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Tutorial 5

Tutorial 5: Engineering Applications of Vector Algebra and Analysis

Tutorial QuestionTutorial Solution

  1. The following map shows the location of the docked New Hopes when its skipper decided to navigate to Dunes Beach.

    Figure for Question 1

    a. What is the heading of the route that the skipper should take to Dunes Beach?

    Solution

    Heading 250∘250^\circ

    b. Suppose that as the New Hopes heads for Dunes Beach, a strong wind moves the boat at 0.80.8 knot at a heading of 200∘200^{\circ}. The skipper sticks to the original heading from Part a despite the wind. Make a labelled sketch of the situation that shows the intended boat path, the effects of the wind and the altered path of the boat.

    Solution
    Figure for Solution 1

    c. Determine the speed and heading of the boat on its altered path.

    Solution
    h1=2.4cos⁡(20∘)=2.255v1=2.4sin⁡(20∘)=0.821h2=0.8sin⁡(20∘)=0.274v2=0.8cos⁡(20∘)=0.752H=h1+h2=2.529V=v1+v2=1.573\begin{aligned} h_1&=2.4\cos{(20^\circ)}=2.255\\ v_1&=2.4\sin{(20^\circ)}=0.821\\ h_2&=0.8\sin{(20^\circ)}=0.274\\ v_2&=0.8\cos{(20^\circ)}=0.752\\ H&=h_1+h_2=2.529\\ V&=v_1+v_2=1.573 \end{aligned}

    Altered speed =2.5292+1.5732=2.98=\sqrt{2.529^2+1.573^2}=2.98 knots

    θ=tan⁡−11.5732.529=31.88∘\theta=\tan^{-1}{\frac{1.573}{2.529}}=31.88^\circ

    Altered heading=180∘+(90∘−31.88∘)=238.12∘=180^\circ+(90^\circ-31.88^\circ)=238.12^\circ

  2. a. It takes 12000 J12000 \mathrm{~J} of work to pull a sled 200 m200 \mathrm{~m} with a 150 N150 \mathrm{~N} force. Determine the angle of the rope with the horizontal.

    Solution
    12000=(150)(200)cos⁡θθ=cos⁡−1(12000(150)(200))≈66∘\begin{aligned}12000&=(150)(200)\cos\theta\\\theta&=\cos^{-1}{\left(\frac{12000}{(150)(200)}\right)}\\&\approx66^\circ\end{aligned}

    b. Find the work done by a force F=5i\mathrm{F}=5 \mathrm{i} (magnitude 5 N5 \mathrm{~N} ) in moving an object along the line from the origin to the point (1,1)(1,1) (distance in meters).

    Solution
    P(0,0),Q(1,1),F=5jP(0,0),Q(1,1),\mathbf{F}=5\mathbf{j}
    PQ→=i+j\overrightarrow{PQ}=\mathbf{i}+\mathbf{j}
    W=F⋅PQ→=(5j)(i+j)=5 N⋅m=5 J\mathbf{W}=\mathbf{F}\cdot\overrightarrow{PQ}=(5\mathbf{j})(\mathbf{i}+\mathbf{j})=5\text{ N}\cdot\text{m}=5\text{ J}

    c. The wind passing over a boat's sail exerted a 1000-lb (pound) magnitude force F\mathbf{F} as shown here. How much work did the wind perform in moving the boat forward 1 mile? Answer in foot-pounds where 1 mile =5280=5280 foots.

    Figure for Question 2
    Solution
    W=∣F∣∣PQ→∣cos⁡θ=(1000)(5280)(cos⁡60∘)=2,640,000 ft⋅lb\mathbf{W}=|\mathbf{F}|\left|\overrightarrow{PQ}\right|\cos\theta=(1000)(5280)(\cos{60^\circ})=2,640,000\text{ ft}\cdot\text{lb}

  3. a. A bolt is tightened using a 20 N20 \mathrm{~N} force, applied at an angle of 60∘60^{\circ} to the end of a wrench that is 30 cm30 \mathrm{~cm} long. Calculate the magnitude of the torque about its point of rotation.

    Solution
    ∣τ⃗∣=(0.3)(20)sin⁡60∘≈5.2 J‾‾|\vec{\tau}|=(0.3)(20)\sin{60^\circ}\approx\underline{\underline{5.2\text{ J}}}

    b. Explain the difference between both pictures

    Figure for Question 3
    Solution

    Both have the same magnitude but different direction.

  4.   a. Is there a direction u\mathbf{u} in which the rate of change of the temperature function T(x,y,z)=2xy−yzT(x, y, z)=2 x y-y z (temperature in degrees Celsius, distance in feet) at P(1,−1,1)P(1,-1,1) is −3∘C/ft-3^{\circ} \mathrm{C} / f t ?. Give reasons for your answer.

    Solution
    ∇T=2yi+(2x−z)j−yk\nabla T=2y\textbf{i}+(2x-z)\textbf{j}-y\textbf{k}
    ∇T(1,−1,1)=−2i+j+k\nabla T(1,-1,1)=-2\mathbf{i}+\mathbf{j}+\mathbf{k}
    ∣∇T(1,−1,1)∣=(−2)2+12+12=6|\nabla T(1,-1,1)|=\sqrt{(-2)^2+1^2+1^2}=\sqrt{6}

    ⇒\Rightarrow No. The minimum rate of change is −6-\sqrt{6}.

    b. A paraboloid of revolution has equation of 2z=x2+y22 z=x^{2}+y^{2}. Find the unit normal vector to the surface at the point (1,3,5)(1,3,5) and normal and tangent line plane to the surface at the same point.

