Week 1

Tutorial 1: Vector Analysis


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Slide version: Slides
MATLAB: Tutorial 1

  1. Given vectors with
    1. P1=(3,2,2)P_1=(3, 2, -2), P2=(1,0,2)P_2 = (-1, 0, -2), and P3=(2,1,5)P_3 = (-2, 1, 5)
      Solve the components and length of the vector for:
      a. P2P1P_2-P_1 and P3P1P_3-P_1
      b. 3P1+2P3|3P_1+2P_3|
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      Solution
      Question (a)
      P2P1=(1,0,2)(3,2,2)=(4,2,0)P_2-P_1=(-1,0,-2)-(3,2,-2)=\underline{\underline{(-4,-2,0)}}
      P2P1=(4)2+(2)2+02=20|P_2-P_1|=\sqrt{(-4)^2+(-2)^2+0^2}=\underline{\underline{\sqrt{20}}}
      P3P1=(2,1,5)(3,2,2)=(5,1,7)P_3-P_1=(-2,1,5)-(3,2,-2)=\underline{\underline{(-5,-1,7)}}
      P3P1=(5)2+(1)2+(7)2=75=53|P_3-P_1|=\sqrt{(-5)^2+(-1)^2+(7)^2}=\sqrt{75}=\underline{\underline{5\sqrt{3}}}
      Question (b)
      3P1+2P3=3(3,2,2)+2(2,1,5)=(9,6,6)+(4,2,10)=(5,8,4)=52+82+42=105\begin{aligned}|3P_1+2P_3|&=|3(3,2,-2)+2(-2,1,5)|\\&=|(9,6,-6)+(-4,2,10)|\\&=|(5,8,4)|\\&=\sqrt{5^2+8^2+4^2}\\&=\underline{\underline{\sqrt{105}}}\end{aligned}
  1. Find uvu\cdot v and the angle between vector uu and vv for
    1. a. u=2𝑖2𝑗+𝑘u = 2𝑖 − 2𝑗 + 𝑘, 𝑣=3𝑖+4𝑘𝑣 = 3𝑖 + 4𝑘
      b. u=3𝑖7𝑗u=\sqrt{3}𝑖 − 7𝑗, 𝑣=3𝑖+𝑗2𝑘𝑣 =\sqrt{3}𝑖 + 𝑗 − 2𝑘
      c. 𝑢=2𝑖+𝑗𝑢=2𝑖+𝑗, 𝑣=𝑖+2𝑗𝑘𝑣=𝑖+2𝑗−𝑘
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      Solution
      Question (a)
      θ=cos1(uvuv)=cos1((2)(3)+(2)(0)+(1)(4)22+(2)2+1232+02+42)=cos1(10925)=cos1(23)0.84 rad\begin{aligned}\theta&=\cos^{-1}\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)\\&=\cos^{-1}\left(\frac{(2)(3)+(-2)(0)+(1)(4)}{\sqrt{2^2+(-2)^2+1^2}\sqrt{3^2+0^2+4^2}}\right)\\&=\cos^{-1}\left(\frac{10}{\sqrt{9}\sqrt{25}}\right)\\&=\cos^{-1}\left(\frac{2}{3}\right)\\&\approx\underline{\underline{0.84\text{ rad}}}\end{aligned}
      Question (b)
      θ=cos1(uvuv)=cos1((3)(3)+(7)(1)+(0)(2)(3)2+(7)2+02(3)2+12+(2)2)=cos1(37528)=cos1(126)1.77 rad\begin{aligned}\theta&=\cos^{-1}\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)\\&=\cos^{-1}\left(\frac{(\sqrt{3})(\sqrt{3})+(-7)(1)+(0)(-2)}{\sqrt{(\sqrt{3})^2+(-7)^2+0^2}\sqrt{(\sqrt{3})^2+1^2+(-2)^2}}\right)\\&=\cos^{-1}\left(\frac{3-7}{\sqrt{52}\sqrt{8}}\right)\\&=\cos^{-1}\left(\frac{-1}{\sqrt{26}}\right)\\&\approx\underline{\underline{1.77\text{ rad}}}\end{aligned}
      Question (c)
