# Calculus/Multivariable and differential calculus:Exercises

 ← Partial differential equations Calculus Extensions → Multivariable and differential calculus:Exercises

## Parametric Equations

1. Find parametric equations describing the line segment from P(0,0) to Q(7,17).

x=7t and y=17t, where 0 ≤ t ≤ 1

2. Find parametric equations describing the line segment from ${\displaystyle P(x_{1},y_{1})}$ to ${\displaystyle Q(x_{2},y_{2})}$.

${\displaystyle x=x_{1}+(x_{2}-x_{1})t{\mbox{ and }}y=y_{1}+(y_{2}-y_{1})t,{\mbox{ where }}0\leq t\leq 1}$

3. Find parametric equations describing the ellipse centered at the origin with major axis of length 6 along the x-axis and the minor axis of length 3 along the y-axis, generated clockwise.

${\displaystyle x=3\cos(-t),\ y=1.5\sin(-t)}$

## Polar Coordinates

20. Convert the equation into Cartesian coordinates: ${\displaystyle r=\sin(\theta )\sec ^{2}(\theta ).}$

Making the substitutions ${\displaystyle \sin(\theta )={\frac {y}{r}}}$ and ${\displaystyle \sec(\theta )={\frac {r}{x}}}$ gives ${\displaystyle r={\frac {y}{r}}\left({\frac {r}{x}}\right)^{2}}$ ${\displaystyle \iff r={\frac {ry}{x^{2}}}}$ ${\displaystyle \iff y=x^{2}}$

21. Find an equation of the line y=mx+b in polar coordinates.

Making the substitutions ${\displaystyle x=r\cos(\theta )}$ and ${\displaystyle y=r\sin(\theta )}$ gives ${\displaystyle r\sin(\theta )=mr\cos(\theta )+b}$ ${\displaystyle \iff (\sin(\theta )-m\cos(\theta ))r=b}$ ${\displaystyle \iff r={\frac {b}{\sin(\theta )-m\cos(\theta )}}}$

Sketch the following polar curves without using a computer.

22. ${\displaystyle r=2-2\sin(\theta )}$
23. ${\displaystyle r^{2}=4\cos(\theta )}$
24. ${\displaystyle r=2\sin(5\theta )}$

Sketch the following sets of points.

25. ${\displaystyle \{(r,\theta ):\theta =2\pi /3\}}$
26. ${\displaystyle \{(r,\theta ):|\theta |\leq \pi /3{\mbox{ and }}|r|<3\}}$

## Calculus in Polar Coordinates

Find points where the following curves have vertical or horizontal tangents.

40. ${\displaystyle r=4\cos(\theta )}$

Horizontal tangents occur at points where ${\displaystyle {\frac {dy}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\sin(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0}$.

Vertical tangents occur at points where ${\displaystyle {\frac {dx}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\cos(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0}$.

${\displaystyle {\frac {dr}{d\theta }}=-4\sin(\theta )}$

The condition for a horizontal tangent gives: ${\displaystyle (-4\sin(\theta ))\sin(\theta )+(4\cos(\theta ))\cos(\theta )=0}$ ${\displaystyle \iff \cos ^{2}(\theta )-\sin ^{2}(\theta )=0}$ ${\displaystyle \iff \cos(2\theta )=0}$ ${\displaystyle \iff 2\theta =\pm \pi /2}$ ${\displaystyle \iff \theta =\pm \pi /4}$

Horizontal tangents occur at ${\displaystyle (r,\theta )=(2{\sqrt {2}},\pi /4),(2{\sqrt {2}},-\pi /4)}$ which correspond to the Cartesian points ${\displaystyle (2,2)}$ and ${\displaystyle (2,-2)}$.

The condition for a vertical tangent gives: ${\displaystyle (-4\sin(\theta ))\cos(\theta )-(4\cos(\theta ))\sin(\theta )=0}$ ${\displaystyle \iff 2\sin(\theta )\cos(\theta )=0}$ ${\displaystyle \iff \sin(2\theta )=0}$ ${\displaystyle \iff 2\theta =0,\pi }$ ${\displaystyle \iff \theta =0,\pi /2}$

Vertical tangents occur at ${\displaystyle (r,\theta )=(0,\pi /2),(4,0)}$ which correspond to the Cartesian points ${\displaystyle (0,0)}$ and ${\displaystyle (4,0)}$.

Horizontal tangents at (2,2) and (2,-2); vertical tangents at (0,0) and (4,0)

41. ${\displaystyle r=2+2\sin(\theta )}$

Horizontal tangents occur at points where ${\displaystyle {\frac {dy}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\sin(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0}$.

Vertical tangents occur at points where ${\displaystyle {\frac {dx}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\cos(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0}$.

${\displaystyle {\frac {dr}{d\theta }}=2\cos(\theta )}$

The condition for a horizontal tangent gives: ${\displaystyle (2\cos(\theta ))\sin(\theta )+(2+2\sin(\theta ))\cos(\theta )=0}$ ${\displaystyle \iff 4\sin(\theta )\cos(\theta )+2\cos(\theta )=0}$ ${\displaystyle \iff \cos(\theta )(2\sin(\theta )+1)=0}$ ${\displaystyle \iff \cos(\theta )=0\;{\text{or}}\;\sin(\theta )=-{\frac {1}{2}}}$ ${\displaystyle \iff \theta =-5\pi /6,-\pi /2,-\pi /6,\pi /2}$

Horizontal tangents occur at ${\displaystyle (r,\theta )=(1,-5\pi /6),(0,-\pi /2),(1,-\pi /6),(4,\pi /2)}$ which correspond to the Cartesian points ${\displaystyle (-{\sqrt {3}}/2,-1/2)}$, ${\displaystyle (0,0)}$, ${\displaystyle ({\sqrt {3}}/2,-1/2)}$, and ${\displaystyle (0,4)}$. Point ${\displaystyle (0,0)}$ corresponds to a vertical cusp however and should be excluded leaving ${\displaystyle (-{\sqrt {3}}/2,-1/2)}$, ${\displaystyle ({\sqrt {3}}/2,-1/2)}$, and ${\displaystyle (0,4)}$.

The condition for a vertical tangent gives: ${\displaystyle (2\cos(\theta ))\cos(\theta )-(2+2\sin(\theta ))\sin(\theta )=0}$ ${\displaystyle \iff 2\cos ^{2}(\theta )-2\sin ^{2}(\theta )-2\sin(\theta )=0}$ ${\displaystyle \iff 2(1-\sin ^{2}(\theta ))-2\sin ^{2}(\theta )-2\sin(\theta )=0}$ ${\displaystyle \iff -4\sin ^{2}(\theta )-2\sin(\theta )+2=0}$ ${\displaystyle \iff 2\sin ^{2}(\theta )+\sin(\theta )-1=0}$ ${\displaystyle \iff (2\sin(\theta )-1)(\sin(\theta )+1)=0}$ ${\displaystyle \iff \sin(\theta )=-1,1/2}$ ${\displaystyle \iff \theta =-\pi /2,\pi /6,5\pi /6}$

Vertical tangents occur at ${\displaystyle (r,\theta )=(0,-\pi /2),(3,\pi /6),(3,5\pi /6)}$ which correspond to the Cartesian points ${\displaystyle (0,0)}$, ${\displaystyle (3{\sqrt {3}}/2,3/2)}$, and ${\displaystyle (-3{\sqrt {3}}/2,3/2)}$.

Horizontal tangents at (r,θ) = (4,π/2), (1,7π/6) and (1,-π/6); vertical tangents at (r,θ) = (3,π/6), (3,5π/6), and (0,3π/2)

Sketch the region and find its area.

42. The region inside the limaçon ${\displaystyle 2+\cos(\theta )}$

Given an infinitesimal wedge with angle ${\displaystyle d\theta }$ and radius ${\displaystyle r}$, the area is ${\displaystyle {\frac {1}{2}}r^{2}d\theta }$. The total area is therefore ${\displaystyle A=\int _{\theta =0}^{2\pi }{\frac {1}{2}}r^{2}d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }{\frac {1}{2}}(2+\cos(\theta ))^{2}d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{2}}\cos ^{2}(\theta ))d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{4}}(\cos(2\theta )+1))d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }({\frac {9}{4}}+2\cos(\theta )+{\frac {1}{4}}\cos(2\theta ))d\theta }$ ${\displaystyle =({\frac {9}{4}}\theta +2\sin(\theta )+{\frac {1}{8}}\sin(2\theta )){\bigg |}_{\theta =0}^{2\pi }}$ ${\displaystyle ={\frac {9}{2}}\pi }$.

