Homework_2.tex

\documentclass{article}
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\title{MATH202 Homework 2}
\author{Dryden Bryson}
\date{\today}

\begin{document}

\maketitle
\newpage


\section*{Q1.}
To find the shortest distance between our point $S$ and our line $L_{1}$, we apply the following formula:
$$
\frac{\left\vert \vec{PS} \times  \vec{v} \right\vert }{|\vec{v}|}
$$
First let us form $\vec{PS}$: $$\vec{PS}=\vec{S}-\vec{P}=\langle 3,12,-5 \rangle-\langle -1,2,3 \rangle  = \langle 4,10,-8 \rangle $$
Then we can compute the cross product $\vec{PS}\times \vec{v}$: 
$$\begin{aligned}
    \langle 4,10,-8 \rangle \times  \langle 2,-4,-6 \rangle = 
    \begin{vmatrix}
    i & j & k\\
    4 & 10 & -8\\
    2 & -4 & -6
    \end{vmatrix}&= i \begin{vmatrix}
     10 & -8 \\
     -4 & -6
    \end{vmatrix}-j \begin{vmatrix}
        4 & -8\\
        2 & -6\\
    \end{vmatrix}+k \begin{vmatrix}
        4 & 10\\
        2 & -4
    \end{vmatrix}
    \\&= i\left[ (10)(-6)-(-8)(-4) \right] - j\left[ (4)(-6)-(-8)(2) \right] + k\left[ (4)(-4)-(10)(2) \right]\\
    &= i(-60-32)-j(-24+16)+k(-16-20)\\
    &= -92i+8j-36k\\
    &= \langle -92, 8, -36 \rangle
\end{aligned}$$
Finally we can compute the magnitudes of $\vec{PS} \times \vec{v}$ and $\vec{v}$:
$$|\vec{PS} \times \vec{v}| = \sqrt{(-92)^2 + 8^2 + (-36)^2} = \sqrt{8464 + 64 + 1296} = \sqrt{9824}$$
$$|\vec{v}| = \sqrt{2^2 + (-4)^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56}$$
Thus we can substitute these values into our distance formula:
$$\frac{\left\vert \vec{PS} \times  \vec{v} \right\vert }{|\vec{v}|}=\frac{\sqrt{9824}}{\sqrt{56}} \approx 13.24494\dots$$
Meaning that we have an exact distance of $\frac{\sqrt{9824}}{\sqrt{56}}$ units, or an approximate distance of $13.24494\dots$ units.
\newpage

\section*{Q2.}
To compute $\vec{y}=\vec{v}\times \vec{w}$, we do the following:
$$\begin{aligned}
    \langle 6,-3,12 \rangle \times \langle -6, 4, 2 \rangle = \begin{vmatrix}
    i & j & k\\
    6 & -3 & 12\\
    -6 & 4 & 2
    \end{vmatrix}&= i \begin{vmatrix}
     -3 & 12 \\
     4 & 2
    \end{vmatrix}-j \begin{vmatrix}
     6 & 12\\
     -6 & 2\\
    \end{vmatrix}+k \begin{vmatrix}
     6 & -3\\
     -6 & 4
    \end{vmatrix}\\
    &= i\left[ (-3)(2)-(12)(4) \right] - j\left[ (6)(2)-(12)(-6) \right] + k\left[ (6)(4)-(-3)(-6) \right]\\
    &= i(-6-48)-j(12+72)+k(24-18)\\
    &= -54i-84j+6k\\
    &= \langle -54, -84, 6 \rangle
\end{aligned}$$
Thus we have that $\vec{v}\times \vec{w}=\vec{y}=\langle -54, -84, 6 \rangle$. To compute the volume of the parallelipiped, we can use the following formula:
$$V=|\vec{y}\cdot (\vec{v}\times \vec{w})|$$Since we have that $\vec{y}=\vec{v}\times \vec{w}$ we can perform a substitution and it suffices to find the absolute value of the dot product of $\vec{y}$ with itself:
$$V=|\vec{y}\cdot \vec{y}|=|(-54)^2+(-84)^2+6^2|=|2916+7056+36|=|10008|=10008$$
Thus we have that the volume of the parallelipiped is $10008$ units.