      θ=cos1(uvuv)=cos1((2)(1)+(1)(2)+(0)(1)22+12+0212+22+(1)2)=cos1(456)=cos1(430)0.75 rad\begin{aligned}\theta&=\cos^{-1}\left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)\\&=\cos^{-1}\left(\frac{(2)(1)+(1)(2)+(0)(-1)}{\sqrt{2^2+1^2+0^2}\sqrt{1^2+2^2+(-1)^2}}\right)\\&=\cos^{-1}\left(\frac{4}{\sqrt{5}\sqrt{6}}\right)\\&=\cos^{-1}\left(\frac{4}{\sqrt{30}}\right)\\&\approx\underline{\underline{0.75\text{ rad}}}\end{aligned}
  1. Find the area of the triangle determined by the points PP, QQ, and RR and a unit vector perpendicular to plane PQRPQR for
    1. a. P(1,1,2)P (1, -1, 2), Q(2,0,1)Q (2, 0, 1), R(0,2,1)R (0, 2, 1)
      b. P(2,2,1)P (2, -2, 1), Q(3,1,2)Q (3, -1, 2), R(3,1,1)R (3, -1, 1)
      c. P(2,2,0)P (-2, 2, 0), Q(0,1,1)Q (0, 1, -1), R(1,2,2)R (-1, 2, -2)
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      Solution
      Question (a)
      PQ=(2,0,1)(1,1,2)=(1,1,1)\overrightarrow{PQ}=(2,0,1)-(1,-1,2)=(1,1,-1)
      PR=(0,2,1)(1,1,2)=(1,3,1)\overrightarrow{PR}=(0,2,1)-(1,-1,2)=(-1,3,-1)
      PQ×PR=ijk111131=2i+2j+4k\overrightarrow{PQ}\times\overrightarrow{PR}=\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&-1\\-1&3&-1\end{array}\right|=2\mathbf{i}+2\mathbf{j}+4\mathbf{k}
      Area=12PQ×PR=124+4+16=6\text{Area}=\frac{1}{2}\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|=\frac{1}{2}\sqrt{4+4+16}=\underline{\underline{\sqrt{6}}}
      u=PQ×PRPQ×PR=16(i+j+2k)\mathbf{u}=\frac{\overrightarrow{PQ}\times\overrightarrow{PR}}{\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|}=\underline{\underline{\frac{1}{\sqrt{6}}(\mathbf{i}+\mathbf{j}+2\mathbf{k})}}
      Question (b)
      PQ=(3,1,2)(2,2,1)=(1,1,1)\overrightarrow{PQ}=(3,-1,2)-(2,-2,1)=(1,1,1)
      PR=(3,1,1)(2,2,1)=(1,1,0)\overrightarrow{PR}=(3,-1,1)-(2,-2,1)=(1,1,0)
      PQ×PR=ijk111110=i+j\overrightarrow{PQ}\times\overrightarrow{PR}=\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&1\\1&1&0\end{array}\right|=-\mathbf{i}+\mathbf{j}
      Area=12PQ×PR=121+1=22\text{Area}=\frac{1}{2}\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|=\frac{1}{2}\sqrt{1+1}=\underline{\underline{\frac{\sqrt{2}}{2}}}
      u=PQ×PRPQ×PR=12(i+j)=12(ij)\mathbf{u}=\frac{\overrightarrow{PQ}\times\overrightarrow{PR}}{\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|}=-\frac{1}{\sqrt{2}}(-\mathbf{i}+\mathbf{j})=\underline{\underline{-\frac{1}{\sqrt{2}}(\mathbf{i}-\mathbf{j})}}
      Question (c)
      PQ=(0,1,1)(2,2,0)=(2,1,1)\overrightarrow{PQ}=(0,1,-1)-(-2,2,0)=(2,-1,-1)
      PR=(1,2,2)(2,2,0)=(1,0,2)\overrightarrow{PR}=(-1,2,-2)-(-2,2,0)=(1,0,-2)