    Solution
    x2+y2−2z=0x^2+y^2-2z=0
    ∇f(x,y,z)=2xi^+2yj^−2zk^\nabla f(x,y,z)=2x\hat{i}+2y\hat{j}-2z\hat{k}
    ∇f(1,3,5)=2i^+6j^−2k^\nabla f(1,3,5)=2\hat{i}+6\hat{j}-2\hat{k}

    Tangent line:

    2(x−1)+6(y−3)−2(z−5)=02x−6y−2z−10=0\begin{aligned}2(x-1)+6(y-3)-2(z-5)&=0\\ 2x-6y-2z-10=0 \end{aligned}

    Normal line:

    x=1+2t;y=3+6t;z=5−2t or x−12=y−36=z−5−2=tx=1+2t;\quad y=3+6t;\quad z=5-2t\qquad\text{ or }\qquad \frac{x-1}{2}=\frac{y-3}{6}=\frac{z-5}{-2}=t

    c. Find the equations of the tangent plane and normal line to the surfaces

    i. 2x2+y2−z2=−32 x^{2}+y^{2}-z^{2}=-3 at (1,2,3)(1,2,3)

    Solution
    ∇f(x,y,z)=4xi^+2yj^−2zk^\nabla f(x,y,z)=4x\hat{i}+2y\hat{j}-2z\hat{k}
    ∇f(1,2,3)=4i^+4j^−6k^\nabla f(1,2,3)=4\hat{i}+4\hat{j}-6\hat{k}

    Tangent line:

    4(x−1)+4(y−2)−6(z−3)=04x+4y−6z+6=0\begin{aligned} 4(x-1)+4(y-2)-6(z-3)&=0\\ 4x+4y-6z+6=0 \end{aligned}

    Normal line:

    x=1+4t;y=2+4t;z=3−6tx=1+4t;\quad y=2+4t;\quad z=3-6t

    ii. 30−y2−z2=x230-y^{2}-z^{2}=x^{2} at (1,−2,5)(1,-2,5)

    Solution
    x2+y2+z2−30=0x^2+y^2+z^2-30=0
    ∇f(x,y,z)=2xi^+2yj^+2zk^\nabla f(x,y,z)=2x\hat{i}+2y\hat{j}+2z\hat{k}
    ∇f(1,−2,5)=2i^−4j^+10k^\nabla f(1,-2,5)=2\hat{i}-4\hat{j}+10\hat{k}

    Tangent line:

    2(x−1)−4(y+2)+10(z−5)=02x−4y+10z−60=0\begin{aligned} 2(x-1)-4(y+2)+10(z-5)&=0\\ 2x-4y+10z-60=0 \end{aligned}

    Normal line:

    x=1+2t;y=−2−4t;z=5+10tx=1+2t;\quad y=-2-4t;\quad z=5+10t

  5.   a. Determine the divergence of the vector field F(x,y)=x/yi+(2x−3y)jF(x, y)=x / y i+(2 x-3 y) j together with its physical meaning.

    Solution
    ∇⋅F(x,y)=∂F1∂x+∂F2∂y=∂∂x(xy)+∂∂y(2x−3y)=1y−3‾‾\begin{aligned} \nabla\cdot\mathbf{F}(x,y)&=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}\\ &=\frac{\partial}{\partial x}\left(\frac{x}{y}\right)+\frac{\partial}{\partial y}(2x-3y)\\ &=\underline{\underline{\frac{1}{y}-3}}\end{aligned}

    The divergence of the vector is a scalar, thus it is either expanding or compressing.

    b. Determine the curl of the vector field F(x,y,z)=xi−yj+zkF(x, y, z)=x \boldsymbol{i}-y \boldsymbol{j}+z \boldsymbol{k} together with its physical meaning.

    Solution
    ∇×F=(∂F3∂y−∂F2∂z)i−(∂F3∂x−∂F1∂z)j+(∂F2∂x−∂F1∂y)k=(∂(z)∂y−∂(−y)∂z)i−(∂(z)∂x−∂(x)∂z)j+(∂(−y)∂x−∂(x)∂y)k=0i−0j+0k=0\begin{aligned} \nabla\times\mathbf{F}&=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\mathbf{i}-\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)\mathbf{j}+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\mathbf{k}\\ &=\left(\frac{\partial (z)}{\partial y}-\frac{\partial (-y)}{\partial z}\right)\mathbf{i}-\left(\frac{\partial (z)}{\partial x}-\frac{\partial (x)}{\partial z}\right)\mathbf{j}+\left(\frac{\partial (-y)}{\partial x}-\frac{\partial (x)}{\partial y}\right)\mathbf{k}\\ &=0\mathbf{i}-0\mathbf{j}+0\mathbf{k}=0 \end{aligned}

    Therefore the vector field F=xi−yj+zk\mathbf{F}=x\mathbf{i}-y\mathbf{j}+z\mathbf{k} is an irrotational vector field.

  6. A force of 2.5 N2.5 \mathrm{~N} is applied perpendicular to the handle of a spanner with length of 15 cm to tighten a bolt. Find the torque exerted by the force about the center of the bolt and the direction of the torque. (Ans: 37.5×10−2Nm37.5 \times 10^{-2} \mathrm{Nm} )

    Solution

    Angle between rr and FF, θ=90∘\theta=90^\circ

    τ=∣F∣∣r∣sin⁡θ=2.5N×0.015m×sin⁡90=37.5×10−2Nm\tau=|F||r|\sin\theta=2.5\text{N}\times0.015\text{m}\times\sin90=37.5\times10^{-2}\text{Nm}

    Direction: anticlockwise

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