The following 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?
Heading 250 ∘ 250^\circ 25 0 ∘
b. Suppose that as the New Hopes heads for Dunes Beach, a strong wind moves the boat at 0.8 0.8 0.8 200 ∘ 200^{\circ} 20 0 ∘
c. Determine the speed and heading of the boat on its altered path.
h 1 = 2.4 cos ( 20 ∘ ) = 2.255 v 1 = 2.4 sin ( 20 ∘ ) = 0.821 h 2 = 0.8 sin ( 20 ∘ ) = 0.274 v 2 = 0.8 cos ( 20 ∘ ) = 0.752 H = h 1 + h 2 = 2.529 V = v 1 + v 2 = 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} h 1 v 1 h 2 v 2 H V = 2.4 cos ( 2 0 ∘ ) = 2.255 = 2.4 sin ( 2 0 ∘ ) = 0.821 = 0.8 sin ( 2 0 ∘ ) = 0.274 = 0.8 cos ( 2 0 ∘ ) = 0.752 = h 1 + h 2 = 2.529 = v 1 + v 2 = 1.573 Altered speed = 2.529 2 + 1.573 2 = 2.98 =\sqrt{2.529^2+1.573^2}=2.98 = 2.52 9 2 + 1.57 3 2 = 2.98
θ = tan − 1 1.573 2.529 = 31.88 ∘ \theta=\tan^{-1}{\frac{1.573}{2.529}}=31.88^\circ θ = tan − 1 2.529 1.573 = 31.8 8 ∘ Altered heading= 180 ∘ + ( 90 ∘ − 31.88 ∘ ) = 238.12 ∘ =180^\circ+(90^\circ-31.88^\circ)=238.12^\circ = 18 0 ∘ + ( 9 0 ∘ − 31.8 8 ∘ ) = 238.1 2 ∘
a. It takes 12000 J 12000 \mathrm{~J} 12000 J 200 m 200 \mathrm{~m} 200 m 150 N 150 \mathrm{~N} 150 N
12000 = ( 150 ) ( 200 ) cos θ θ = 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} 12000 θ = ( 150 ) ( 200 ) cos θ = cos − 1 ( ( 150 ) ( 200 ) 12000 ) ≈ 6 6 ∘ b. Find the work done by a force F = 5 i \mathrm{F}=5 \mathrm{i} F = 5 i 5 N 5 \mathrm{~N} 5 N ( 1 , 1 ) (1,1) ( 1 , 1 )
P ( 0 , 0 ) , Q ( 1 , 1 ) , F = 5 i P(0,0),Q(1,1),\mathbf{F}=5\mathbf{i} P ( 0 , 0 ) , Q ( 1 , 1 ) , F = 5 i P Q → = i + j \overrightarrow{PQ}=\mathbf{i}+\mathbf{j} P Q = i + j W = F ⋅ P Q → = ( 5 i ) ( i + j ) = 5 N ⋅ m = 5 J \mathbf{W}=\mathbf{F}\cdot\overrightarrow{PQ}=(5\mathbf{i})(\mathbf{i}+\mathbf{j})=5\text{ N}\cdot\text{m}=5\text{ J} W = F ⋅ P Q = ( 5 i ) ( i + j ) = 5 N ⋅ m = 5 J c. The wind passing over a boat's sail exerted a 1000-lb (pound) magnitude force F \mathbf{F} F = 5280 =5280 = 5280
W = ∣ F ∣ ∣ P Q → ∣ 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} W = ∣ F ∣ P Q cos θ = ( 1000 ) ( 5280 ) ( cos 6 0 ∘ ) = 2 , 640 , 000 ft ⋅ lb
a. A bolt is tightened using a 20 N 20 \mathrm{~N} 20 N 60 ∘ 60^{\circ} 6 0 ∘ 30 c m 30 \mathrm{~cm} 30 cm
∣ τ ⃗ ∣ = ( 0.3 ) ( 20 ) sin 60 ∘ ≈ 5.2 N ⋅ m ‾ ‾ |\vec{\tau}|=(0.3)(20)\sin{60^\circ}\approx\underline{\underline{5.2\text{ N}\cdot\text{m}}} ∣ τ ∣ = ( 0.3 ) ( 20 ) sin 6 0 ∘ ≈ 5.2 N ⋅ m b. Explain the difference between both pictures
Both have the same magnitude but different direction.