9π/2
43. The region inside the petals of the rose ${\displaystyle 4\cos(2\theta )}$ and outside the circle ${\displaystyle r=2}$

There are 4 petals, as seen in the image below. The area of just one of the petals needs to be computed and the multiplied by ${\displaystyle 4}$.

It is first necessary to compute the angular limits of one of the petals. The petals start and end at points where ${\displaystyle 4\cos(2\theta )=2}$ ${\displaystyle \iff \cos(2\theta )={\frac {1}{2}}}$ ${\displaystyle \Leftarrow 2\theta =\pm {\frac {\pi }{3}}}$ ${\displaystyle \iff \theta =\pm {\frac {\pi }{6}}}$. The bounds on one of the petals are ${\displaystyle \left[-{\frac {\pi }{6}},+{\frac {\pi }{6}}\right]}$.

Given an annular wedge with angle ${\displaystyle d\theta }$, inner radius ${\displaystyle r_{1}}$, and an outer radius of ${\displaystyle r_{2}}$, the area is ${\displaystyle {\frac {1}{2}}(r_{2}^{2}-r_{1}^{2})d\theta }$. The total area of all 4 petals is therefore ${\displaystyle A=4\int _{\theta =-\pi /6}^{\pi /6}{\frac {1}{2}}((4\cos(2\theta ))^{2}-2^{2})d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(8\cos ^{2}(2\theta )-2)d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(4(\cos(4\theta )+1)-2)d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(2+4\cos(4\theta ))d\theta }$ ${\displaystyle =4(2\theta +\sin(4\theta )){\bigg |}_{\theta =-\pi /6}^{\pi /6}}$ ${\displaystyle =4({\frac {2\pi }{3}}+{\sqrt {3}})}$ ${\displaystyle ={\frac {8\pi }{3}}+4{\sqrt {3}}}$.

${\displaystyle 8\pi /3+4{\sqrt {3}}}$

## Vectors and Dot Product

60. Find an equation of the sphere with center (1,2,0) passing through the point (3,4,5)

The general equation for a sphere is ${\displaystyle (x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=r^{2}}$ where ${\displaystyle (x_{0},y_{0},z_{0})}$ is the location of the sphere's center and ${\displaystyle r}$ is the sphere's radius.

It is already known that the sphere's center is ${\displaystyle (x_{0},y_{0},z_{0})=(1,2,0)}$. The sphere's radius is the distance between (1,2,0) and (3,4,5) which is ${\displaystyle r={\sqrt {(3-1)^{2}+(4-2)^{2}+(5-0)^{2}}}={\sqrt {4+4+25}}={\sqrt {33}}}$.

Therefore the sphere's equation is: ${\displaystyle (x-1)^{2}+(y-2)^{2}+z^{2}=33}$.

61. Sketch the plane passing through the points (2,0,0), (0,3,0), and (0,0,4)
62. Find the value of ${\displaystyle |\mathbf {u} +3\mathbf {v} |}$ if ${\displaystyle \mathbf {u} =\langle 1,3,0\rangle }$ and ${\displaystyle \mathbf {v} =\langle 3,0,2\rangle }$

${\displaystyle \mathbf {u} +3\mathbf {v} =\langle 1,3,0\rangle +3\langle 3,0,2\rangle }$ ${\displaystyle =\langle 1,3,0\rangle +\langle 9,0,6\rangle }$ ${\displaystyle =\langle 10,3,6\rangle }$.

Therefore: ${\displaystyle |\mathbf {u} +3\mathbf {v} |={\sqrt {10^{2}+3^{2}+6^{2}}}={\sqrt {100+9+36}}={\sqrt {145}}}$.

${\displaystyle {\sqrt {145}}}$

63. Find all unit vectors parallel to ${\displaystyle \langle 1,2,3\rangle }$

The length of ${\displaystyle \langle 1,2,3\rangle }$ is ${\displaystyle {\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}}$. Therefore ${\displaystyle {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ is a unit vector that points in the same direction as ${\displaystyle \langle 1,2,3\rangle }$, and ${\displaystyle -{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ is a unit vector that points in the opposite direction as ${\displaystyle \langle 1,2,3\rangle }$.

${\displaystyle \pm {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ are the unit vectors that are parallel to ${\displaystyle \langle 1,2,3\rangle }$.

64. Prove one of the distributive properties for vectors in ${\displaystyle \mathbb {R} ^{3}}$: ${\displaystyle c(\mathbf {u} +\mathbf {v} )=c\mathbf {u} +c\mathbf {v} }$

${\displaystyle c(\mathbf {u} +\mathbf {v} )=c(\langle u_{1},u_{2},u_{3}\rangle +\langle v_{1},v_{2},v_{3}\rangle )}$ ${\displaystyle =c\langle u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}\rangle }$ ${\displaystyle =\langle c(u_{1}+v_{1}),c(u_{2}+v_{2}),c(u_{3}+v_{3})\rangle }$ ${\displaystyle =\langle cu_{1}+cv_{1},cu_{2}+cv_{2},cu_{3}+cv_{3}\rangle }$ ${\displaystyle =\langle cu_{1},cu_{2},cu_{3}\rangle +\langle cv_{1},cv_{2},cv_{3}\rangle }$ ${\displaystyle =c\langle u_{1},u_{2},u_{3}\rangle +c\langle v_{1},v_{2},v_{3}\rangle }$ ${\displaystyle =c\mathbf {u} +c\mathbf {v} }$.

65. Find all unit vectors orthogonal to ${\displaystyle 3\mathbf {i} +4\mathbf {j} }$ in ${\displaystyle \mathbb {R} ^{2}}$

Rotating ${\displaystyle \mathbf {u} =\langle 3,4\rangle }$ ${\displaystyle 90^{\circ }}$ counterclockwise gives ${\displaystyle \mathbf {u} ^{\perp }=\langle -4,3\rangle }$. ${\displaystyle \mathbf {u} ^{\perp }}$ is orthogonal to ${\displaystyle \mathbf {u} }$, and the normalization of ${\displaystyle \mathbf {u} ^{\perp }}$ and its negative are the only unit vectors that are orthogonal to ${\displaystyle \mathbf {u} }$.

The magnitude of ${\displaystyle \mathbf {u} ^{\perp }}$ is ${\displaystyle {\sqrt {(-4)^{2}+3^{2}}}={\sqrt {16+9}}={\sqrt {25}}=5}$ so the only unit vectors that are orthogonal to ${\displaystyle \mathbf {u} }$ are ${\displaystyle \pm \left\langle {\frac {-4}{5}},{\frac {3}{5}}\right\rangle }$.

66. Find all unit vectors orthogonal to ${\displaystyle 3\mathbf {i} +4\mathbf {j} }$ in ${\displaystyle \mathbb {R} ^{3}}$

All vectors ${\displaystyle \mathbf {u} =\langle u_{1},u_{2},u_{3}\rangle }$ that are orthogonal to ${\displaystyle \langle 3,4,0\rangle }$ must satisfy ${\displaystyle \langle 3,4,0\rangle \cdot \mathbf {u} =0}$ ${\displaystyle \iff 3u_{1}+4u_{2}=0}$ ${\displaystyle \iff u_{2}=-{\frac {3}{4}}u_{1}}$.

The set of possible values of ${\displaystyle \mathbf {u} }$ is ${\displaystyle \left\{\left\langle x,-{\frac {3}{4}}x,z\right\rangle {\bigg |}x,z\in \mathbb {R} \right\}}$. The restriction that ${\displaystyle |\mathbf {u} |=1}$ becomes ${\displaystyle {\sqrt {u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}}=1}$ ${\displaystyle \iff x^{2}+{\frac {9}{16}}x^{2}+z^{2}=1}$ ${\displaystyle \iff {\frac {25}{16}}x^{2}+z^{2}=1}$ ${\displaystyle \iff \left({\frac {x}{4/5}}\right)^{2}+z^{2}=1}$.

The set of possible ${\displaystyle x}$ and ${\displaystyle z}$ is an ellipse with radii ${\displaystyle 4/5}$ and ${\displaystyle 1}$. One possible parameterization of ${\displaystyle x}$ and ${\displaystyle z}$ is ${\displaystyle x={\frac {4}{5}}\cos(\theta )}$ and ${\displaystyle z=\sin(\theta )}$ where ${\displaystyle \theta \in \mathbb {R} }$. This parameterization yields ${\displaystyle \mathbf {u} =\left\langle {\frac {4}{5}}\cos(\theta ),-{\frac {3}{5}}\cos(\theta ),\sin(\theta )\right\rangle }$ where ${\displaystyle \theta \in \mathbb {R} }$ as the complete set of unit vectors that are orthogonal to ${\displaystyle \langle 3,4,0\rangle }$.