      PQ×PR=ijk211102=2i+3j+k\overrightarrow{PQ}\times\overrightarrow{PR}=\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-1&-1\\1&0&-2\end{array}\right|=2\mathbf{i}+3\mathbf{j}+\mathbf{k}
      Area=12PQ×PR=124+9+1=142\text{Area}=\frac{1}{2}\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|=\frac{1}{2}\sqrt{4+9+1}=\underline{\underline{\frac{\sqrt{14}}{2}}}
      u=PQ×PRPQ×PR=114(2i+3j+k)\mathbf{u}=\frac{\overrightarrow{PQ}\times\overrightarrow{PR}}{\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|}=\underline{\underline{\frac{1}{\sqrt{14}}(2\mathbf{i}+3\mathbf{j}+\mathbf{k})}}
  1. Find directional derivative of
    1. a. f(x,y,z)=x2yzf(x,y,z)=x^2yz in the direction of 4i^3k^4\hat{i}-3\hat{k} at point (1,1,1)(1,-1,1).
      b. f(x,y,z)=xyzf(x,y,z)=\sqrt{xyz} in the direction of v^=(4,2,2)\hat{v}=(4,-2,2) at point (3,2,6)(3,2,6).
      c. h(x,y,z)=xy+yz+zxh(x,y,z)=xy+yz+zx in the direction of v^=(3,6,2)\hat{v}=(3,6,-2) at point (1,1,2)(1,-1,2).
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      Solution
      Question (a)
      Duf(x,y,z)=f(x,y,z)uD_uf(x,y,z)=\nabla f(x,y,z)\cdot \mathbf{u}
      f(x,y,z)=2xyzi^+x2zj^+x2yk^f(1,1,1)=2i^+j^k^\begin{aligned}\nabla f(x,y,z)&=2xyz\hat{i}+x^2z\hat{j}+x^2y\hat{k}\\\nabla f(1,-1,1)&=-2\hat{i}+\hat{j}-\hat{k}\end{aligned}
      u=uu=4i^3k^16+9=4i^3j^5\mathbf{u}=\frac{\overrightarrow{u}}{|u|}=\frac{4\hat{i}-3\hat{k}}{\sqrt{16+9}}=\frac{4\hat{i}-3\hat{j}}{5}
      Du=(2i^+j^k^)(4i^3j^5)=(2)(45)+0+(1)(35)=85+35=1\begin{aligned}D_u&=(-2\hat{i}+\hat{j}-\hat{k})\left(\frac{4\hat{i}-3\hat{j}}{5}\right)\\&=(-2)\left(\frac{4}{5}\right)+0+(1)\cdot\left(\frac{3}{5}\right)\\&=-\frac{8}{5}+\frac{3}{5}\\&=\underline{\underline{-1}}\end{aligned}
      Question (b)
      Duf(x,y,z)=f(x,y,z)uD_uf(x,y,z)=\nabla f(x,y,z)\cdot \mathbf{u}
      f(x,y,z)=yz2xyzi^+xz2xyzj^+xy2xyzk^f(3,2,6)=1212i^+1812j^+612k^=i^+32j^+12k^\begin{aligned}\nabla f(x,y,z)&=\frac{yz}{2\sqrt{xyz}}\hat{i}+\frac{xz}{2\sqrt{xyz}}\hat{j}+\frac{xy}{2\sqrt{xyz}}\hat{k}\\\nabla f(3,2,6)&=\frac{12}{12}\hat{i}+\frac{18}{12}\hat{j}+\frac{6}{12}\hat{k}=\hat{i}+\frac{3}{2}\hat{j}+\frac{1}{2}\hat{k}\end{aligned}
      u=uu=4i^2j^+2k^16+4+4=4i^2j^+2k^26\mathbf{u}=\frac{\overrightarrow{u}}{|u|}=\frac{4\hat{i}-2\hat{j}+2\hat{k}}{\sqrt{16+4+4}}=\frac{4\hat{i}-2\hat{j}+2\hat{k}}{2\sqrt{6}}
      Du=(i^+32j^+12k^)(4i^2j^+2k^26)=126[1(4)+32(2)+12(2)]=66\begin{aligned}D_u&=(\hat{i}+\frac{3}{2}\hat{j}+\frac{1}{2}\hat{k})\left(\frac{4\hat{i}-2\hat{j}+2\hat{k}}{2\sqrt{6}}\right)\\&=\frac{1}{2\sqrt{6}}\left[1(4)+\frac{3}{2}(-2)+\frac{1}{2}(2)\right]\\&=\underline{\underline{\frac{\sqrt{6}}{6}}}\end{aligned}