\newpage

\section*{Q3.}
First we will find the parametric equation of our line that passes through the point $P(-1,2,3)$ with the direction $\vec{v}=\langle -4,6,2 \rangle$:
$$(x,y,z)=(-1-4t, 2+6t, 3+2t)$$
Then we can substitute the equation of the line into the equation of the plane, $-x+3y- 2z=6$, we have:
$$\begin{aligned}
    -(-1-4t)+3(2+6t)-2(3+2t)&=6\\
    1+4t+6+18t-6-4t&=6\\
    1+18t&=6\\
    18t&=5\\
    t&=\frac{5}{18}
\end{aligned}$$
Then we can substitute $t=\frac{5}{18}$ back into the parametric equation of the line to find the point of intersection:
$$\begin{aligned}
    (x,y,z)&=\left(-1-4\left(\frac{5}{18}\right), 2+6\left(\frac{5}{18}\right), 3+2\left(\frac{5}{18}\right)\right)\\
    &=(-1-\frac{20}{18}, 2+\frac{30}{18}, 3+\frac{10}{18})\\
    &=\left(-\frac{38}{18}, \frac{66}{18}, \frac{64}{18}\right)\\
    &=\left(-\frac{19}{9}, \frac{33}{9}, \frac{32}{9}\right)
\end{aligned}$$
Thus the point at which the line $L_{1}$ intersects the plane is $\left(-\frac{19}{9}, \frac{33}{9}, \frac{32}{9}\right)$.
\newpage

\section*{Q4.}
It just so happens that the first given plane $(x-y=1)$ already satisfies the condition of being perpendicular to the plane $(x+y-2z=0)$ since the dot product of their normal vectors is $0$, as seen below:
$$\langle 1,-1,0 \rangle \cdot \langle 1,1,-2 \rangle = (1)(1)+(-1)(1)+(0)(-2)=1-1+0=0$$
And since the defining plane $(x-y=1)$ already satisfies the condition of containing the line of intersection between $x-y=1$ and $y+2z=3$, then the problem is trivially solved with the solution plane being: $$x-y=1$$
\newpage

\section*{Q5.}
The $\arcsin$ function is defined only for values in the range $[-1,1]$. We have that the domain of the function is: $$
-1 \leq x^{2}+y^{2}-\frac{3}{2} \leq 1
$$
We can simplify this inequality as such: $$
\frac{1}{2} \leq x^{2}+y^{2} \leq \frac{5}{2}
$$
Recognizing that $x^{2}+y^{2}=r^{2}$ in polar coordinates, we can rewrite this as: $$
\frac{1}{2} \leq r^{2} \leq \frac{5}{2}=> \sqrt{\frac{1}{2}} \leq r \leq \sqrt{\frac{5}{2}}
$$
Thus we simply need to sketch an anullus with inner radius $\sqrt{\frac{1}{2}}$ and outer radius $\sqrt{\frac{5}{2}}$:

\begin{figure}[h]
    \centering
    \includegraphics[width=0.6\textwidth]{hw2q5}
    \caption{An annulus with inner radius $\sqrt{\frac{1}{2}}$ and outer radius $\sqrt{\frac{5}{2}}$.}
\end{figure}

\newpage

\section*{Q6.}
To find the shortest distance between the lines $L_{1}$ and $L_{2}$, we need to find the magnitutude of the shortest line segment that connects both lines. This line segment will be perpendicular to both lines, thus we can use the cross product of the direction vectors of both lines to find a vector that is perpendicular to both lines. Let us first find the direction vectors of both lines:
\begin{itemize}
    \item $L_{1}: L1: x(t) = 4+2t, y(t) = -5+4t, z(t) = 1+3t $\; thus $\vec{v_{1}} = \langle 2, 4, 3 \rangle$
    \item $L2: x(s) = 2+s, y(s) = -1+3s,
z(s) = 2s$\; thus $\vec{v_{2}} = \langle 1, 3, 2 \rangle$
\end{itemize}
Then we need to compute the cross product of these two direction vectors:
$$\begin{aligned}
    \langle 2,4,3 \rangle \times  \langle 1,3,2 \rangle =
    \begin{vmatrix}
    i & j & k\\
    2 & 4 & 3\\
    1 & 3 & 2
    \end{vmatrix}&= i \begin{vmatrix}
        4 & 3 \\
        3 & 2
    \end{vmatrix}
    -j \begin{vmatrix}
        2 & 3\\
        1 & 2\\
    \end{vmatrix}
    +k \begin{vmatrix}
        2 & 4\\
        1 & 3
    \end{vmatrix}\\
    &= i\left[ (4)(2)-(3)(3) \right] - j\left[ (2)(2)-(3)(1) \right] + k\left[ (2)(3)-(4)(1) \right]\\
    &= i(8-9)-j(4-3)+k(6-4)\\
    &= -1i-1j+2k\\
    &= \langle -1, -1, 2 \rangle 
\end{aligned}$$
We now know that the line segment which has the shortest distance between $L_{1}$ and $L_{2}$ has direction $\langle -1, -1, 2 \rangle$. We then need to find a point which this line passes through.\\\\
To find the position in space that this line pases through, we can pick any point on $L_{1}$ and any point on $L_{2}$ and create a directed line segment between the two, then we project this line segment in the direction of our direction vector $\langle -1, -1, 2 \rangle$. We can pick $t=0$ for $L_{1}$ and $s=0$ for $L_{2}$ to get the points:
\begin{itemize}
    \item $P_{1}=\langle 4,-5,1 \rangle $
    \item $P_{2}=\langle 2,-1,0 \rangle $
\end{itemize}
And thus we can form the directed line segment $\vec{P_{1}P_{2}}$:
$$\vec{P_{1}P_{2}}=\vec{P_{2}}-\vec{P_{1}}=\langle 2,-1,0 \rangle -\langle 4,-5,1 \rangle = \langle -2,4,-1 \rangle $$
Now we need to perform the following projection:
$$\begin{aligned}
    \text{proj}_{\vec{v_{1}}\times \vec{v_{2}}}\vec{P_{1}P_{2}} &= \frac{\langle -2,4,1 \rangle \cdot \langle -1,-1,2 \rangle }{\langle -1,-1,2 \rangle \cdot  \langle -1,-1,2 \rangle}\langle -1,-1,2 \rangle\\
    &= \frac{2-4-2}{1+1+4}\langle -1,-1,2 \rangle \\
    &= \frac{-4}{6}\langle -1,-1,2 \rangle \\
    &= \langle \frac{2}{3}, \frac{2}{3}, -\frac{4}{3} \rangle
\end{aligned}$$
The result of this projection is the directed line segment with the shortest distance between $L_{1}$ and $L_{2}$. To find the magnitude of this line segment, we compute: $$\left\vert \langle \frac{2}{3},\frac{2}{3},-\frac{4}{3} \rangle  \right\vert = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(-\frac{4}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{16}{9}} = \sqrt{\frac{24}{9}} = \sqrt{\frac{8}{3}}=1.63299\dots$$
\newpage