Re-parameterizing by letting ${\displaystyle c=\cos(\theta )}$ gives the set ${\displaystyle \left\langle {\frac {4}{5}}c,-{\frac {3}{5}}c,\pm {\sqrt {1-c^{2}}}\right\rangle ,\ c\in [-1,1]}$

67. Find all unit vectors that make an angle of ${\displaystyle \pi /3}$ with the vector ${\displaystyle \langle 1,2\rangle }$

The angle that ${\displaystyle \langle 1,2\rangle }$ makes with the x-axis is ${\displaystyle \arctan(2)}$ counterclockwise.

Making a both a clockwise and a counterclockwise rotation of ${\displaystyle \pi /3}$ gives ${\displaystyle \langle \cos(\arctan(2)\mp \pi /3),\sin(\arctan(2)\mp \pi /3)\rangle }$ ${\displaystyle =\langle \cos(\arctan(2))\cos(\mp \pi /3)-\sin(\arctan(2))\sin(\mp \pi /3),\sin(\arctan(2))\cos(\mp \pi /3)+\cos(\arctan(2))\sin(\mp \pi /3)\rangle }$ ${\displaystyle =\left\langle {\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}-{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}},{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}+{\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}}\right\rangle }$ ${\displaystyle =\left\langle {\frac {1}{2{\sqrt {5}}}}\pm {\frac {\sqrt {3}}{\sqrt {5}}},{\frac {1}{\sqrt {5}}}\mp {\frac {\sqrt {3}}{2{\sqrt {5}}}}\right\rangle }$ ${\displaystyle ={\frac {1}{2{\sqrt {5}}}}\left\langle 1\pm 2{\sqrt {3}},2\mp {\sqrt {3}}\right\rangle }$ ${\displaystyle ={\frac {\sqrt {5}}{10}}\left\langle 1\pm 2{\sqrt {3}},\ 2\mp {\sqrt {3}}\right\rangle }$

## Cross Product

Find ${\displaystyle \mathbf {u} \times \mathbf {v} }$ and ${\displaystyle \mathbf {v} \times \mathbf {u} }$

80. ${\displaystyle \mathbf {u} =\langle -4,1,1\rangle }$ and ${\displaystyle \mathbf {v} =\langle 0,1,-1\rangle }$

${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (1)(-1)-(1)(1),(1)(0)-(-4)(-1),(-4)(1)-(1)(0)\rangle =\langle -2,-4,-4\rangle }$

${\displaystyle \mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle 2,4,4\rangle }$

81. ${\displaystyle \mathbf {u} =\langle 1,2,-1\rangle }$ and ${\displaystyle \mathbf {v} =\langle 3,-4,6\rangle }$

${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (2)(6)-(-1)(-4),(-1)(3)-(1)(6),(1)(-4)-(2)(3)\rangle =\langle 8,-9,-10\rangle }$

${\displaystyle \mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle -8,9,10\rangle }$

Find the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$.

82. ${\displaystyle \mathbf {u} =\langle -3,0,2\rangle }$ and ${\displaystyle \mathbf {v} =\langle 1,1,1\rangle }$

The cross product of vectors ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ is a vector with length ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ where ${\displaystyle \theta }$ is the angle between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$. ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ is the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$, so this area is found by computing ${\displaystyle |\mathbf {u} \times \mathbf {v} |}$.

${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (0)(1)-(2)(1),(2)(1)-(-3)(1),(-3)(1)-(0)(1)\rangle =\langle -2,5,-3\rangle }$

${\displaystyle A=|\mathbf {u} \times \mathbf {v} |={\sqrt {(-2)^{2}+5^{2}+(-3)^{2}}}={\sqrt {4+25+9}}={\sqrt {38}}}$

83. ${\displaystyle \mathbf {u} =\langle 8,2,-3\rangle }$ and ${\displaystyle \mathbf {v} =\langle 2,4,-4\rangle }$

The cross product of vectors ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ is a vector with length ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ where ${\displaystyle \theta }$ is the angle between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$. ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ is the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$, so this area is found by computing ${\displaystyle |\mathbf {u} \times \mathbf {v} |}$.

${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (2)(-4)-(-3)(4),(-3)(2)-(8)(-4),(8)(4)-(2)(2)\rangle =\langle 4,26,28\rangle }$

${\displaystyle A=|\mathbf {u} \times \mathbf {v} |={\sqrt {4^{2}+26^{2}+28^{2}}}={\sqrt {2^{2}(2^{2}+13^{2}+14^{2})}}=2{\sqrt {369}}=6{\sqrt {41}}}$

84. Find all vectors that satisfy the equation ${\displaystyle \langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle }$

The cross product is orthogonal to both multiplicand vectors. ${\displaystyle \langle 0,1,1\rangle }$ should be orthogonal to both ${\displaystyle \langle 1,1,1\rangle }$ and ${\displaystyle \mathbf {u} }$. However, ${\displaystyle \langle 1,1,1\rangle \cdot \langle 0,1,1\rangle =2\neq 0}$ so ${\displaystyle \langle 1,1,1\rangle }$ and ${\displaystyle \langle 0,1,1\rangle }$ are not orthogonal. The equation ${\displaystyle \langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle }$ is never true, and therefore the set of vectors ${\displaystyle \mathbf {u} }$ that satisfy the equation is ${\displaystyle \emptyset =}$"None".

85. Find the volume of the parallelepiped with edges given by position vectors ${\displaystyle \langle 5,0,0\rangle }$, ${\displaystyle \langle 1,4,0\rangle }$, and ${\displaystyle \langle 2,2,7\rangle }$

The volume of a parallelepiped with edges defined by the vectors ${\displaystyle \mathbf {u} }$, ${\displaystyle \mathbf {v} }$, and ${\displaystyle \mathbf {w} }$ is the absolute value of the scalar triple product: ${\displaystyle |(\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w} |}$.

${\displaystyle |(\langle 5,0,0\rangle \times \langle 1,4,0\rangle )\cdot \langle 2,2,7\rangle |}$ ${\displaystyle =|\langle (0)(0)-(0)(4),(0)(1)-(5)(0),(5)(4)-(0)(1)\rangle \cdot \langle 2,2,7\rangle |}$ ${\displaystyle =|\langle 0,0,20\rangle \cdot \langle 2,2,7\rangle |}$ ${\displaystyle =140}$

86. A wrench has a pivot at the origin and extends along the x-axis. Find the magnitude and the direction of the torque at the pivot when the force ${\displaystyle \mathbf {F} =\langle 1,2,3\rangle }$ is applied to the wrench n units away from the origin.

The moment arm is ${\displaystyle \mathbf {r} =\langle n,0,0\rangle }$, so the torque applied is ${\displaystyle \tau =\mathbf {r} \times \mathbf {F} =\langle n,0,0\rangle \times \langle 1,2,3\rangle =\langle (0)(3)-(0)(2),(0)(1)-(n)(3),(n)(2)-(0)(1)\rangle =\langle 0,-3n,2n\rangle }$

The magnitude of the torque is ${\displaystyle |\tau |={\sqrt {0^{2}+(-3n)^{2}+(2n)^{2}}}=n{\sqrt {13}}}$. The torque's direction is ${\displaystyle {\hat {\tau }}={\frac {\tau }{|\tau |}}=\langle 0,-3/{\sqrt {13}},2/{\sqrt {13}}\rangle }$.

Prove the following identities or show them false by giving a counterexample.

87. ${\displaystyle \mathbf {u} \times (\mathbf {u} \times \mathbf {v} )=\mathbf {0} }$

False: ${\displaystyle \mathbf {i} \times (\mathbf {i} \times \mathbf {j} )=-\mathbf {j} }$

88. ${\displaystyle \mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )=\mathbf {w} \cdot (\mathbf {u} \times \mathbf {v} )}$

True: Once expressed in component form, both sides evaluate to ${\displaystyle u_{1}v_{2}w_{3}-u_{1}v_{3}w_{2}+u_{2}v_{3}w_{1}-u_{2}v_{1}w_{3}+u_{3}v_{1}w_{2}-u_{3}v_{2}w_{1}}$

89. ${\displaystyle (\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=2(\mathbf {u} \times \mathbf {v} )}$

True: ${\displaystyle (\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=(\mathbf {u} -\mathbf {v} )\times \mathbf {u} +(\mathbf {u} -\mathbf {v} )\times \mathbf {v} }$${\displaystyle =\mathbf {u} \times \mathbf {u} -\mathbf {v} \times \mathbf {u} +\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {v} }$${\displaystyle =\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {u} }$${\displaystyle =\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {v} }$${\displaystyle =2(\mathbf {u} \times \mathbf {v} )}$

## Calculus of Vector-Valued Functions

100. Differentiate ${\displaystyle \mathbf {r} (t)=\langle te^{-t},t\ln t,t\cos(t)\rangle }$.

${\displaystyle \langle e^{-t}-te^{-t},\ln(t)+1,\cos(t)-t\sin(t)\rangle }$

101. Find a tangent vector for the curve ${\displaystyle \mathbf {r} (t)=\langle 2t^{4},6t^{3/2},10/t\rangle }$ at the point ${\displaystyle t=1}$.

