:

\(\oint A.ds=\iiint{\left( \nabla .A \right)dv}\)

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

Concept:

The divergence of the given vector field,

\(div\vec F = \nabla \cdot \vec F{{\;\;\;\;\;}} \ldots \left( 1 \right)\)

Calculation:

Given:

\(\vec F = {x^2}z\hat i + xy\hat j - y{z^2}\hat k\)

Using equation (1),

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

\(\Rightarrow\nabla \cdot \vec F= 2xz + x – 2yz\)

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

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 \(\triangledown f\), and we conclude that \(\triangledown f = 0\)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:

\(\mathop \smallint \limits_0^1 \left[ {2\left( {4t} \right)\hat i + \left( {2t} \right)\hat j} \right].\left( {\hat idx + \hat jdy} \right)\)

\(= \mathop \smallint \limits_0^1 8tdx + 2tdy\)

\(= \mathop \smallint \limits_0^1 16tdt + 8tdt\)

\(= \mathop \smallint \limits_0^1 24tdt = 12\)

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 =  \(\begin{bmatrix}1&0&0\\\ 0 &\cos \theta&-\sin \theta\\\ 0&\sin \theta&\cos \theta\end{bmatrix}\)

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,

\(\rm\begin{bmatrix}x'\\\ y'\\\ z'\end{bmatrix}=\begin{bmatrix}1&0&0\\\ 0 &\cos \theta&-\sin \theta\\\ 0&\sin \theta&\cos \theta\end{bmatrix}\rm\begin{bmatrix}1\\\ 1\\\ 1\end{bmatrix}\)

= \(\begin{bmatrix}1&0&0\\\ 0 &\frac{1}{\sqrt2}&-\frac{1}{\sqrt2}\\\ 0&\frac{1}{\sqrt2}&\frac{1}{\sqrt2}\end{bmatrix}\rm\begin{bmatrix}1\\\ 1\\\ 1\end{bmatrix}=\begin{bmatrix}1\\\ 0\\\ \sqrt2\end{bmatrix}\)

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 \(\vec{A} = aî + bĵ + ck̂\) and B = \(\vec{B}=pî + qĵ + rk̂\)

Then anyone diagonal of the rhombus is given by \(D_1=\vec{A}+\vec{B}\)

The other diagonal is given by \(D_2=\vec{B}-\vec{A}\)

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, \(\vec{A} = 2î + 4ĵ - 5k̂\) and B = \(\vec{B}=1î + 2ĵ + 3k̂\)

Then anyone diagonal of the rhombus is given by \(D_1=\vec{A}+\vec{B}\)

\(D_1=\vec{A}+\vec{B}\)

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 \(D_2=\vec{B}-\vec{A}\)

\(D_2=\vec{B}-\vec{A}\)

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:

\(DD = \nabla f.\frac{{\vec u}}{{\left| u \right|}}\)

where \(\frac{{\vec u}}{{\left| u \right|}}={\hat{u}}\)

grad f or ∇ f is defined by the equation,

\(grad\;f = \nabla f = i\frac{{\partial f}}{{\partial x}} + j\frac{{\partial f}}{{\partial y}} + k\frac{{\partial f}}{{\partial z}}\)

Calculation:

Given:

z = y²e^2x

\(gradz\;= \nabla z \;=\; 2{y^2}{e^{2x}}\hat i + 2y{e^{2x}}\hat j\)

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

Unit vector \(\vec b = \alpha \hat i + \beta \hat j\)

Directional derivative \( = \nabla z.\vec b = \left( {2{e^4}\hat i - 2{e^4}\hat j} \right).(\alpha \hat i + \beta \hat j)\) = 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  \(\vec v = \frac{1}{3}\left( { - {y^3}\hat i + {x^3}\hat j + {z^3}\hat k} \right)\)

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

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

We have to find

\(I = \smallint \left( {\nabla \times u} \right).\hat nds\)

⇒ from stokes theorem, we have

\(\mathop \int\!\!\!\int \nolimits_s \left( {\nabla \times u} \right).\hat nds = \mathop \oint \nolimits_c u.dr\)

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

Now, changing the integral with substitution

x = cos θ ⇒ dx = -sin θ dθ

y = sin θ ⇒ dy = cos θ dθ

\(I = \frac{1}{3}\mathop \smallint \nolimits_0^{2\pi } {\sin ^4}\theta d\theta + \mathop \smallint \nolimits_0^{2\pi } \frac{1}{3}{\cos ^4}\theta d\theta \)