BPHCT-133 Solved Assignment 2024

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BPHCT-133 Solved Assignment 2024 Sample Solution
bphct-133-solved-assignment-2024–ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

bphct-133-solved-assignment-2024–ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

BPHCT-133 Solved Assignment 2024 SS
PART A
  1. a) Determine the direction in which the scalar field f ( x , y ) = 2 x 2 − y 2 + x y f ( x , y ) = 2 x 2 − y 2 + x y f(x,y)=2x^(2)-y^(2)+xyf(x, y)=2 x^2-y^2+x yf(x,y)=2x2−y2+xy increases most rapidly at the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1).
Answer:
The direction in which a scalar field f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) increases most rapidly at a given point is given by the gradient of the field at that point. The gradient is a vector that points in the direction of the greatest increase of the function and its magnitude is the rate of increase in that direction.
The gradient of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is given by:
∇ f ( x , y ) = ( ∂ f ∂ x , ∂ f ∂ y ) ∇ f ( x , y ) = ∂ f ∂ x , ∂ f ∂ y grad f(x,y)=((del f)/(del x),(del f)/(del y))\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)∇f(x,y)=(∂f∂x,∂f∂y)
For the given scalar field f ( x , y ) = 2 x 2 − y 2 + x y f ( x , y ) = 2 x 2 − y 2 + x y f(x,y)=2x^(2)-y^(2)+xyf(x, y) = 2x^2 – y^2 + xyf(x,y)=2x2−y2+xy, the partial derivatives are:
∂ f ∂ x = 4 x + y ∂ f ∂ x = 4 x + y (del f)/(del x)=4x+y\frac{\partial f}{\partial x} = 4x + y∂f∂x=4x+y
∂ f ∂ y = − 2 y + x ∂ f ∂ y = − 2 y + x (del f)/(del y)=-2y+x\frac{\partial f}{\partial y} = -2y + x∂f∂y=−2y+x
At the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1, 1)(1,1):
∂ f ∂ x | ( 1 , 1 ) = 4 ( 1 ) + 1 = 5 ∂ f ∂ x | ( 1 , 1 ) = 4 ( 1 ) + 1 = 5 (del f)/(del x)|_((1,1))=4(1)+1=5\frac{\partial f}{\partial x}\bigg|_{(1, 1)} = 4(1) + 1 = 5∂f∂x|(1,1)=4(1)+1=5
∂ f ∂ y | ( 1 , 1 ) = − 2 ( 1 ) + 1 = − 1 ∂ f ∂ y | ( 1 , 1 ) = − 2 ( 1 ) + 1 = − 1 (del f)/(del y)|_((1,1))=-2(1)+1=-1\frac{\partial f}{\partial y}\bigg|_{(1, 1)} = -2(1) + 1 = -1∂f∂y|(1,1)=−2(1)+1=−1
So, the gradient of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) at the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1, 1)(1,1) is:
∇ f ( 1 , 1 ) = ( 5 , − 1 ) ∇ f ( 1 , 1 ) = ( 5 , − 1 ) grad f(1,1)=(5,-1)\nabla f(1, 1) = (5, -1)∇f(1,1)=(5,−1)
Therefore, the scalar field f ( x , y ) = 2 x 2 − y 2 + x y f ( x , y ) = 2 x 2 − y 2 + x y f(x,y)=2x^(2)-y^(2)+xyf(x, y) = 2x^2 – y^2 + xyf(x,y)=2x2−y2+xy increases most rapidly at the point ( 1 , 1 ) ( 1 , 1 ) (1,1)(1, 1)(1,1) in the direction of the vector ( 5 , − 1 ) ( 5 , − 1 ) (5,-1)(5, -1)(5,−1).
b) Calculate the work done by a force F → = 3 x i ^ + 2 y j ^ F → = 3 x i ^ + 2 y j ^ vec(F)=3x hat(i)+2y hat(j)\overrightarrow{\mathbf{F}}=3 x \hat{\mathbf{i}}+2 y \hat{\mathbf{j}}F→=3xi^+2yj^ in moving a particle once counter-clockwise along the ellipse x 2 4 + y 2 9 = 1 x 2 4 + y 2 9 = 1 (x^(2))/(4)+(y^(2))/(9)=1\frac{x^2}{4}+\frac{y^2}{9}=1x24+y29=1.