${\displaystyle {\dot {\mathbf {r} }}(t)=\langle 8t^{3},9t^{1/2},-10/t^{2}\rangle }$ so a possible a tangent vector at ${\displaystyle t=1}$ is ${\displaystyle {\dot {\mathbf {r} }}(1)=\langle 8,9,-10\rangle }$

102. Find the unit tangent vector for the curve ${\displaystyle \mathbf {r} (t)=\langle t,2,2/t\rangle ,\ t\neq 0}$.

${\displaystyle {\dot {\mathbf {r} }}(t)=\langle 1,0,-2/t^{2}\rangle }$

${\displaystyle |{\dot {\mathbf {r} }}(t)|={\sqrt {1^{2}+0^{2}+(-2/t^{2})^{2}}}={\sqrt {1+4/t^{4}}}={\frac {\sqrt {t^{4}+4}}{t^{2}}}}$ so the unit tangent vector is ${\displaystyle \mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle t^{2},0,-2\rangle }{\sqrt {t^{4}+4}}}}$

103. Find the unit tangent vector for the curve ${\displaystyle \mathbf {r} (t)=\langle \sin(t),\cos(t),e^{-t}\rangle ,\ t\in [0,\pi ]}$ at the point ${\displaystyle t=0}$.

${\displaystyle {\dot {\mathbf {r} }}(t)=\langle \cos(t),-\sin(t),-e^{-t}\rangle }$

${\displaystyle |{\dot {\mathbf {r} }}(t)|={\sqrt {(\cos(t))^{2}+(-\sin(t))^{2}+(-e^{-t})^{2}}}={\sqrt {1+e^{-2t}}}}$ so the unit tangent vector is ${\displaystyle \mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle \cos(t),-\sin(t),-e^{-t}\rangle }{\sqrt {1+e^{-2t}}}}}$

At ${\displaystyle t=0}$: ${\displaystyle \mathbf {T} (0)={\frac {\langle 1,0,-1\rangle }{\sqrt {2}}}}$

104. Find ${\displaystyle \mathbf {r} }$ if ${\displaystyle \mathbf {r} '(t)=\langle {\sqrt {t}},\cos(\pi t),4/t\rangle }$ and ${\displaystyle \mathbf {r} (1)=\langle 2,3,4\rangle }$.

For an arbitrary ${\displaystyle t>0}$ the position ${\displaystyle \mathbf {r} (t)}$ can be computed by the integral ${\displaystyle \mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du}$.

${\displaystyle \mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du}$ ${\displaystyle =\langle 2,3,4\rangle +\int _{u=1}^{t}\langle {\sqrt {u}},\cos(\pi u),4/u\rangle du}$ ${\displaystyle =\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}u^{3/2},{\frac {\sin(\pi u)}{\pi }},4\ln(u)\right\rangle {\Bigg |}_{u=1}^{t}}$ ${\displaystyle =\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}(t^{3/2}-1),{\frac {\sin(\pi t)}{\pi }},4\ln(t)\right\rangle }$ ${\displaystyle =\left\langle {\frac {2t^{3/2}+4}{3}},{\frac {\sin(\pi t)}{\pi }}+3,4\ln(t)+4\right\rangle }$

105. Evaluate ${\displaystyle \displaystyle \int _{0}^{\ln 2}(e^{-t}\mathbf {i} +2e^{2t}\mathbf {j} -4e^{t}\mathbf {k} )dt}$

${\displaystyle \int _{t=0}^{\ln 2}\langle e^{-t},2e^{2t},-4e^{t}\rangle dt}$ ${\displaystyle =\langle -e^{-t},e^{2t},-4e^{t}\rangle {\bigg |}_{t=0}^{\ln 2}}$ ${\displaystyle =\langle -1/2-(-1),4-1,-8-(-4)\rangle }$ ${\displaystyle =\langle 1/2,3,-4\rangle }$

## Motion in Space

120. Find velocity, speed, and acceleration of an object if the position is given by ${\displaystyle \mathbf {r} (t)=\langle 3\sin(t),5\cos(t),4\sin(t)\rangle }$.

${\displaystyle \mathbf {v} =\langle 3\cos(t),-5\sin(t),4\cos(t)\rangle }$, ${\displaystyle |\mathbf {v} |=5}$, ${\displaystyle \mathbf {a} =\langle -3\sin(t),-5\cos(t),-4\sin(t)\rangle }$

121. Find the velocity and the position vectors for ${\displaystyle t\geq 0}$ if the acceleration is given by ${\displaystyle \mathbf {a} (t)=\langle e^{-t},1\rangle ,\ \mathbf {v} (0)=\langle 1,0\rangle ,\ \mathbf {r} (0)=\langle 0,0\rangle }$.

${\displaystyle \mathbf {v} (t)=\mathbf {v} (0)+\int _{u=0}^{t}\mathbf {a} (u)du}$ ${\displaystyle =\langle 1,0\rangle +\int _{u=0}^{t}\langle e^{-u},1\rangle du}$ ${\displaystyle =\langle 1,0\rangle +\langle -e^{-u},u\rangle {\bigg |}_{u=0}^{t}}$ ${\displaystyle =\langle 1,0\rangle +\langle -e^{-t}-(-1),t-0\rangle }$ ${\displaystyle =\langle 2-e^{-t},t\rangle }$

${\displaystyle \mathbf {r} (t)=\mathbf {r} (0)+\int _{u=0}^{t}\mathbf {v} (u)du}$ ${\displaystyle =\langle 0,0\rangle +\int _{u=0}^{t}\langle 2-e^{-u},u\rangle du}$ ${\displaystyle =\langle 2u+e^{-u},{\frac {1}{2}}u^{2}\rangle {\bigg |}_{u=0}^{t}}$ ${\displaystyle =\langle (2t+e^{-t})-1,{\frac {1}{2}}t^{2}-0\rangle }$ ${\displaystyle =\langle e^{-t}+2t-1,t^{2}/2\rangle }$

## Length of Curves

Find the length of the following curves.

140. ${\displaystyle \mathbf {r} (t)=\langle 4\cos(3t),4\sin(3t)\rangle ,\ t\in [0,2\pi /3].}$

For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately ${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle -12\sin(3t),12\cos(3t)\rangle |dt=12dt}$.

The total length is therefore: ${\displaystyle L=\int _{t=0}^{2\pi /3}ds=\int _{t=0}^{2\pi /3}12dt=12t{\Bigg |}_{t=0}^{2\pi /3}=8\pi }$

141. ${\displaystyle \mathbf {r} (t)=\langle 2+3t,1-4t,3t-4\rangle ,\ t\in [1,6].}$

For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately ${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 3,-4,3\rangle |dt={\sqrt {9+16+9}}\cdot dt={\sqrt {34}}\cdot dt}$.

The total length is therefore: ${\displaystyle L=\int _{t=1}^{6}ds=\int _{t=1}^{6}{\sqrt {34}}\cdot dt={\sqrt {34}}\cdot t{\Bigg |}_{t=1}^{6}=5{\sqrt {34}}}$

## Parametrization and Normal Vectors

142. Find a description of the curve that uses arc length as a parameter: ${\displaystyle \mathbf {r} (t)=\langle t^{2},2t^{2},4t^{2}\rangle \ t\in [1,4].}$

For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately ${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 2t,4t,8t\rangle |dt=|t|{\sqrt {4+16+64}}\cdot dt={\sqrt {84}}\cdot tdt=2{\sqrt {21}}\cdot tdt}$

Given an upper bound of ${\displaystyle u\in [1,4]}$, the arc length swept out from ${\displaystyle t=1}$ to ${\displaystyle t=u}$ is: ${\displaystyle s=\int _{t=1}^{u}ds=\int _{t=1}^{u}2{\sqrt {21}}\cdot tdt={\sqrt {21}}\cdot t^{2}{\bigg |}_{t=1}^{u}={\sqrt {21}}(u^{2}-1)}$

The arc length spans a range from ${\displaystyle 0}$ to ${\displaystyle 15{\sqrt {21}}}$. For an arc length of ${\displaystyle s\in [0,15{\sqrt {21}}]}$, the upper bound on ${\displaystyle t}$ that generates an arc length of ${\displaystyle s}$ is ${\displaystyle u={\sqrt {{\frac {s}{\sqrt {21}}}+1}}}$, and the point at which this upper bound occurs is: ${\displaystyle \displaystyle \mathbf {r} (s)=\left({\frac {s}{\sqrt {21}}}+1\right)\langle 1,2,4\rangle }$

143. Find the unit tangent vector T and the principal unit normal vector N for the curve ${\displaystyle \mathbf {r} (t)=\langle t^{2},t\rangle .}$ Check that TN=0.