a. Is there a direction u \mathbf{u} u T ( x , y , z ) = 2 x y − y z T(x, y, z)=2 x y-y z T ( x , y , z ) = 2 x y − y z P ( 1 , − 1 , 1 ) P(1,-1,1) P ( 1 , − 1 , 1 ) − 3 ∘ C / f t -3^{\circ} \mathrm{C} / f t − 3 ∘ C / f t
∇ T = 2 y i + ( 2 x − z ) j − y k \nabla T=2y\textbf{i}+(2x-z)\textbf{j}-y\textbf{k} ∇ T = 2 y i + ( 2 x − z ) j − y k ∇ T ( 1 , − 1 , 1 ) = − 2 i + j + k \nabla T(1,-1,1)=-2\mathbf{i}+\mathbf{j}+\mathbf{k} ∇ T ( 1 , − 1 , 1 ) = − 2 i + j + k ∣ ∇ T ( 1 , − 1 , 1 ) ∣ = ( − 2 ) 2 + 1 2 + 1 2 = 6 |\nabla T(1,-1,1)|=\sqrt{(-2)^2+1^2+1^2}=\sqrt{6} ∣∇ T ( 1 , − 1 , 1 ) ∣ = ( − 2 ) 2 + 1 2 + 1 2 = 6 ⇒ \Rightarrow ⇒ − 6 -\sqrt{6} − 6
b. A paraboloid of revolution has equation of 2 z = x 2 + y 2 2 z=x^{2}+y^{2} 2 z = x 2 + y 2 ( 1 , 3 , 5 ) (1,3,5) ( 1 , 3 , 5 )
x 2 + y 2 − 2 z = 0 x^2+y^2-2z=0 x 2 + y 2 − 2 z = 0 ∇ f ( x , y , z ) = 2 x i ^ + 2 y j ^ − 2 k ^ \nabla f(x,y,z)=2x\hat{i}+2y\hat{j}-2\hat{k} ∇ f ( x , y , z ) = 2 x i ^ + 2 y j ^ − 2 k ^ ∇ f ( 1 , 3 , 5 ) = 2 i ^ + 6 j ^ − 2 k ^ \nabla f(1,3,5)=2\hat{i}+6\hat{j}-2\hat{k} ∇ f ( 1 , 3 , 5 ) = 2 i ^ + 6 j ^ − 2 k ^ Unit normal vector:
n ^ = ∇ f ( 1 , 3 , 5 ) ∣ ∇ f ( 1 , 3 , 5 ) ∣ = ⟨ 2 , 6 , − 2 ⟩ 2 2 + 6 2 + ( − 2 ) 2 = ⟨ 2 , 6 , − 2 ⟩ 2 11 = ⟨ 1 11 , 3 11 , − 1 11 ⟩ \hat{n}=\frac{\nabla f(1,3,5)}{|\nabla f(1,3,5)|}
=\frac{\langle 2,6,-2\rangle}{\sqrt{2^{2}+6^{2}+(-2)^{2}}}
=\frac{\langle 2,6,-2\rangle}{2\sqrt{11}}
=\left\langle \frac{1}{\sqrt{11}},\frac{3}{\sqrt{11}},-\frac{1}{\sqrt{11}}\right\rangle n ^ = ∣∇ f ( 1 , 3 , 5 ) ∣ ∇ f ( 1 , 3 , 5 ) = 2 2 + 6 2 + ( − 2 ) 2 ⟨ 2 , 6 , − 2 ⟩ = 2 11 ⟨ 2 , 6 , − 2 ⟩ = ⟨ 11 1 , 11 3 , − 11 1 ⟩ Tangent plane:
2 ( x − 1 ) + 6 ( y − 3 ) − 2 ( z − 5 ) = 0 2 x + 6 y − 2 z − 10 = 0 \begin{aligned}2(x-1)+6(y-3)-2(z-5)&=0\\
2x+6y-2z-10&=0
\end{aligned} 2 ( x − 1 ) + 6 ( y − 3 ) − 2 ( z − 5 ) 2 x + 6 y − 2 z − 10 = 0 = 0 Normal line:
x = 1 + 2 t ; y = 3 + 6 t ; z = 5 − 2 t or x − 1 2 = y − 3 6 = z − 5 − 2 = t x=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 x = 1 + 2 t ; y = 3 + 6 t ; z = 5 − 2 t or 2 x − 1 = 6 y − 3 = − 2 z − 5 = t c. Find the equations of the tangent plane and normal line to the surfaces
i. 2 x 2 + y 2 − z 2 = − 3 2 x^{2}+y^{2}-z^{2}=-3 2 x 2 + y 2 − z 2 = − 3 ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 )
∇ f ( x , y , z ) = 4 x i ^ + 2 y j ^ − 2 z k ^ \nabla f(x,y,z)=4x\hat{i}+2y\hat{j}-2z\hat{k} ∇ f ( x , y , z ) = 4 x i ^ + 2 y j ^ − 2 z k ^ ∇ f ( 1 , 2 , 3 ) = 4 i ^ + 4 j ^ − 6 k ^ \nabla f(1,2,3)=4\hat{i}+4\hat{j}-6\hat{k} ∇ f ( 1 , 2 , 3 ) = 4 i ^ + 4 j ^ − 6 k ^ Tangent plane:
4 ( x − 1 ) + 4 ( y − 2 ) − 6 ( z − 3 ) = 0 4 x + 4 y − 6 z + 6 = 0 \begin{aligned}
4(x-1)+4(y-2)-6(z-3)&=0\\
