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Vector Integral and Differential Calculus (ACM 20150) – Assignment 4 Issue Date: Monday 17th October 2016 Due Date: Wednesday 26th October 2016 1. (a) If u = x3 x̂−xy ŷ +yz 2 ẑ and φ = zx−z 3 y, compute u·∇φ and u×∇φ. Evaluate your results at the point (1, −1, 1). (b) Let u = (yz + 4xy)x̂ + (xz + 2x2 − 3z 2 )ŷ + (xy − 6yz)ẑ. Compute ∇ × u. Find ψ(x, y, z) such that u = ∇ψ. (c) Find the unit outward drawn normal to the surface (x − 2)2 + (y − 1)2 + z 2 = 25 at the point (−2, 4, 0). (d) Find the equation for the tangent plane and the equation of the normal line to the surface xz 2 + x2 y = z − 1 at the point (1, −1, 1). (a) We have, ∇φ = z x̂ − z 3 ŷ + (x − 3z 2 y)ẑ. hence u · ∇φ = x3 x̂ − xy ŷ + yz 2 ẑ · z x̂ − z 3 ŷ + (x − 3z 2 y)ẑ , = zx3 + xyz 3 + yz 2 (x − 3z 2 y). Evaluating at (1, −1, 1), this is (1)(1) + (1)(−1)(1) + (−1)(1) [1 − 3(−1)] = −4. Moreover, x3 x̂ − xy ŷ + yz 2 ẑ × z x̂ − z 3 ŷ + (x − 3z 2 y)ẑ , x̂ ŷ ẑ 3 2 , yz = x −xy z −z 3 (x − 3z 2 y) = x̂ −xy x − 3z 2 y + yz 5 − ŷ x3 x − 3z 2 y − yz 3 + ẑ −x3 z 3 + xyz . u × ∇φ = Evaluating at (1, −1, 1), this is x̂ [(1)(1 + 3) − 1] − ŷ [(1 + 3) + 1] + ẑ (−1 − 1) = 3x̂ − 5ŷ − 2ẑ. 1 Vector Integral and Differential Calculus Vector calculus operations (b) We have x̂ ŷ ẑ ∂x ∂y ∂z ∇×u= yz + 4xy xz + 2x2 − 3z 2 xy − 6yz . Hence, ∇ × u = x̂ ∂y (xy − 6yz) − ∂z xz + 2x2 − 3z 2 − ŷ [∂x (xy − 6yz) − ∂z (yz + 4xy)] ẑ ∂x xz + 2x2 − 3z 2 − ∂y (yz + 4xy) , = x̂ [x − 6z − (x − 6z)] − ŷ [y − y] + ẑ [z + 4x − (z + 4x)] , = 0. We now work out the scalar field ψ such that u = ∇ψ: ψx = yz + 4xy ψy = xz + 2x2 − 3z 2 ψz = xy − 6yz =⇒ =⇒ =⇒ ψ = xyz + 2x2 y + f (y, z), ψ = xyz − 3yz 2 + 2x2 y + g(x, z), ψ = xyz − 3yz 2 + h(x, y). Simply by inspection, h(x, y) = 2x2 y, hence ψ = xyz − 3yz 2 + 2x2 y + Const. Check: ψx = yz + 4xy, ψy = xz − 3z 2 + 2x2 , ψz = xy − 6yz. (c) Let φ(x, y, z) = (x − 2)2 + (y − 1)2 + z 2 − 25. Hence, φ = 0 describes a sphere of radius 5 centred at (2, 1, 0). A normal to the sphere is ∇φ = 2(x − 2)x̂ + 2(y − 1)ŷ + 2z ẑ, hence n̂ = ∇φ 2(x − 2)x̂ + 2(y − 1)ŷ + 2z =p . |∇φ| 4(x − 2)2 + 4(y − 1)2 + 4z 2 Note that φ(−2, 4, 0) = (−4)2 + 32 + 0 − 25 = 0, hence (−2, 4, 0) lies on the sphere. At this point, ∇φ = 2(−4)x̂ + 2(3)ŷ = −8x̂ + 6ŷ, and n̂ = ∇φ = |∇φ| q 1 100 (−8x̂ + 6ŷ) = 1 10 (−8x̂ + 6ŷ) = 51 (−4x̂ + 3ŷ) . We know from class notes that this normal vector points outwards. 