A tangent vector is ${\displaystyle {\frac {d\mathbf {r} }{dt}}=\langle 2t,1\rangle }$. Normalizing this vector to get the unit tangent vector gives: ${\displaystyle \mathbf {T} (t)={\frac {d\mathbf {r} }{ds}}={\frac {d\mathbf {r} /dt}{|d\mathbf {r} /dt|}}=\displaystyle {\frac {\langle 2t,1\rangle }{\sqrt {4t^{2}+1}}}}$

A vector that has the direction of the principal unit normal vector is ${\displaystyle {\frac {d\mathbf {T} }{dt}}=\left\langle {\frac {2}{\sqrt {4t^{2}+1}}}-{\frac {8t^{2}}{(4t^{2}+1)^{3/2}}},{\frac {-4t}{(4t^{2}+1)^{3/2}}}\right\rangle }$ ${\displaystyle ={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2(4t^{2}+1)-8t^{2},-4t\rangle }$ ${\displaystyle ={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2,-4t\rangle }$ ${\displaystyle ={\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }$

Normalizing ${\displaystyle {\frac {d\mathbf {T} }{dt}}}$ gives the principal unit normal vector: ${\displaystyle \mathbf {N} (t)}$ ${\displaystyle ={\frac {d\mathbf {T} /dt}{|d\mathbf {T} /dt|}}}$ ${\displaystyle ={\frac {{\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }{{\frac {2}{(4t^{2}+1)^{3/2}}}{\sqrt {4t^{2}+1}}}}}$ ${\displaystyle =\displaystyle {\frac {\langle 1,-2t\rangle }{\sqrt {4t^{2}+1}}}}$

## Equations of Lines And Planes

160. Find an equation of a plane passing through points ${\displaystyle (1,1,2),\ (1,2,2),\ (-1,0,1).}$

Let ${\displaystyle P}$ denote a plane that contains points ${\displaystyle (1,1,2)}$, ${\displaystyle (1,2,2)}$, and ${\displaystyle (-1,0,1)}$. Let ${\displaystyle \mathbf {N} }$ denote an arbitrary vector that is orthogonal to ${\displaystyle P}$, and ${\displaystyle \mathbf {r} _{0}}$ denote the position vector of an arbitrary point contained by ${\displaystyle P}$. A point at position vector ${\displaystyle \mathbf {r} }$ is contained by ${\displaystyle P}$ if and only if the displacement from ${\displaystyle \mathbf {r} _{0}}$ is orthogonal to ${\displaystyle \mathbf {N} }$. This yields the equation ${\displaystyle \mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0}$ ${\displaystyle \iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$.

The displacement from ${\displaystyle (1,1,2)}$ to ${\displaystyle (1,2,2)}$, which is ${\displaystyle \langle 0,1,0\rangle }$, and the displacement from ${\displaystyle (1,1,2)}$ to ${\displaystyle (-1,0,1)}$, which is ${\displaystyle \langle -2,-1,-1\rangle }$, are both contained by ${\displaystyle P}$ so the cross product of these two displacements forms a candidate ${\displaystyle \mathbf {N} }$: ${\displaystyle \mathbf {N} =\langle 0,1,0\rangle \times \langle -2,-1,-1\rangle }$ ${\displaystyle =\langle (1)(-1)-(0)(-1),(0)(-2)-(0)(-1),(0)(-1)-(1)(-2)\rangle }$ ${\displaystyle =\langle -1,0,2\rangle }$

Any of ${\displaystyle (1,1,2)}$, ${\displaystyle (1,2,2)}$, and ${\displaystyle (-1,0,1)}$ is a candidate ${\displaystyle \mathbf {r} _{0}}$. Let ${\displaystyle \mathbf {r} _{0}=\langle 1,1,2\rangle }$

The equation becomes ${\displaystyle \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$ ${\displaystyle \iff \langle -1,0,2\rangle \cdot \langle x,y,z\rangle -\langle -1,0,2\rangle \cdot \langle 1,1,2\rangle }$ ${\displaystyle \iff -x+2z-3=0}$ ${\displaystyle \iff x-2z+3=0}$

161. Find an equation of a plane parallel to the plane 2xy+z=1 passing through the point (0,2,-2)

Let ${\displaystyle P}$ denote a plane that is parallel to the plane ${\displaystyle 2x-y+z=1}$ and contains the point ${\displaystyle (0,2,-2)}$. Let ${\displaystyle \mathbf {N} }$ denote an arbitrary vector that is orthogonal to ${\displaystyle P}$, and ${\displaystyle \mathbf {r} _{0}}$ denote the position vector of an arbitrary point contained by ${\displaystyle P}$. A point at position vector ${\displaystyle \mathbf {r} }$ is contained by ${\displaystyle P}$ if and only if the displacement from ${\displaystyle \mathbf {r} _{0}}$ is orthogonal to ${\displaystyle \mathbf {N} }$. This yields the equation ${\displaystyle \mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0}$ ${\displaystyle \iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$.

Any vector ${\displaystyle \mathbf {N} }$ that is orthogonal to ${\displaystyle P}$ is also orthogonal to ${\displaystyle 2x-y+z=1}$ and vice versa. Since ${\displaystyle 2x-y+z=1}$ ${\displaystyle \iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle =1}$, the coefficient vector ${\displaystyle \langle 2,-1,1\rangle }$ is orthogonal to ${\displaystyle 2x-y+z=1}$, so a candidate ${\displaystyle \mathbf {N} }$ is ${\displaystyle \mathbf {N} =\langle 2,-1,1\rangle }$.

Since point ${\displaystyle (0,2,-2)}$ is contained by ${\displaystyle P}$, let ${\displaystyle \mathbf {r} _{0}=\langle 0,2,-2\rangle }$.

The equation becomes ${\displaystyle \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$ ${\displaystyle \iff \langle 2,-1,1\rangle \cdot \langle x,y,z\rangle -\langle 2,-1,1\rangle \cdot \langle 0,2,-2\rangle =0}$ ${\displaystyle \iff 2x-y+z+4=0}$

162. Find an equation of the line perpendicular to the plane x+y+2z=4 passing through the point (5,5,5).

Let ${\displaystyle P}$ denote an arbitrary plane. Let ${\displaystyle \mathbf {N} }$ denote an arbitrary vector that is orthogonal to ${\displaystyle P}$, and ${\displaystyle \mathbf {r} _{0}}$ denote the position vector of an arbitrary point contained by ${\displaystyle P}$. A point at position vector ${\displaystyle \mathbf {r} }$ is contained by ${\displaystyle P}$ if and only if the displacement from ${\displaystyle \mathbf {r} _{0}}$ is orthogonal to ${\displaystyle \mathbf {N} }$. This yields the equation ${\displaystyle \mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0}$ ${\displaystyle \iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$. Therefore the equation that defines ${\displaystyle P}$ is ${\displaystyle \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$.

The equation ${\displaystyle x+y+2z=4}$ is equivalent to ${\displaystyle \langle 1,1,2\rangle \cdot \langle x,y,z\rangle -4=0}$. This implies that the coefficient vector ${\displaystyle \mathbf {N} =\langle 1,1,2\rangle }$ is orthogonal to the plane defined by ${\displaystyle x+y+2z=4}$. A line that passes through point ${\displaystyle (5,5,5)}$ and is parallel to ${\displaystyle \mathbf {N} =\langle 1,1,2\rangle }$ is parameterized by: ${\displaystyle \mathbf {r} (t)=\langle 5,5,5\rangle +t\langle 1,1,2\rangle =\langle 5+t,5+t,5+2t\rangle }$

163. Find an equation of the line where planes x+2yz=1 and x+y+z=1 intersect.

To describe the line that forms the intersection between ${\displaystyle x+2y-z=1}$ and ${\displaystyle x+y+z=1}$, both ${\displaystyle x}$ and ${\displaystyle y}$ will be expressed as functions of ${\displaystyle z}$ (it can also be the case that ${\displaystyle y}$ and ${\displaystyle z}$ are functions of ${\displaystyle x}$, etc.). Let ${\displaystyle E_{1}}$ denote equation ${\displaystyle x+2y-z=1}$ and ${\displaystyle E_{2}}$ denote equation ${\displaystyle x+y+z=1}$.