Answer:
The work done by a force F → F → vec(F)\vec{F}F→ along a path C C CCC is given by the line integral of the force along the path:
W = ∫ C F → ⋅ d r → W = ∫ C   F → ⋅ d r → W=int _(C) vec(F)*d vec(r)W = \int_C \vec{F} \cdot d\vec{r}W=∫CF→⋅dr→
where d r → d r → d vec(r)d\vec{r}dr→ is a differential element of the path.
Given the force F → = 3 x i ^ + 2 y j ^ F → = 3 x i ^ + 2 y j ^ vec(F)=3x hat(i)+2y hat(j)\vec{F} = 3x \hat{i} + 2y \hat{j}F→=3xi^+2yj^, we can express this in terms of the parameterization of the ellipse. The ellipse x 2 4 + y 2 9 = 1 x 2 4 + y 2 9 = 1 (x^(2))/(4)+(y^(2))/(9)=1\frac{x^2}{4} + \frac{y^2}{9} = 1x24+y29=1 can be parameterized as:
x ( t ) = 2 cos ⁡ ( t ) x ( t ) = 2 cos ⁡ ( t ) x(t)=2cos(t)x(t) = 2 \cos(t)x(t)=2cos⁡(t)
y ( t ) = 3 sin ⁡ ( t ) y ( t ) = 3 sin ⁡ ( t ) y(t)=3sin(t)y(t) = 3 \sin(t)y(t)=3sin⁡(t)
where t t ttt ranges from 0 0 000 to 2 π 2 π 2pi2\pi2π for one complete counter-clockwise traversal of the ellipse.
The differential element of the path in terms of t t ttt is:
d r → = d x d t d t i ^ + d y d t d t j ^ = − 2 sin ⁡ ( t ) d t i ^ + 3 cos ⁡ ( t ) d t j ^ d r → = d x d t d t i ^ + d y d t d t j ^ = − 2 sin ⁡ ( t ) d t i ^ + 3 cos ⁡ ( t ) d t j ^ d vec(r)=(dx)/(dt)dt hat(i)+(dy)/(dt)dt hat(j)=-2sin(t)dt hat(i)+3cos(t)dt hat(j)d\vec{r} = \frac{dx}{dt} dt \hat{i} + \frac{dy}{dt} dt \hat{j} = -2 \sin(t) dt \hat{i} + 3 \cos(t) dt \hat{j}dr→=dxdtdti^+dydtdtj^=−2sin⁡(t)dti^+3cos⁡(t)dtj^
Substituting the expressions for x ( t ) x ( t ) x(t)x(t)x(t), y ( t ) y ( t ) y(t)y(t)y(t), and d r → d r → d vec(r)d\vec{r}dr→ into the integral for work, we get:
W = ∫ 0 2 π ( 3 ⋅ 2 cos ⁡ ( t ) i ^ + 2 ⋅ 3 sin ⁡ ( t ) j ^ ) ⋅ ( − 2 sin ⁡ ( t ) d t i ^ + 3 cos ⁡ ( t ) d t j ^ ) W = ∫ 0 2 π   ( 3 ⋅ 2 cos ⁡ ( t ) i ^ + 2 ⋅ 3 sin ⁡ ( t ) j ^ ) ⋅ ( − 2 sin ⁡ ( t ) d t i ^ + 3 cos ⁡ ( t ) d t j ^ ) W=int_(0)^(2pi)(3*2cos(t) hat(i)+2*3sin(t) hat(j))*(-2sin(t)dt hat(i)+3cos(t)dt hat(j))W = \int_0^{2\pi} (3 \cdot 2 \cos(t) \hat{i} + 2 \cdot 3 \sin(t) \hat{j}) \cdot (-2 \sin(t) dt \hat{i} + 3 \cos(t) dt \hat{j})W=∫02π(3⋅2cos⁡(t)i^+2⋅3sin⁡(t)j^)⋅(−2sin⁡(t)dti^+3cos⁡(t)dtj^)