2 Vector Integral and Differential Calculus Vector calculus operations (d) Define φ = xz 2 + x2 y − z + 1. The equation φ = 0 defines a surface with outward-pointing normal n̂ = ∇φ/|∇φ|. Hence, x̂(z 2 + 2xy) + ŷx2 + ẑ(2zx − 1) . n̂ = p (z 2 + 2xy)2 + x4 + (2zx − 1)2 Note, φ(1, −1, 1) = 0, hence (1, −1, 1) lives in the surface. At this point, n̂ = x̂(−1) + ŷ + ẑ −x̂ + ŷ + ẑ √ √ = . ··· 3 Tangent plane: Point on plane: r0 = (1, −1, 1); general point r = (x, y, z); general outward-pointing normal: (−1, 1, 1). Equation: (r − r0 ) · n = 0, or (x−1)nx +(y+1)ny +(z−1)nz = 0 =⇒ (x−1)(−1)+(y+1)(1)+(z−1)(1) = 0. Tidy up: x − y − z = 1, or z = x − y − 1. Normal line: r(t) = r0 + tn̂, = (1, −1, 1) + (−1, 1, 1)t, or −x + 1 = y + 1 = z − 1. 3 Vector Integral and Differential Calculus Vector calculus operations 2. (a) Find ∇rn , where n is an integer. (b) Compute ∇f (r). (c) Compute ∇2 f (r). Here r = (x, y, z), r = |r|, and f (·) is a smooth functions of a single variable. Answer clue for (c): if f (r) = ln r, then ∇2 f (r) = 1/r2 . First, consider p x2 + y 2 + z 2 , ∂ x̂ (x2 + y 2 + z 2 )1/2 + · · · , ∂x 2x 1 x̂ 2 2 + ··· , (x + y 2 + z 2 )1/2 x̂x + ··· , |r| r . |r| ∇|r| = ∇ = = = = Now, using the ordinary rules of calculus, (a) ∇|r|n = n|r|n−1 ∇|r| = n|r|n−1 (r/|r|) = nr|r|n−2 . (b) Extend this result to r ∇f (r) = f 0 (r) . r (c) We have the following string of results: f 0 (r) r , ∇ · (∇f (r)) = ∇ · r f 0 (r) f 0 (r) (∇ · r) + r · ∇ , r r 0 0 3f 0 (r) f (r) r , +r· r r r 00 3f 0 (r) rf (r) − f 0 (r) +r , r r2 3f 0 (r) 1 + f 00 (r) − f 0 (r), r r 2f 0 (r) + f 00 (r). r 2 ∇ f (r) = = = = = = Check: f (r) = ln r, f 0 (r) = 1/r, f 00 (r) = −1/r2 , hence ∇2 ln r = 2 1 1 − 2 = 2. 2 r r r 4 Vector Integral and Differential Calculus Vector calculus operations y (1,1) x Figure 1: H 3. (a) Evaluate A · dx around the closed curve in Figure 1 if A = (x − y)x̂ + (x + y)ŷ. R (b) Evaluate A · dx along the curve x2 + y 2 = 1, z = 1, in the positive direction from (0, 1, 1) to (1, 0, 1), if A = (yz + 2x)x̂ + xz ŷ + (xy + 2z)ẑ. Hint: Use the following parametrization of the curve: (cos ϕ, sin ϕ, 1). x(ϕ) = (a) Break up closed curve into segments I and II. On segment I, we have xI (t) = (x, y) = (x, x2 ) = (t, t2 ), hence dxI = dxI dt = (1, 2t)dt. dt We have A · dxI = [(x − y), (x + y)] · (1, 2t)dt, = (x − y)dt + (x + y)2t dt, = (t − t2 )dt + 2t(t + t2 )dt. The parameter t goes from t = 0 to t = 1. Hence, Z Z 1 A · dx = (t − t2 ) + 2t(t + t2 ) dt. I 0 5 Vector Integral and Differential Calculus Vector calculus operations This is a straightforward integral that evaluates to 4/3. Next, we do segment II, whereon we hhave xII (t) = (x, y) = (x, x1/2 ) = (t, t1/2 ), hence dxII = dxII dt = 1, 12 t−1/2 dt. dt We have A · dxII = [(x − y), (x + y)] 1, 21 t−1/2 dt, = (x − y) + 12 (x + y)t−1/2 , = (t − t1/2 )dt + 21 (t + t1/2 )t−1/2 dt, = tdt − t1/2 dt + 12 t1/2 dt + 12 dt, = tdt − 12 t1/2 dt + 21 dt. The parameter t goes from t = 1 to t = 0. However, this is the same as going from t = 0 to t = 1 provided we put an overall minus sign in front of the