\section*{Q7.}
\subsection*{a)}
We can compute this limit by using polar coordinates, i.e where: 
\begin{itemize}
    \item $x=r \cos \theta$
    \item $y=r \sin \theta$
\end{itemize}
Let us substitute these values into our limit:
\begin{table}[htp]
\centering
\begin{tabular}{ccll}
 $lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}y}{x^{2}+y^{2}}$  & $=$  & $lim_{r\rightarrow 0}\frac{(r\cos\theta)^{2}(r\sin\theta)}{(r\cos\theta)^{2}+(r\sin\theta)^{2}}$  &   \\
   & $=$  & $lim_{r\rightarrow 0}\frac{r^{2}\cos^{2}\theta r\sin\theta}{r^{2}\cos^{2}\theta +r^{2}\sin^{2}\theta}$  &  \text{Distributivity of exponentiation} \\
   & $=$  & $lim_{r\rightarrow 0}\frac{r^{3}\cos^{2}\theta \sin\theta}{r^{2}(\cos^{2}\theta +\sin^{2}\theta)}$  &   \text{Simplification \& factoring}\\
   & $=$  & $lim_{r\rightarrow 0}\frac{r\cancel{^{3}}\cos^{2}\theta \sin\theta}{\cancel{r^{2}}(1)}$  & \text{Known trig identity}   \\
    & $=$  & $lim_{r\rightarrow 0}r\cos^{2}\theta \sin\theta$  &  \text{Simplification}\\
\end{tabular}
\end{table}\\
We can now solve the limit by direction substitution of $r=0$ which gives us:
$$0\cdot \cos^{2}\theta \sin\theta = 0$$
Which means that our limit is $0$ for all values of $\theta$, since $r\rightarrow 0$ dominates.
\subsection*{b)}
We can show that this limit does not exist by the fact that it approaches different values when approached along different paths:
\begin{itemize}
    \item $y=0$: $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}y}{x^{4}+y^{2}}=\lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}(0)}{x^{4}+(0)^{2}}=0$$
    \item $y=x^{2}$: $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}y}{x^{4}+y^{2}}=\lim_{(x,y)\rightarrow(0,0)}\frac{x^{2}(x^{2})}{x^{4}+(x^{2})^{2}}=\lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+x^{4}}=\lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{2x^{4}}=\frac{1}{2}$$
\end{itemize}
Since there exists 2 paths approaching $(0,0)$ with two different limits, we can conclude that the limit does not exist.


\newpage

\section*{Q8.}
We can simply factor the numerator of $f(x,y)$ in order to remove the "hole" in $f(x,y)$ which makes it non-continuous:
$$f(x,y)=\frac{x^{3}-y^{3}}{x-y}=\frac{(x-y)(x^{2}+xy+y^{2})}{x-y} = x^{2}+xy+y^{2}$$
Thuse we can define a new piecewise function $h(x,y)$ that is continuous at all points including:
$$h(x,y)=\begin{cases}
\frac{x^{3}-y^{3}}{x-y} & \text{if } x \neq y \\
x^{2}+xy+y^{2} & \text{if } x = y
\end{cases}$$

\end{document}