W = ∫ 0 2 π ( − 12 cos ⁡ ( t ) sin ⁡ ( t ) + 18 sin ⁡ ( t ) cos ⁡ ( t ) ) d t W = ∫ 0 2 π   ( − 12 cos ⁡ ( t ) sin ⁡ ( t ) + 18 sin ⁡ ( t ) cos ⁡ ( t ) ) d t W=int_(0)^(2pi)(-12 cos(t)sin(t)+18 sin(t)cos(t))dtW = \int_0^{2\pi} (-12 \cos(t) \sin(t) + 18 \sin(t) \cos(t)) dtW=∫02π(−12cos⁡(t)sin⁡(t)+18sin⁡(t)cos⁡(t))dt
W = ∫ 0 2 π 6 sin ⁡ ( t ) cos ⁡ ( t ) d t W = ∫ 0 2 π   6 sin ⁡ ( t ) cos ⁡ ( t ) d t W=int_(0)^(2pi)6sin(t)cos(t)dtW = \int_0^{2\pi} 6 \sin(t) \cos(t) dtW=∫02π6sin⁡(t)cos⁡(t)dt
To integrate this, we can use the trigonometric identity sin ⁡ ( 2 t ) = 2 sin ⁡ ( t ) cos ⁡ ( t ) sin ⁡ ( 2 t ) = 2 sin ⁡ ( t ) cos ⁡ ( t ) sin(2t)=2sin(t)cos(t)\sin(2t) = 2 \sin(t) \cos(t)sin⁡(2t)=2sin⁡(t)cos⁡(t), so sin ⁡ ( t ) cos ⁡ ( t ) = 1 2 sin ⁡ ( 2 t ) sin ⁡ ( t ) cos ⁡ ( t ) = 1 2 sin ⁡ ( 2 t ) sin(t)cos(t)=(1)/(2)sin(2t)\sin(t) \cos(t) = \frac{1}{2} \sin(2t)sin⁡(t)cos⁡(t)=12sin⁡(2t):
W = ∫ 0 2 π 6 ⋅ 1 2 sin ⁡ ( 2 t ) d t W = ∫ 0 2 π   6 ⋅ 1 2 sin ⁡ ( 2 t ) d t W=int_(0)^(2pi)6*(1)/(2)sin(2t)dtW = \int_0^{2\pi} 6 \cdot \frac{1}{2} \sin(2t) dtW=∫02π6⋅12sin⁡(2t)dt
W = 3 ∫ 0 2 π sin ⁡ ( 2 t ) d t W = 3 ∫ 0 2 π   sin ⁡ ( 2 t ) d t W=3int_(0)^(2pi)sin(2t)dtW = 3 \int_0^{2\pi} \sin(2t) dtW=3∫02πsin⁡(2t)dt
The integral of sin ⁡ ( 2 t ) sin ⁡ ( 2 t ) sin(2t)\sin(2t)sin⁡(2t) over a complete period (from 0 0 000 to 2 π 2 π 2pi2\pi2π) is zero, because the positive and negative areas cancel each other out:
W = 3 ⋅ 0 = 0 W = 3 ⋅ 0 = 0 W=3*0=0W = 3 \cdot 0 = 0W=3⋅0=0
Therefore, the work done by the force F → = 3 x i ^ + 2 y j ^ F → = 3 x i ^ + 2 y j ^ vec(F)=3x hat(i)+2y hat(j)\vec{F} = 3x \hat{i} + 2y \hat{j}F→=3xi^+2yj^ in moving a particle once counter-clockwise along the ellipse x 2 4 + y 2 9 = 1 x 2 4 + y 2 9 = 1 (x^(2))/(4)+(y^(2))/(9)=1\frac{x^2}{4} + \frac{y^2}{9} = 1x24+y29=1 is 0 0 000 joules.