To find ${\displaystyle x}$ as a function of ${\displaystyle z}$ subtract 2 times ${\displaystyle E_{2}}$ from ${\displaystyle E_{1}}$ to get ${\displaystyle E_{1}-2E_{2}:(x+2y-z)-2(x+y+z)=1-2(1)}$ ${\displaystyle \iff -x-3z=-1}$ ${\displaystyle \iff x=1-3z}$

To find ${\displaystyle y}$ as a function of ${\displaystyle z}$ subtract ${\displaystyle E_{2}}$ from ${\displaystyle E_{1}}$ to get ${\displaystyle E_{1}-E_{2}:(x+2y-z)-(x+y+z)=1-1}$ ${\displaystyle \iff y-2z=0}$ ${\displaystyle \iff y=2z}$

Parameterizing ${\displaystyle z}$ with ${\displaystyle t}$ gives the parameterization ${\displaystyle \mathbf {r} (t)=\langle 1-3t,2t,t\rangle }$

164. Find the angle between the planes x+2yz=1 and x+y+z=1.

Let ${\displaystyle P}$ denote an arbitrary plane. Let ${\displaystyle \mathbf {N} }$ denote an arbitrary vector that is orthogonal to ${\displaystyle P}$, and ${\displaystyle \mathbf {r} _{0}}$ denote the position vector of an arbitrary point contained by ${\displaystyle P}$. A point at position vector ${\displaystyle \mathbf {r} }$ is contained by ${\displaystyle P}$ if and only if the displacement from ${\displaystyle \mathbf {r} _{0}}$ is orthogonal to ${\displaystyle \mathbf {N} }$. This yields the equation ${\displaystyle \mathbf {N} \cdot (\mathbf {r} -\mathbf {r} _{0})=0}$ ${\displaystyle \iff \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$. Therefore the equation that defines ${\displaystyle P}$ is ${\displaystyle \mathbf {N} \cdot \mathbf {r} -\mathbf {N} \cdot \mathbf {r} _{0}=0}$.

Let ${\displaystyle P_{1}}$ be the plane described by ${\displaystyle x+2y-z=1}$ and ${\displaystyle P_{2}}$ be the plane described by ${\displaystyle x+y+z=1}$

Since ${\displaystyle x+2y-z=1\iff \langle 1,2,-1\rangle \cdot \langle x,y,z\rangle -1=0}$, the coefficient vector ${\displaystyle \mathbf {N} _{1}=\langle 1,2,-1\rangle }$ is orthogonal to ${\displaystyle P_{1}}$.

Since ${\displaystyle x+y+z=1\iff \langle 1,1,1\rangle \cdot \langle x,y,z\rangle -1=0}$, the coefficient vector ${\displaystyle \mathbf {N} _{2}=\langle 1,1,1\rangle }$ is orthogonal to ${\displaystyle P_{2}}$.

The angle ${\displaystyle \theta }$ between ${\displaystyle P_{1}}$ and ${\displaystyle P_{2}}$ is equivalent to the angle between ${\displaystyle \mathbf {N} _{1}}$ and ${\displaystyle \mathbf {N} _{2}}$: ${\displaystyle \theta =\arccos \left({\frac {\mathbf {N} _{1}\cdot \mathbf {N} _{2}}{|\mathbf {N} _{1}||\mathbf {N} _{2}|}}\right)}$ ${\displaystyle =\arccos \left({\frac {\langle 1,2,-1\rangle \cdot \langle 1,1,1\rangle }{|\langle 1,2,-1\rangle ||\langle 1,1,1\rangle |}}\right)}$ ${\displaystyle =\arccos \left({\frac {1+2-1}{{\sqrt {1+4+1}}{\sqrt {1+1+1}}}}\right)}$ ${\displaystyle =\arccos {\frac {2}{\sqrt {18}}}}$

165. Find the distance from the point (3,4,5) to the plane x+y+z=1.

Given a unit length vector ${\displaystyle {\hat {\mathbf {n} }}}$, consider an axis ${\displaystyle w}$ oriented in the direction of ${\displaystyle {\hat {\mathbf {n} }}}$. The "${\displaystyle w}$ coordinate" is determined by orthogonally projecting points onto the ${\displaystyle w}$ axis. Given a position vector ${\displaystyle \mathbf {r} }$, the expression ${\displaystyle {\hat {\mathbf {n} }}\cdot \mathbf {r} }$ computes the ${\displaystyle w}$ coordinate.

The equation ${\displaystyle x+y+z=1}$ is equivalent to ${\displaystyle \langle 1,1,1\rangle \cdot \langle x,y,z\rangle =1}$ ${\displaystyle \iff {\frac {\langle 1,1,1\rangle }{|\langle 1,1,1\rangle |}}\cdot \langle x,y,z\rangle ={\frac {1}{|\langle 1,1,1\rangle |}}}$ ${\displaystyle \iff \langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle \cdot \langle x,y,z\rangle =1/{\sqrt {3}}}$

Letting ${\displaystyle {\hat {\mathbf {n} }}=\langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle }$, the plane ${\displaystyle x+y+z=1}$ consists of all points whose ${\displaystyle w}$ coordinate is ${\displaystyle 1/{\sqrt {3}}}$. The ${\displaystyle w}$ coordinate of ${\displaystyle (3,4,5)}$ is ${\displaystyle \langle 1/{\sqrt {3}},1/{\sqrt {3}},1/{\sqrt {3}}\rangle \cdot \langle 3,4,5\rangle ={\frac {12}{\sqrt {3}}}=4{\sqrt {3}}}$.

The distance between the plane ${\displaystyle x+y+z=1}$ and the point ${\displaystyle (3,4,5)}$ along the ${\displaystyle w}$ axis is ${\displaystyle \left|4{\sqrt {3}}-{\frac {1}{\sqrt {3}}}\right|=\left|{\frac {12{\sqrt {3}}-{\sqrt {3}}}{3}}\right|}$ ${\displaystyle ={\frac {11}{3}}{\sqrt {3}}}$

The distance ${\displaystyle {\frac {11}{3}}{\sqrt {3}}}$ is the distance between the point and plane along a direction that is orthogonal to the plane, and is hence the shortest distance.

## Limits And Continuity

Evaluate the following limits.

180. ${\displaystyle \displaystyle \lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}}$

${\displaystyle \lim _{(x,y)\rightarrow (1,-2)}{\frac {y^{2}+2xy}{y+2x}}=\lim _{(x,y)\rightarrow (1,-2)}{\frac {y(y+2x)}{y+2x}}}$ ${\displaystyle =\lim _{(x,y)\rightarrow (1,-2)}y}$ ${\displaystyle =-2}$

181. ${\displaystyle \displaystyle \lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}}$

${\displaystyle \lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{x+y-9}}=\lim _{(x,y)\rightarrow (4,5)}{\frac {{\sqrt {x+y}}-3}{({\sqrt {x+y}}-3)({\sqrt {x+y}}+3)}}}$ ${\displaystyle =\lim _{(x,y)\rightarrow (4,5)}{\frac {1}{{\sqrt {x+y}}+3}}}$ ${\displaystyle ={\frac {1}{{\sqrt {4+5}}+3}}=1/6}$

At what points is the function f continuous?

182. ${\displaystyle f(x,y)=\ln |x-y|}$

${\displaystyle \{(x,y)\mid x\neq y\}}$

183. ${\displaystyle f(x,y)=\displaystyle {\frac {\ln(x^{2}+y^{2})}{x-y+1}}}$

All points (x,y) except for (0,0) and the line y=x+1

Use the two-path test to show that the following limits do not exist. (A path does not have to be a straight line.)