4x+4y-6z+6=0
\end{aligned} 4 ( x − 1 ) + 4 ( y − 2 ) − 6 ( z − 3 ) 4 x + 4 y − 6 z + 6 = 0 = 0 Normal line:
x = 1 + 4 t ; y = 2 + 4 t ; z = 3 − 6 t x=1+4t;\quad y=2+4t;\quad z=3-6t x = 1 + 4 t ; y = 2 + 4 t ; z = 3 − 6 t ii. 30 − y 2 − z 2 = x 2 30-y^{2}-z^{2}=x^{2} 30 − y 2 − z 2 = x 2 ( 1 , − 2 , 5 ) (1,-2,5) ( 1 , − 2 , 5 )
x 2 + y 2 + z 2 − 30 = 0 x^2+y^2+z^2-30=0 x 2 + y 2 + z 2 − 30 = 0 ∇ f ( x , y , z ) = 2 x i ^ + 2 y j ^ + 2 z k ^ \nabla f(x,y,z)=2x\hat{i}+2y\hat{j}+2z\hat{k} ∇ f ( x , y , z ) = 2 x i ^ + 2 y j ^ + 2 z k ^ ∇ f ( 1 , − 2 , 5 ) = 2 i ^ − 4 j ^ + 10 k ^ \nabla f(1,-2,5)=2\hat{i}-4\hat{j}+10\hat{k} ∇ f ( 1 , − 2 , 5 ) = 2 i ^ − 4 j ^ + 10 k ^ Tangent plane:
2 ( x − 1 ) − 4 ( y + 2 ) + 10 ( z − 5 ) = 0 2 x − 4 y + 10 z − 60 = 0 \begin{aligned}
2(x-1)-4(y+2)+10(z-5)&=0\\
2x-4y+10z-60=0
\end{aligned} 2 ( x − 1 ) − 4 ( y + 2 ) + 10 ( z − 5 ) 2 x − 4 y + 10 z − 60 = 0 = 0 Normal line:
x = 1 + 2 t ; y = − 2 − 4 t ; z = 5 + 10 t x=1+2t;\quad y=-2-4t;\quad z=5+10t x = 1 + 2 t ; y = − 2 − 4 t ; z = 5 + 10 t
a. Determine the divergence of the vector field F ( x , y ) = x / y i + ( 2 x − 3 y ) j F(x, y)=x / y i+(2 x-3 y) j F ( x , y ) = x / y i + ( 2 x − 3 y ) j
∇ ⋅ F ( x , y ) = ∂ F 1 ∂ x + ∂ F 2 ∂ y = ∂ ∂ x ( x y ) + ∂ ∂ y ( 2 x − 3 y ) = 1 y − 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} ∇ ⋅ F ( x , y ) = ∂ x ∂ F 1 + ∂ y ∂ F 2 = ∂ x ∂ ( y x ) + ∂ y ∂ ( 2 x − 3 y ) = y 1 − 3 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 ) = x i − y j + z k F(x, y, z)=x \boldsymbol{i}-y \boldsymbol{j}+z \boldsymbol{k} F ( x , y , z ) = x i − y j + z k
∇ × F = ( ∂ F 3 ∂ y − ∂ F 2 ∂ z ) i − ( ∂ F 3 ∂ x − ∂ F 1 ∂ z ) j + ( ∂ F 2 ∂ x − ∂ F 1 ∂ y ) k = ( ∂ ( z ) ∂ y − ∂ ( − y ) ∂ z ) i − ( ∂ ( z ) ∂ x − ∂ ( x ) ∂ z ) j + ( ∂ ( − y ) ∂ x − ∂ ( x ) ∂ y ) k = 0 i − 0 j + 0 k = 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} ∇ × F = ( ∂ y ∂ F 3 − ∂ z ∂ F 2 ) i − ( ∂ x ∂ F 3 − ∂ z ∂ F 1 ) j + ( ∂ x ∂ F 2 − ∂ y ∂ F 1 ) k = ( ∂ y ∂ ( z ) − ∂ z ∂ ( − y ) ) i − ( ∂ x ∂ ( z ) − ∂ z ∂ ( x ) ) j + ( ∂ x ∂ ( − y ) − ∂ y ∂ ( x ) ) k = 0 i − 0 j + 0 k = 0 Therefore the vector field F = x i − y j + z k \mathbf{F}=x\mathbf{i}-y\mathbf{j}+z\mathbf{k} F = x i − y j + z k
A force of 2.5 N 2.5 \mathrm{~N} 2.5 N 37.5 × 10 − 2 N m 37.5 \times 10^{-2} \mathrm{Nm} 37.5 × 1 0 − 2 Nm
Angle between r r r F F F θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘
τ = ∣ F ∣ ∣ r ∣ sin θ = 2.5 N × 0.15 m × sin 90 ∘ = 37.5 × 10 − 2 Nm \tau=|F||r|\sin\theta=2.5\text{N}\times0.15\text{m}\times\sin90^\circ=37.5\times10^{-2}\text{Nm} τ = ∣ F ∣∣ r ∣ sin θ = 2.5 N × 0.15 m × sin 9 0 ∘ = 37.5 × 1 0 − 2 Nm Direction: anticlockwise
Solution
Solution