integral at the end: Z Z 1 A · dx = t − 12 t1/2 + 12 dt. II 0 Hence, Z A · dx = − II 1 2 t2 − 1 2 3/2 t 23 1 + 12 t 0 = − 1 − 13 + 12 = − 32 . Put the two results together: Z I Z A · dx = A · dx = A · dx + I II 4 3 − 2 3 = 23 . (b) Solution The parametrization is x(ϕ) = (cos ϕ, sin ϕ, 1). Hence, dx = dx dϕ = (− sin ϕ, cos ϕ, 0)dϕ. dϕ The starting point of the curve is (0, 1, 1), hence (x, y) = (0, 1). But (x, y) = (cos ϕ, sin ϕ), hence ϕ = π/2 at the starting point. Similarly, the end point of the curve is (1, 0, 1), hence (x, y) = (1, 0). But (x, y) = (cos ϕ, sin ϕ), hence ϕ = 2π. Caution: The final parameter value is ϕ = 2π (not ϕ = 0). The reason is that the sense of the integral is in the positive direction. Therefore, when projected into the xy plane, the curve looks like a portion of a circle with an anticlockwise sense, i.e. going from ϕ = π/2 to ϕ = 2π. Thus, A · dx = = = = = [(yz + 2x), xz, (xy + 2z)] · (− sin ϕ, cos ϕ, 0) dϕ, (yz + 2x)(− sin ϕ)dϕ + xz cos ϕdϕ, (sin ϕ + 2 cos ϕ)(− sin ϕ)dϕ + cos2 ϕ dϕ, since z = 1, 2 2 (cos ϕ − sin ϕ)dϕ − 2 cos ϕ sin ϕ dϕ, cos 2ϕ dϕ − sin 2ϕ dϕ. 6 Vector Integral and Differential Calculus Vector calculus operations Hence, Z Z 2π Z cos(2ϕ)dϕ − A · dx = 1 2 sin(2ϕ)dϕ, π/2 π/2 = 2π 2π 2π 1 sin(2ϕ) + 2 cos(2ϕ) , π/2 π/2 = 0 + 21 [cos(4π) − cos(π)] , = 1. 7 Vector Integral and Differential Calculus Vector calculus operations 4. Let u(x, t) be a three-dimensional vector field that also depends on time, t. Assume that u satisfies Euler’s equation, ∂u + u · ∇u = −∇p, ∂t ∇ · u = 0, where p(x, t) is a scalar field. (a) Define ω = ∇ × u and show that Euler’s equation can be re-written as ∂u + ω × u = −∇ p + 21 u2 . ∂t You may assume the following vector identity: (a · ∇) a = (∇ × a) × a + ∇ 1 2 a 2 , for any smooth vector field a. (b) By operating on both sides of the equation in (a) with the curl, show that Euler’s equation can further be re-written as ∂ω + (u · ∇) ω = (ω · ∇) u. ∂t This is the so-called vortex-stretching equation. You may assume the following vector identity: ∇ × (a × b) = (b · ∇) a − (a · ∇) b + a (∇ · b) − b (∇ · a) for any two smooth vector fields a and b. For (a), write (u · ∇)u as ω × u + ∇(u2 /2) as instructed, to obtain ∂u + ω × u = −∇ p + 12 u2 . ∂t For (b), take the curl of this equation, using curl grad = 0 to obtain ∂ω + ∇ × (ω × u) = 0. ∂t Using the vector identity given, namely ∇ × (a × b) = (b · ∇) a − (a · ∇) b + a (∇ · b) − b (∇ · a) , we have ∇ × (ω × u) = (u · ∇) ω − (ω · ∇) u + ω (∇ · u) − u (∇ · ω) . Now, ∇ · u = 0 by assumption. Also, ∇ · ω = div curl u = 0, hence ∇ × (ω × u) = (u · ∇) ω − (ω · ∇) u, 8 Vector Integral and Differential Calculus Vector calculus operations and Euler’s equation is thus ∂ω + (u · ∇) ω = (ω · ∇) u. ∂t 9