Solution to Homework 1. Vector Analysis (P201)

Transcription

1 Solution to Homework 1. Vector Analysis (P1) Q1. A = 3î + ĵ ˆk, B = î + 3ĵ + 4ˆk, C = 4î ĵ 6ˆk. The sides AB, BC and AC are, AB = 4î ĵ 6ˆk AB = 7.48 BC = 5î + 5ĵ + 1ˆk BC = CA = î 3ĵ 4ˆk CA = From which it follows that sum of any two sides is greater than the third. The angles are, AB cos(ab, BC) = BC AB.BC = BC cos(bc, CA) = CA BC.CA = CA cos(ca, AB) = AB CA.AB = 9 15 (AB, BC) = 18 o o = 1.9 o (AB, BC) = 18 o o = 16.1 o 34 6 (AB, BC) = 18 o 7 o = 153 o So the angles add up to 18 o. Or we can directly use A, B, C as sides of the triangle in which case we have A = 14 = 3.74 B = 6 = 5.1 C = = 7.48 We get sum of any two sides is greater than the third. The angles are cos(a, B) = cos(a, C) = cos(b, C) = Again the angles add upto π c or 18 o (A, B) = c = o 14 (A, C) =.67 c = 38.1 o 34 6 (B, C) = (π.67) c =.47 c = 7. o Q. The diagonals on the same face, say on y-plane, connect the corners (,,) to (,1,1) and (,1,) to (,,1). The vectors in this case are f 1 = ĵ + ˆk and f = ĵ + ˆk. Hence the angle θ f is θ f = cos 1 f 1 f = cos = θ f = 9 o. f 1.f The diagonals on the adjacent faces, say on y and x-planes, connect the corners (,,) to (,1,1) and (,,) to (1,,1). The vectors in this case are f 1 = ĵ + ˆk and f = î + ˆk. The angle θ f is θ f = cos 1 f 1 f = cos 1 1 = 1 f 1.f θ f = 6 o. Consider the body diagonals connecting the corners (1,,) to (,1,1) and (,,) to (1,1,1) and the corresponding vectors are b 1 = î + ĵ + ˆk and b = î + ĵ + ˆk. The angle between them θ b is θ b = cos 1 b 1 b = cos 1 1 = 1 b 1.b θ b 7.5 o. 1

2 Q3. The sides of the triangle are AB = î + ĵ, BC = ĵ + 3ˆk and CA = î 3ˆk. Use any two sides, say AB and BS to calculate the area of the triangle formed by AB, BC, CA and unit normal ˆn: area = 1 AB BC = 1 î ĵ ˆk 1 3 = î + 3ĵ + ˆk = = 7 units ˆn = 1 (6î + 3ĵ + ˆk) 7 Q4. Since the vector we are looking for is parallel to xy-plane, it does not depend not and let that vector be v = xî + yĵ. Perpendicular to 4î 3ĵ + ˆk implies, Therefore the unit vector ˆv is (xî + yĵ) (4î 3ĵ + ˆk) = ˆv = 4x 3y = x = 3 4 y 3 4yî + yĵ 3 = 9 16 y + y 5î + 4 5ĵ. Q5. In terms of components, let the vector be A = A x î + A y ĵ + A ˆk. Then the direction cosines are cos α = A x A where, A = A x + A y + A. Hence, cos β = A y A cos γ = A A cos α + cos β + cos γ = A x A x + A y + A A y + A x + A y + A A x + A + A y + A = 1. Q6. The displacement vector is s = ( 3)î + ( 1 )ĵ + (4 + 1)ˆk = î 3ĵ + 5ˆk. Therefore the work done is W = F s = (4î 3ĵ + ˆk) ( î 3ĵ + 5ˆk) = 15 units. Q7. The vector A = A x î + A y ĵ + A ˆk changes to A = A xî + A yĵ + A ˆk under rotation by θ about y-axis, A x cos θ sin θ A x A y = 1 A y. A sin θ cos θ A Then A x = A x cos θ + A sin θ, A y = A y, A = A x sin θ + A cos θ, and therefore, A x + A y + A = A x + A y + A.

3 Q8. The gradient of φ(x, y, ) = e x sin y ln is φ = î ex sin y ln x + ĵ ex sin y ln y = îe x sin y ln + ĵe x cos y ln + ˆk ex sin y φ (,,) = îe sin ln + ĵe cos ln + ˆk e sin = 4.66î.13ĵ ˆk + ˆk ex sin y ln Q9. For φ(x, y, ) = c we have dφ(x, y, ) = φ d l = Then either d l =, which cannot be true because we have some displacement along the surface, or φ = which also cannot be true because it must be a function of at least x or y or, or φ d l which is the only option that implies φ is perpendicular to the surface on which we have the displacement d l. Q1. The unit vector ˆn 1 perpendicular to the surface x + y + = 9 at (,-1,) is (x + y + 9) = xî + yĵ + ˆk (x + y + 9) (, 1,) = 4î ĵ + 4ˆk ˆn 1 = 4 î ĵ + 4 ˆk Similarly, the unit vector ˆn perpendicular to the surface x + y = 3 at (,-1,) is (x + y 3) = xî + yĵ ˆk (x + y 3) (, 1,) = 4î ĵ ˆk ˆn = 4 î ĵ 1 ˆk Hence the angle between the surfaces at (,-1,) is θ = cos 1 (ˆn 1 ˆn ) = cos ( ) = 1.55 c = 88.8 o Q11. The volume element in spherical coordinate is dτ = r sin θdθdφdr and hence the total volume is Volume = R r dr π sin θ dθ π dφ = 4 3 πr3. The surface element of a sphere of radius R is ds = R sin θdθdφ and hence the total surface area of a sphere is, π π Surface = R sin θ dθ dφ = 4πR. 3

