Week 2

Tutorial 2: Engineering Application of Vector Analysis

Application of Vectors

1. A motorboat sets out in the direction of N80°E. The speed of the boat in still water is 20 mph. If the current is flowing directly south and the actual direction of the motorboat is due east, find the speed of the current and the actual speed of the motorboat.
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Solution

Let $\underset{\sim}{u}$ = velocity of motorboat, $\underset{\sim}{v}$ = velocity of current.
$\underset{\sim}{u}=\left[20\cos 10^\circ, 20\sin 10^\circ\right]$
$\underset{\sim}{v}=\left[0,-|\underset{\sim}{v}|\right]$
$\underset{\sim}{u}+\underset{\sim}{v}=\left[20\cos 10^\circ, 20\sin 10^\circ-|\underset{\sim}{v}|\right]$
Actual direction is due east ⇒ no $y$-component in $\underset{\sim}{u}+\underset{\sim}{v}$.
$|\underset{\sim}{v}|=20\sin 10^\circ = \underline{\underline{3.47\text{ mph}}}$
$|\underset{\sim}{u}+\underset{\sim}{v}|=20\cos 10^\circ=\underline{\underline{19.70\text{ mph}}}$

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



a. What is the heading of the route that the skipper should take to Dunes Beach?
b. Suppose that as the New Hopes heads for Dunes Beach with 2.4 knots, a strong wind moves the boat at 0.8 knots at a heading of 200°. The skipper sticks to the original heading from Part a despite the wind. Make a labeled sketch of the situation that shows the intended boat path, the effects of the wind and the altered path of the boat.
c. Determine the speed and heading of the boat on its altered path.
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Solution
Question (a)
Heading 250$^\circ$
Question (b)

Question (c)
$h_1=2.4\cos{(20^\circ)}=2.255$
$v_1=2.4\sin{(20^\circ)}=0.821$
$h_2=0.8\sin{(20^\circ)}=0.274$
$v_2=0.8\cos{(20^\circ)}=0.752$
$H=h_1+h_2=2.529$
$V=v_1+v_2=1.573$
$\text{Altered speed}=\sqrt{2.529^2+1.573^2}=\underline{\underline{2.98\text{ knots}}}$
$\theta=\tan^{-1}{\frac{1.573}{2.529}}=31.88^\circ$
⇒ $\text{Altered heading}=180^\circ+(90^\circ-31.88^\circ)=\underline{\underline{238.12^\circ}}$

Work Done

1. It takes 12 000 J of work to pull a sled 200 m with a 150 N force. Determine the angle of the rope with the horizontal.
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Solution
$\begin{aligned}12000&=(150)(200)\cos\theta\\\theta&=\cos^{-1}{\left(\frac{12000}{(150)(200)}\right)}\\&\approx66^\circ\end{aligned}$

2. Find the work done by a force $\mathbf{F} = 5\mathbf{j}$ (magnitude 5 N) in moving an object along the line from the origin to the point $(1, 1)$ (distance in meters).
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Solution
$P(0,0),Q(1,1),\mathbf{F}=5\mathbf{j}$
$\overrightarrow{PQ}=\mathbf{i}+\mathbf{j}$
$\mathbf{W}=\mathbf{F}\cdot\overrightarrow{PQ}=(5\mathbf{j})(\mathbf{i}+\mathbf{j})=5\text{ N}\cdot\text{m}=\underline{\underline{5\text{ J}}}$

3. The wind passing over a boat’s sail exerted a 1000-lb (pound) magnitude force $\mathbf{F}$ 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 feet.



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Solution
$\mathbf{W}=|\mathbf{F}|\left|\overrightarrow{PQ}\right|\cos\theta=(1000)(5280)(\cos{60^\circ})=\underline{\underline{2,640,000\text{ ft}\cdot\text{lb}}}$

4. Ali turns a crank to lower a bucket of water into a well. Determine the total work done on the bucket if the weight of the bucket is 20 N and the tension force in the rope is 11 N. The bucket lowers a distance of 4.5 m while he is cranking.



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Solution
$\mathbf{W}=\mathbf{F}\cdot\textbf{D}=|\mathbf{F}||\mathbf{D}|\cos\theta$
$\text{Work done by the rope, }W_{rope}= 11(4.5)\cos{180^\circ}=-49.5\text{ J}$
$\text{Work done by the bucket, }W_{bucket}=20(4.5)\cos{0^\circ}=90\text{ J}$
$\begin{aligned}\therefore\text{ Total work done}&=W_{rope}+W_{bucket}\\&=-49.5+90\\&=\underline{\underline{40.5\text{ J}}}\end{aligned}$

Find the Torque / Moment of Force

1. A bolt is tightened using a 20 N force, applied at an angle of 60° to the end of a wrench that is 30 cm long. Calculate the magnitude of the torque about its point of rotation.
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Solution
$|\vec{\tau}|=(0.3)(20)\sin{60^\circ}\approx\underline{\underline{5.2\text{ J}}}$

2. Explain the difference between both pictures below.



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Solution
Both have the same magnitude but different direction.