184. ${\displaystyle \displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {4xy}{3x^{2}+y^{2}}}}$

The limit is 1 along the line y=x, and −1 along the line y=−x

185. ${\displaystyle \displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {y}{\sqrt {x^{2}-y^{2}}}}}$

The limit is 0 along the line y=0, and ${\displaystyle 1/{\sqrt {3}}}$ along the line x=2y

186. ${\displaystyle \displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {x^{3}-y^{2}}{x^{3}+y^{2}}}}$

The limit is 1 along the line y=0, and −1 along the line x=0

187. ${\displaystyle \displaystyle \lim _{(x,y)\rightarrow (0,0)}{\frac {x^{2}y^{2}+y^{6}}{x^{3}}}}$

The limit is 0 along any line of the form y=mx, and 2 along the parabola ${\displaystyle x=y^{2}}$

## Partial Derivatives

200. Find ${\displaystyle \partial z/\partial x}$ if ${\displaystyle \displaystyle z(x,y)={\frac {1}{\ln(xy)}}}$

By repeatedly applying the chain rule: ${\displaystyle {\frac {\partial z}{\partial x}}={\frac {\partial }{\partial x}}\left({\frac {1}{\ln(xy)}}\right)}$ ${\displaystyle ={\frac {-1}{(\ln(xy))^{2}}}{\frac {\partial }{\partial x}}(\ln(xy))}$ ${\displaystyle ={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}{\frac {\partial }{\partial x}}(xy)}$ ${\displaystyle ={\frac {-1}{(\ln(xy))^{2}}}{\frac {1}{xy}}y}$ ${\displaystyle ={\frac {-1}{x(\ln(xy))^{2}}}}$

201. Find all three partial derivatives of the function ${\displaystyle \displaystyle f(x,y,z)=xe^{y^{2}+z}}$

${\displaystyle \displaystyle f_{x}=e^{y^{2}+z},\ f_{y}=2xye^{y^{2}+z},\ f_{z}=f.}$

Find the four second partial derivatives of the following functions.

202. ${\displaystyle f(x,y)=\cos(xy)}$

The first derivatives are: ${\displaystyle f_{x}=-y\sin(xy)}$ and ${\displaystyle f_{y}=-x\sin(xy)}$

The second derivatives are: ${\displaystyle f_{xx}=-y^{2}\cos(xy),\ f_{yy}=-x^{2}\cos(xy),\ f_{xy}=f_{yx}=-\sin(xy)-xy\cos(xy)}$

203. ${\displaystyle f(x,y)=xe^{y}}$

The first derivatives are: ${\displaystyle f_{x}=e^{y}}$ and ${\displaystyle f_{y}=xe^{y}}$

The second derivatives are: ${\displaystyle f_{xx}=0,\ f_{yy}=xe^{y},\ f_{xy}=f_{yx}=e^{y}}$

## Chain Rule

Find ${\displaystyle df/dt.}$

220. ${\displaystyle f(x,y)=x^{2}y-xy^{3},\ x(t)=t^{2},\ y(t)=t^{-2}}$

The single derivatives are: ${\displaystyle {\frac {\partial f}{\partial x}}=2xy-y^{3}}$; ${\displaystyle {\frac {\partial f}{\partial y}}=x^{2}-3xy^{2}}$; ${\displaystyle {\frac {dx}{dt}}=2t}$; and ${\displaystyle {\frac {dy}{dt}}=-2t^{-3}}$

The chain rule gives: ${\displaystyle {\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}}$ ${\displaystyle =(2xy-y^{3})(2t)+(x^{2}-3xy^{2})(-2t^{-3})}$ ${\displaystyle =(2-t^{-6})(2t)+(t^{4}-3t^{-2})(-2t^{-3})}$ ${\displaystyle =(4t-2t^{-5})+(-2t+6t^{-5})}$ ${\displaystyle =2t+4t^{-5}}$

221. ${\displaystyle f(x,y)={\sqrt {x^{2}+y^{2}}},\ x(t)=\cos(2t),\ y(t)=\sin(2t)}$

The single derivatives are: ${\displaystyle {\frac {\partial f}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}}}}}$; ${\displaystyle {\frac {\partial f}{\partial y}}={\frac {y}{\sqrt {x^{2}+y^{2}}}}}$; ${\displaystyle {\frac {dx}{dt}}=-2\sin(2t)}$; and ${\displaystyle {\frac {dy}{dt}}=2\cos(2t)}$

The chain rule gives: ${\displaystyle {\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}}$ ${\displaystyle ={\frac {x}{\sqrt {x^{2}+y^{2}}}}(-2\sin(2t))+{\frac {y}{\sqrt {x^{2}+y^{2}}}}(2\cos(2t))}$ ${\displaystyle ={\frac {\cos(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(-2\sin(2t))+{\frac {\sin(2t)}{\sqrt {\cos ^{2}(2t)+\sin ^{2}(2t)}}}(2\cos(2t))}$ ${\displaystyle =-2\cos(2t)\sin(2t)+2\sin(2t)\cos(2t)}$ ${\displaystyle =0}$

222. ${\displaystyle \displaystyle f(x,y,z)={\frac {x-y}{y+z}},\ x(t)=t,\ \displaystyle y(t)=2t,\ z(t)=3t}$

The single derivatives are: ${\displaystyle {\frac {\partial f}{\partial x}}={\frac {1}{y+z}}}$; ${\displaystyle {\frac {\partial f}{\partial y}}={\frac {(-1)(y+z)-(x-y)(1)}{(y+z)^{2}}}={\frac {-(x+z)}{(y+z)^{2}}}}$; ${\displaystyle {\frac {\partial f}{\partial z}}={\frac {y-x}{(y+z)^{2}}}}$; ${\displaystyle {\frac {dx}{dt}}=1}$; ${\displaystyle {\frac {dy}{dt}}=2}$; and ${\displaystyle {\frac {dz}{dt}}=3}$

The chain rule gives: ${\displaystyle {\frac {df}{dt}}={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}+{\frac {\partial f}{\partial z}}{\frac {dz}{dt}}}$ ${\displaystyle ={\frac {1}{y+z}}(1)+{\frac {-(x+z)}{(y+z)^{2}}}(2)+{\frac {y-x}{(y+z)^{2}}}(3)}$ ${\displaystyle ={\frac {(y+z)-2(x+z)+3(y-x)}{(y+z)^{2}}}}$ ${\displaystyle ={\frac {-5x+4y-z}{(y+z)^{2}}}}$ ${\displaystyle ={\frac {-5t+8t-3t}{(5t)^{2}}}}$ ${\displaystyle =0}$

Find ${\displaystyle f_{s},\ f_{t}.}$

223. ${\displaystyle f(x,y)=\sin(x)\cos(2y),\ x=s+t,\ y=s-t}$

The single derivatives are: ${\displaystyle {\frac {\partial f}{\partial x}}=\cos(x)\cos(2y)}$; ${\displaystyle {\frac {\partial f}{\partial y}}=-2\sin(x)\sin(2y)}$; ${\displaystyle {\frac {\partial x}{\partial s}}=1}$; ${\displaystyle {\frac {\partial x}{\partial t}}=1}$; ${\displaystyle {\frac {\partial y}{\partial s}}=1}$; and ${\displaystyle {\frac {\partial y}{\partial t}}=-1}$

The chain rule gives: ${\displaystyle {\frac {\partial f}{\partial s}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial s}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial s}}}$ ${\displaystyle =(\cos(x)\cos(2y))(1)+(-2\sin(x)\sin(2y))(1)}$ ${\displaystyle =\cos(s+t)\cos(2(s-t))-2\sin(s+t)\sin(2(s-t))}$

and ${\displaystyle {\frac {\partial f}{\partial t}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial t}}}$ ${\displaystyle =(\cos(x)\cos(2y))(1)+(-2\sin(x)\sin(2y))(-1)}$ ${\displaystyle =\cos(s+t)\cos(2(s-t))+2\sin(s+t)\sin(2(s-t))}$

Therefore: ${\displaystyle f_{s}=\cos(s+t)\cos(2s-2t)-2\sin(s+t)\sin(2s-2t)}$ and ${\displaystyle f_{t}=\cos(s+t)\cos(2s-2t)+2\sin(s+t)\sin(2s-2t)}$

224. ${\displaystyle \displaystyle f(x,y,z)={\frac {x-z}{y+z}},\ x(t)=s+t,\ y(t)=st,\ z(t)=s-t}$

The single derivatives are: ${\displaystyle {\frac {\partial f}{\partial x}}={\frac {1}{y+z}}}$; ${\displaystyle {\frac {\partial f}{\partial y}}={\frac {z-x}{(y+z)^{2}}}}$; ${\displaystyle {\frac {\partial f}{\partial z}}={\frac {(-1)(y+z)-(x-z)(1)}{(y+z)^{2}}}={\frac {-(x+y)}{(y+z)^{2}}}}$; ${\displaystyle {\frac {\partial x}{\partial s}}=1}$; ${\displaystyle {\frac {\partial x}{\partial t}}=1}$; ${\displaystyle {\frac {\partial y}{\partial s}}=t}$; ${\displaystyle {\frac {\partial y}{\partial t}}=s}$; ${\displaystyle {\frac {\partial z}{\partial s}}=1}$; and ${\displaystyle {\frac {\partial z}{\partial t}}=-1}$