      Question (c)
      Duf(x,y,z)=f(x,y,z)uD_uf(x,y,z)=\nabla f(x,y,z)\cdot \mathbf{u}
      f(x,y,z)=(y+z)i^+(x+z)j^+(y+x)k^f(1,1,2)=i^+3j^\begin{aligned}\nabla f(x,y,z)&=(y+z)\hat{i}+(x+z)\hat{j}+(y+x)\hat{k}\\\nabla f(1,-1,2)&=\hat{i}+3\hat{j}\end{aligned}
      u=uu=3i^+6j^2k^9+36+4=3i^+6j^2k^7\mathbf{u}=\frac{\overrightarrow{u}}{|u|}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{9+36+4}}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}
      Du=(i^+3j^)(3i^+6j^2k^7)=17[1(3)+3(6)+0(2)]=3\begin{aligned}D_u&=(\hat{i}+3\hat{j})\left(\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}\right)\\&=\frac{1}{7}\left[1(3)+3(6)+0(-2)\right]\\&=\underline{\underline{3}}\end{aligned}
  1. Calculate the divergence of the vector fields F(x,y)F(x, y) and G(x,y)G (x, y)
    1. a. F=y3i+xyjF=y^3\mathbf{i}+xy\mathbf{j}
      b. G=4yx2i+sin(y)j+3kG=\frac{4y}{x^2}\mathbf{i}+\sin{(y)}\mathbf{j}+3\mathbf{k}
      c. G=exi+ln(xy)j+exyzkG=e^x\mathbf{i}+\ln{(xy)}\mathbf{j}+e^{xyz}\mathbf{k}
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      Solution
      Question (a)
      F(x,y)=F1x+F2y=xy3+yxy=0+x=x\begin{aligned}\nabla\cdot F(x,y)&=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}\\&=\frac{\partial}{\partial x}y^3+\frac{\partial}{\partial y}xy\\&=0+x\\&=\underline{\underline{x}}\end{aligned}
      Question (b)
      G=G1x+G2y+G3z=x(4yx2)+ysin(y)+z3=4y×xx2+cos(y)=4y×(2)x21+cos(y)=8yx3+cos(y)=8yx3+cos(y)\begin{aligned}\partial\cdot G&=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z}\\&=\frac{\partial}{\partial x}\left(\frac{4y}{x^2}\right)+\frac{\partial}{\partial y}\sin{(y)}+\frac{\partial}{\partial z}3\\&=4y\times\frac{\partial}{\partial x}x^{-2}+\cos{(y)}\\&=4y \times (-2)x^{-2-1}+\cos{(y)}\\&=-8yx^{-3}+\cos{(y)}\\&=\underline{\underline{-\frac{8y}{x^3}+\cos{(y)}}}\end{aligned}
      Question (c)
      G=G1x+G2y+G3z=xex+yln(xy)+zexyz=ex+y(ln(x)+ln(y))+exyz×z(xyz)=ex+1y+xyexyz\begin{aligned}\nabla\cdot G &=\frac{\partial G_1}{\partial x}+\frac{\partial G_2}{\partial y}+\frac{\partial G_3}{\partial z}\\&=\frac{\partial}{\partial x}e^x+\frac{\partial}{\partial y}\ln{(xy)}+\frac{\partial}{\partial z}e^{xyz}\\&=e^x+\frac{\partial}{\partial y}(\ln{(x)}+\ln{(y)})+e^{xyz}\times\frac{\partial}{\partial z}(xyz)\\&=\underline{\underline{e^x+\frac{1}{y}+xye^{xyz}}}\end{aligned}