c) Use the divergence theorem to calculate the flux of a vector field A → = 2 x i ^ − y j ^ + 3 z k ^ A → = 2 x i ^ − y j ^ + 3 z k ^ vec(A)=2x hat(i)-y hat(j)+3z hat(k)\overrightarrow{\mathbf{A}}=2 x \hat{\mathbf{i}}-y \hat{\mathbf{j}}+3 z \hat{\mathbf{k}}A→=2xi^−yj^+3zk^ over a cube of side 2 a 2 a 2a2 a2a which has its vertices at ( ± a , ± a , ± a ) ( ± a , ± a , ± a ) (+-a,+-a,+-a)( \pm a, \pm a, \pm a)(±a,±a,±a).
Answer:
The divergence theorem states that the flux of a vector field A → A → vec(A)\vec{A}A→ through a closed surface S S SSS is equal to the volume integral of the divergence of A → A → vec(A)\vec{A}A→ over the volume V V VVV enclosed by S S SSS:
∬ S A → ⋅ d S → = ∭ V ( ∇ ⋅ A → ) d V ∬ S   A → ⋅ d S → = ∭ V   ( ∇ ⋅ A → ) d V ∬_(S) vec(A)*d vec(S)=∭_(V)(grad* vec(A))dV\iint_S \vec{A} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{A}) dV∬SA→⋅dS→=∭V(∇⋅A→)dV
Given the vector field A → = 2 x i ^ − y j ^ + 3 z k ^ A → = 2 x i ^ − y j ^ + 3 z k ^ vec(A)=2x hat(i)-y hat(j)+3z hat(k)\vec{A} = 2x \hat{i} – y \hat{j} + 3z \hat{k}A→=2xi^−yj^+3zk^, we first need to calculate its divergence:
∇ ⋅ A → = ∂ ∂ x ( 2 x ) + ∂ ∂ y ( − y ) + ∂ ∂ z ( 3 z ) = 2 − 1 + 3 = 4 ∇ ⋅ A → = ∂ ∂ x ( 2 x ) + ∂ ∂ y ( − y ) + ∂ ∂ z ( 3 z ) = 2 − 1 + 3 = 4 grad* vec(A)=(del)/(del x)(2x)+(del)/(del y)(-y)+(del)/(del z)(3z)=2-1+3=4\nabla \cdot \vec{A} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(-y) + \frac{\partial}{\partial z}(3z) = 2 – 1 + 3 = 4∇⋅A→=∂∂x(2x)+∂∂y(−y)+∂∂z(3z)=2−1+3=4
Since the divergence is a constant, the volume integral becomes:
∭ V ( ∇ ⋅ A → ) d V = 4 ∭ V d V ∭ V   ( ∇ ⋅ A → ) d V = 4 ∭ V   d V ∭_(V)(grad* vec(A))dV=4∭_(V)dV\iiint_V (\nabla \cdot \vec{A}) dV = 4 \iiint_V dV∭V(∇⋅A→)dV=4∭VdV
The volume V V VVV is a cube with side length 2 a 2 a 2a2a2a, so its volume is ( 2 a ) 3 = 8 a 3 ( 2 a ) 3 = 8 a 3 (2a)^(3)=8a^(3)(2a)^3 = 8a^3(2a)3=8a3. Therefore, the flux of the vector field A → A → vec(A)\vec{A}A→ over the cube is:
∬ S A → ⋅ d S → = 4 × 8 a 3 = 32 a 3 ∬ S   A → ⋅ d S → = 4 × 8 a 3 = 32 a 3 ∬_(S) vec(A)*d vec(S)=4xx8a^(3)=32a^(3)\iint_S \vec{A} \cdot d\vec{S} = 4 \times 8a^3 = 32a^3∬SA→⋅dS→=4×8a3=32a3
Hence, the flux of the vector field A → A → vec(A)\vec{A}A→ over the cube of side 2 a 2 a 2a2a2a with vertices at ( ± a , ± a , ± a ) ( ± a , ± a , ± a ) (+-a,+-a,+-a)(\pm a, \pm a, \pm a)(±a,±a,±a) is 32 a 3 32 a 3 32a^(3)32a^332a3.
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