3. Physical therapists use torque to analyze triceps exercise shown in the diagram below. The triceps muscle exerts a force on the elbow-end of the forearm, affecting the rotation of the forearm. If the triceps exerts a force of 15 N, what is the magnitude of torque applied to the forearm?



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Solution
$\text{Magnitude of torque}=(2.5\text{ cm})(15\text{ N})\sin{75^\circ}=\underline{\underline{36.22\text{ cm}\cdot\text{N}}}$

Gradient: Directional Derivatives, Tangent Plane and Normal Plane

1. Is there a direction $\mathbf{u}$ in which the rate of change of the temperature function $𝑇(𝑥, 𝑦, 𝑧) = 2𝑥𝑦 − 𝑦𝑧$ (temperature in degrees Celsius, distance in feet) at $𝑃(1, −1,1)$is −3℃/ft?. Give reasons for your answer.
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Solution
$\nabla T=2y\textbf{i}+(2x-z)\textbf{j}-y\textbf{k}$
$\nabla T(1,-1,1)=-2\mathbf{i}+\mathbf{j}+\mathbf{k}$
$|\nabla T(1,-1,1)|=\sqrt{(-2)^2+1^2+1^2}=\sqrt{6}$
⇒ No. The minimum rate of change is $-\sqrt{6}$.

2. A paraboloid of revolution has equation of $2z=x^2+y^2$. Find the unit normal vector to the surface at the point $(1, 3, 5)$ and normal and tangent line plane to the surface at the same point.
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Solution
$x^2+y^2-2z=0$
$\nabla f(x,y,z)=2x\hat{i}+2y\hat{j}-2z\hat{k}$
$\nabla f(1,3,5)=2\hat{i}+6\hat{j}-2\hat{k}$
Tangent line:
$\begin{aligned}2(x-1)+6(y-3)-2(z-5)&=0\\\underline{\underline{2x-6y-2z-10=0}}\end{aligned}$
Normal line:
$\underline{\underline{x=1+2t;\quad y=3+6t;\quad z=5-2t}}$ or $\underline{\underline{\frac{x-1}{2}=\frac{y-3}{6}=\frac{z-5}{-2}=t}}$

3. Find the equations of the tangent plane and normal line to the surfaces
a. $2x^2+y^2-z^2=3$ at $(1,2,3)$
b. $30-y^2-z^2=x^2$ at $(1,-2,5)$
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Solution
Question (a)
$\nabla f(x,y,z)=4x\hat{i}+2y\hat{j}-2z\hat{k}$
$\nabla f(1,2,3)=4\hat{i}+4\hat{j}-6\hat{k}$
Tangent line:
$\begin{aligned}4(x-1)+4(y-2)-6(z-3)&=0\\\underline{\underline{4x+4y-6z+6=0}}\end{aligned}$
Normal line:
$\underline{\underline{x=1+4t;\quad y=2+4t;\quad z=3-6t}}$
Question (b)
$x^2+y^2+z^2-30=0$
$\nabla f(x,y,z)=2x\hat{i}+2y\hat{j}+2z\hat{k}$
$\nabla f(1,-2,5)=2\hat{i}-4\hat{j}+10\hat{k}$
Tangent line:
$\begin{aligned}2(x-1)-4(y+2)+10(z-5)&=0\\\underline{\underline{2x-4y+10z-60=0}}\end{aligned}$
Normal line:
$\underline{\underline{x=1+2t;\quad y=-2-4t;\quad z=5+10t}}$

Divergence and Curl

1. Determine the divergence of the following vector field $\mathbf{F}(x,y)=\frac{x}{y}\mathbf{i}+(2x-3y)\mathbf{j}$ together its physical meaning.
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Solution
$\begin{aligned}\nabla\cdot\mathbf{F}(x,y)&=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}\\&=\frac{\partial}{\partial x}\left(\frac{x}{y}\right)+\frac{\partial}{\partial y}(2x-3y)\\&=\underline{\underline{\frac{1}{y}-3}}\end{aligned}$
The divergence of the vector is a scalar, thus it is either expanding or compressing.

2. Determine the curl of the following vector field $\mathbf{F}(x,y,z)=x\mathbf{i}-y\mathbf{j}+z\mathbf{k}$ together its physical meaning.
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Solution
$\begin{aligned}\nabla\times\mathbf{F}&=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)\mathbf{i}-\left(\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z}\right)\mathbf{j}+\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\mathbf{k}\\&=\left(\frac{\partial (z)}{\partial y}-\frac{\partial (-y)}{\partial z}\right)\mathbf{i}-\left(\frac{\partial (z)}{\partial x}-\frac{\partial (x)}{\partial z}\right)\mathbf{j}+\left(\frac{\partial (-y)}{\partial x}-\frac{\partial (x)}{\partial y}\right)\mathbf{k}\\&=0\mathbf{i}-0\mathbf{j}+0\mathbf{k}=\underline{\underline{0}}\end{aligned}$
Therefore the vector field $\mathbf{F}=x\mathbf{i}-y\mathbf{j}+z\mathbf{k}$ is an irrotational vector field.