4 Q1. The gradient operator in spherical polar coordinate is ˆr + ˆθ 1 r θ + ˆφ 1 r sin θ φ. (a) ln r = ˆr r ln r = ˆr r (b) r n = ˆr r rn = nr n 1ˆr (c) f(r) = ˆr r f(r) = ˆrf (r) Q13. The divergence is A = x x3 + y ( x y) + y4 = 3 x + 8y 3 Q14. The divergence operator in spherical polar coordinate is A = 1 1 r sin θ θ (sin θa θ) + 1 r sin θ r r (r A r ) + A φ φ. Here our A r = r cos θ, A θ = r sin θ, A φ = r sin θ cos φ. So, A = 1 (r 3 cos θ) r + 1 (r sin θ) + 1 r r sin θ θ r sin θ = 3 cos θ + cos θ sin φ = 5 cos θ sin φ. (r sin θ cos φ) φ Q15. and Q16. are standard stuff, hence the calculations are not provided here. Consult any text book containing vector analysis. Q17. The φ = x + 4xy + y 3 evaluated at a (,, ) and b (1, 1, 1) are φ a = and φ b = 7. So, φ b φ a = 7. To integrate b a φ d l, we first calculate φ d l = [(x + 4y)î + (4x + 3 )ĵ + (6y )ˆk] [dxî + dyĵ + dˆk] = (x + 4y) dx + (4x + 3 ) dy + 6y d For path (,,) to (1,,) the dy = d = and y = = since y and do not change and y and are always along that path, hence 1 (x + 4y) dx = 1 x dx = x 1 = 1 For path (1,,) to (1,1,) the dx = d = and x = 1, = along this path, hence 1 (4x + 3 ) dy = 1 4 dy = 4y 1 = 4 Finally, for path (1,1,) to (1,1,1) we have dx = dy = and x = y = 1 along this path, hence 1 1 6y d = 6 d = 3 1 = Summing all the path contributions, b a φ d l = 7 = φ b φ a. Q18. To do the surface integral, we would project the surface of the hemisphere x + y + = a on xy-plane. For that first we calculate the unit normal to the hemisphere at some 4

5 arbitrary point (x, y, ) which is (x + y + a ) = xî + yĵ + ˆk. Therefore, ˆn = x 4x + 4y + 4 î + y 4x + 4y + 4 ĵ + 4x + 4y + 4 ˆk = x a î + y aĵ + a ˆk The projection of an element of surface area ds on hemisphere surface onto xy-plane ds is ds cos θ = ds where θ is the angle between ˆn and ˆk which is ds dx dy = ds cos θ = ds (ˆn ˆk) = ds a ds = Then the surface integral becomes A d s = A ˆnds = dx dy a ( x (xî + yĵ ˆk) a î + y aĵ + a ˆk ) dx dy a The limit of the integrals are: a x a and a x y a x. Note that = on xy-plane and we could have expressed x = ± a y etc. A d s = x + y dx dy = x=a y= a x x= a 3(x + y ) a y= a x a x y dx dy To fecilitate integration we can change over to polar coordinate as, x = r cos φ, y = r sin φ, dx dy = r dr dφ and the limits are r a and φ π (please convince yourself). The answer, therefore, is A d s = φ=π r=a φ= r= 3r a r dr dφ = a r You could have changed the coordinates (x, y, ) to (r, θ, φ) at the very begining and done the calculation. Q19. The divergence of v = xyî + yĵ + 3xˆk is v = x (xy) + y (y) + (3x) = y + + 3x The volume integral is, ( v) dτ = dx ydy d + dx dy d + 3 xdx dy d = 48 5

6 Next we do the surface integrals v d s on the faces of the cube xy plane at = xy plane at = y plane at x = y plane at x = x plane at y = x plane at y = 3x dx dy = 3x dx dy = 4 xy dy d = xy dy d = 8 y d dx = y d dx = 16 Threfore, v d s = 48 = v dτ, the divergence theorem is verified. Q(a) The curl of v = xyî + yĵ + 3xˆk and the area element on shaded area are, v = yî 3ĵ xˆk and d s = dy d î Since y + =, the surface integral is ( v) d s = ydy d = [ ] y dy d = 8 3 Next we do the line integrals, where the line element is d l = dxî + dyĵ + dˆk, (,, ) (,, ) v d l = y dy = (,, ) (,, ) (,, ) (,, ) v d l = v d l = (y dy + 3x d) = 3x d = Therefore, ( v) d s = 8 3 = v d l, the Stoke s theorem is verified. y( y) dy = 8 3 Q(b) The equation of the inclined surface is x + y + = 8 and unit vector on the surface is ˆn = 3î + 1 3ĵ + ˆk. 3 The surface area element written in terms of area element of the projected surface on xy-plane is ds = dxdy = 3 ˆn ˆk dxdy. The curl of v is x î + (x xy)ĵ. The surface integral is, [ ( v) ˆnds = [x î+(x xy)ĵ] 3î + 3ĵ + 1 ] 3 ˆk 3 dx dy = [x + 1 (x xy)] dx dy To perform the above integral, we note that x and y are related by x + y = 8. [ 8 8 y ] [ ( v) ˆnds 8 8 y ] = x 1 dx dy + (x xy)dx dy = = 3 3 6

7 Next we consider the line integral v d l = (x dx y dy + x y d), (4,, ) (, 8, ) (, 8, ) (,, 4) (,, 4) (4,, ) (x dx y dy + x y d) = (x dx y dy + x y d) = (x dx y dy + x y d) = y dy = 3 y dy = 3 x(4 x) dx = 3 3 Therefore, ( v) d s = 3 3 = v d l, the Stoke s theorem is verified. 7