  1. Calculate the curl of the following vector fields F(x,y,z)F(x, y, z).
    1. a. F=3x2i+2zjxkF=3x^2\mathbf{i}+2z\mathbf{j}-x\mathbf{k}
      b. F=y3i+xyjzkF=y^3\mathbf{i}+xy\mathbf{j}-z\mathbf{k}
      c. F=(1+y+z2)i+(exyz)j(xyz)kF=(1+y+z^2)\mathbf{i}+(e^{xyz})\mathbf{j}-(xyz)\mathbf{k}
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      Solution
      Question (a)
      ×F=ijkxyz3x22zx=((x)y(2z)z)i((x)x(3x2)z)j+((2z)x(3x2)y)k=(02)i(10)j+(0+0)k=2i+j\begin{aligned}\nabla \times F&=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3x^2&2z&-x\end{array}\right|\\&=\left(\frac{\partial(-x)}{\partial y}-\frac{\partial(2z)}{\partial z}\right)\vec{i}-\left(\frac{\partial(-x)}{\partial x}-\frac{\partial(3x^2)}{\partial z}\right)\vec{j}+\left(\frac{\partial(2z)}{\partial x}-\frac{\partial(3x^2)}{\partial y}\right)\vec{k}\\&=(0-2)\vec{i}-(-1-0)\vec{j}+(0+0)\vec{k}\\&=\underline{\underline{-2\vec{i}+\vec{j}}}\end{aligned}
      Question (b)
      ×F=(F3yF2z)i(F3xF1z)j+(F2xF1y)k=((z)y(xy)z)i((z)x(y3)z)j+((xy)x(y3)y)k=0i0j+(y3y2)k=(y3y2)k\begin{aligned}\nabla\times F&=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\vec{i}-\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)\vec{j}+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\vec{k}\\&=\left(\frac{\partial (-z)}{\partial y}-\frac{\partial (xy)}{\partial z}\right)\vec{i}-\left(\frac{\partial (-z)}{\partial x}-\frac{\partial (y^3)}{\partial z}\right)\vec{j}+\left(\frac{\partial (xy)}{\partial x}-\frac{\partial (y^3)}{\partial y}\right)\vec{k}\\&=0\vec{i}-0\vec{j}+(y-3y^2)\vec{k}\\&=\underline{\underline{(y-3y^2)\vec{k}}}\end{aligned}
      ⇒ the curl vector is in the kk direction
       
      Question (c)
      ×F=((xyz)y(exyz)z)i+((1+y+z2)z(xyz)x)j+((exyz)z(1+y+z2)y)k=(xzxyexyz)i+(2z+yz)j+(yzexyz1)k\begin{aligned}\nabla\times F&=\left(\frac{\partial(-xyz)}{\partial y}-\frac{\partial (e^{xyz})}{\partial z}\right)\vec{i}+\left(\frac{\partial (1+y+z^2)}{\partial z}-\frac{\partial (-xyz)}{\partial x}\right)\vec{j}+\left(\frac{\partial (e^{xyz})}{\partial z}-\frac{\partial (1+y+z^2)}{\partial y}\right)\vec{k}\\&=\underline{\underline{(-xz-xye^{xyz})\vec{i}+(2z+yz)\vec{j}+(yze^{xyz}-1)\vec{k}}}\end{aligned}