The chain rule gives: ${\displaystyle {\frac {\partial f}{\partial s}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial s}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial s}}+{\frac {\partial f}{\partial z}}{\frac {\partial z}{\partial s}}}$ ${\displaystyle ={\frac {1}{y+z}}(1)+{\frac {z-x}{(y+z)^{2}}}(t)+{\frac {-(x+y)}{(y+z)^{2}}}(1)}$ ${\displaystyle ={\frac {(y+z)+t(z-x)-(x+y)}{(y+z)^{2}}}}$ ${\displaystyle ={\frac {(st+s-t)+t(-2t)-(s+t+st)}{(st+s-t)^{2}}}}$ ${\displaystyle ={\frac {-2t-2t^{2}}{(st+s-t)^{2}}}}$ ${\displaystyle ={\frac {-2t(t+1)}{(st+s-t)^{2}}}}$

and ${\displaystyle {\frac {\partial f}{\partial t}}={\frac {\partial f}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {\partial y}{\partial t}}+{\frac {\partial f}{\partial z}}{\frac {\partial z}{\partial t}}}$ ${\displaystyle ={\frac {1}{y+z}}(1)+{\frac {z-x}{(y+z)^{2}}}(s)+{\frac {-(x+y)}{(y+z)^{2}}}(-1)}$ ${\displaystyle ={\frac {(y+z)+s(z-x)+(x+y)}{(y+z)^{2}}}}$ ${\displaystyle ={\frac {(st+s-t)+s(-2t)+(s+t+st)}{(st+s-t)^{2}}}}$ ${\displaystyle ={\frac {2s}{(st+s-t)^{2}}}}$

Therefore: ${\displaystyle f_{s}={\frac {-2t(t+1)}{(st+s-t)^{2}}}}$ and ${\displaystyle f_{t}={\frac {2s}{(st+s-t)^{2}}}}$

225. The volume of a pyramid with a square base is ${\displaystyle V={\frac {1}{3}}x^{2}h}$, where x is the side of the square base and h is the height of the pyramid. Suppose that ${\displaystyle \displaystyle x(t)={\frac {t}{t+1}}}$ and ${\displaystyle \displaystyle h(t)={\frac {1}{t+1}}}$ for ${\displaystyle t\geq 0.}$ Find ${\displaystyle V'(t).}$

The single derivatives are: ${\displaystyle {\frac {\partial V}{\partial x}}={\frac {2}{3}}xh}$; ${\displaystyle {\frac {\partial V}{\partial h}}={\frac {1}{3}}x^{2}}$; ${\displaystyle x'(t)={\frac {(1)(t+1)-t(1)}{(t+1)^{2}}}={\frac {1}{(t+1)^{2}}}}$; and ${\displaystyle h'(t)={\frac {-1}{(t+1)^{2}}}}$

The chain rule gives: ${\displaystyle V'(t)={\frac {\partial V}{\partial x}}x'(t)+{\frac {\partial V}{\partial h}}h'(t)}$ ${\displaystyle ={\frac {2}{3}}xh{\frac {1}{(t+1)^{2}}}+{\frac {1}{3}}x^{2}{\frac {-1}{(t+1)^{2}}}}$ ${\displaystyle ={\frac {2t}{3(t+1)^{4}}}-{\frac {t^{2}}{3(t+1)^{4}}}}$ ${\displaystyle ={\frac {2t-t^{2}}{3(t+1)^{4}}}}$

## Tangent Planes

Find an equation of a plane tangent to the given surface at the given point(s).

240. ${\displaystyle xy\sin(z)=1,\ (1,2,\pi /6),\ (-1,-2,5\pi /6).}$

Start with a point ${\displaystyle (x_{0},y_{0},z_{0})}$ that is on the surface. Perturbing the ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ coordinates by infinitesimal amounts ${\displaystyle dx}$, ${\displaystyle dy}$, and ${\displaystyle dz}$ respectively, changes the value of ${\displaystyle xy\sin(z)}$ by the infinitesimal amount ${\displaystyle d(xy\sin(z))=(dx)y_{0}\sin(z_{0})+x_{0}(dy)\sin(z_{0})+x_{0}y_{0}d(\sin(z))=(y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz}$, and the value of ${\displaystyle 1}$ by ${\displaystyle d(1)=0}$. To remain in the surface it must be the case that ${\displaystyle d(xy\sin(z))=d(1)}$ ${\displaystyle \iff (y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0}$.

To linearly extrapolate the condition ${\displaystyle (y_{0}\sin(z_{0}))dx+(x_{0}\sin(z_{0}))dy+(x_{0}y_{0}\cos(z_{0}))dz=0}$ to a tangent plane at ${\displaystyle (x_{0},y_{0},z_{0})}$, replace the infinitesimal perturbations ${\displaystyle dx}$, ${\displaystyle dy}$, and ${\displaystyle dz}$ with large perturbations ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$, and ${\displaystyle \Delta z}$ to get ${\displaystyle (y_{0}\sin(z_{0}))\Delta x+(x_{0}\sin(z_{0}))\Delta y+(x_{0}y_{0}\cos(z_{0}))\Delta z=0}$. Any point in the tangent plane at ${\displaystyle (x_{0},y_{0},z_{0})}$ can be reached by an appropriate choice of ${\displaystyle \Delta x}$, ${\displaystyle \Delta y}$, and ${\displaystyle \Delta z}$ where ${\displaystyle (y_{0}\sin(z_{0}))\Delta x+(x_{0}\sin(z_{0}))\Delta y+(x_{0}y_{0}\cos(z_{0}))\Delta z=0}$. Any point ${\displaystyle (x,y,z)}$ in the tangent plane at ${\displaystyle (x_{0},y_{0},z_{0})}$ must satisfy ${\displaystyle (y_{0}\sin(z_{0}))(x-x_{0})+(x_{0}\sin(z_{0}))(y-y_{0})+(x_{0}y_{0}\cos(z_{0}))(z-z_{0})=0}$.

The point ${\displaystyle (x_{0},y_{0},z_{0})=(1,2,\pi /6)}$ lies in the surface, and the tangent plane is ${\displaystyle (x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0}$.

The point ${\displaystyle (x_{0},y_{0},z_{0})=(-1,-2,5\pi /6)}$ lies in the surface, and the tangent plane is ${\displaystyle -(x+1)-{\frac {1}{2}}(y+2)-{\sqrt {3}}(z-5\pi /6)=0}$.

The tangent planes are therefore: ${\displaystyle (x-1)+{\frac {1}{2}}(y-2)+{\sqrt {3}}(z-\pi /6)=0,\ (x+1)+{\frac {1}{2}}(y+2)+{\sqrt {3}}(z-5\pi /6)=0}$

241. ${\displaystyle z=x^{2}e^{x-y},\ (2,2,4),\ (-1,-1,1).}$

Start with a point ${\displaystyle (x_{0},y_{0},z_{0})}$ that is on the surface. Perturbing the ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ coordinates by infinitesimal amounts ${\displaystyle dx}$, ${\displaystyle dy}$, and ${\displaystyle dz}$ respectively, changes the value of ${\displaystyle z}$ by the infinitesimal amount ${\displaystyle dz}$, and the value of ${\displaystyle x^{2}e^{x-y}}$ by ${\displaystyle d(x^{2}e^{x-y})=d(x^{2})e^{x_{0}-y_{0}}+x_{0}^{2}d(e^{x-y})}$ ${\displaystyle =(2x_{0}dx)e^{x_{0}-y_{0}}+x_{0}^{2}(e^{x_{0}-y_{0}}d(x-y))}$ ${\displaystyle =(2x_{0}e^{x_{0}-y_{0}})dx+x_{0}^{2}(e^{x_{0}-y_{0}}(dx-dy))}$ ${\displaystyle =(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy}$. To remain in the surface it must be the case that ${\displaystyle dz=d(x^{2}e^{x-y})}$ ${\displaystyle \iff dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy}$.

To linearly extrapolate the condition ${\displaystyle dz=(2x_{0}+x_{0}^{2})e^{x_{0}-y_{0}}dx-x_{0}^{2}e^{x_{0}-y_{0}}dy}$ to a tangent plane at