Prelim 1 Practice Problems

Textbook

Problem 1.6.10

• My Solution

  First, choose $z_0 = 0$ and $q(u) = u^8 + u^4 + u^2+1$. Flag pole is $1$. The position of the man is $u^2 + 1$. The leash is $u^8 + u^4$. $b_j = 1$. Now, we take $B= \max{1,1,1,1} = 1$. We take a $

  u

  =\rho$ and $\rho$ less than $\min{\frac{1}{(8-2)(1)}, \frac{1}{1}^{1/2}, 1}$. Our minimum must occur when $u < 1/6$.

  For example, let’s try $u = 1/10(1+i)$. $q(1/10(1+i)) =0.999600 + 0.02i$. $

  q(u)

  = 0.9998$.

Problem 1.8.7

• My Solution

  Let $L(t) = f(\gamma (t))$.

  \[[DL(t)] = [Df\begin{pmatrix}t\\t^2\\\vdots\\t^n\end{pmatrix}][D\gamma (t)]\]

  By Theorem 1.8.1, the derivative of a vector-valued function is the derivative of each entry in the matrix is given by its single variable derivative

  \[[D\gamma (t)] = \begin{bmatrix}1\\2t\\3t^2\\ \vdots\\nt^{n-1}\end{bmatrix}.\]

  The derivative of $f$ is the $1\times n$ matrix of partial derivatives $D_i$.

  \[[Df(x, \dots, x_n)] = \begin{bmatrix} D_1\sum_{i=1}^{n-1}x_ix_{i+1}&\dots& D_n\sum_{i=1}^{n-1}x_ix_{i+1} \end{bmatrix}\] \[=\begin{bmatrix} x_2& x_1+x_3& x_2+x_4&\dots& x_{n-2} + x_n& x_{n-1} \end{bmatrix}\]

  Taking the dot product of these two matrices, we find that $[DL(t)]$ is

  \[t^2 + 2t(t+t^3)+3t^2(t^2+t^4)+\dots+(n-1)t^{n-2}(t^{n-2}+t^n)+nt^{n-1}(t^{n-1})\] \[=t^2 + \sum_{i=2}^{n-1}it^{i-1} (t^{i-1}+t^{i+1})+nt^{n-1}(t^{n-1})\]

Problem 1.24

• My Solution

  The set $A_m$ converges only when $\theta = 2\pi$. If so, then $A_m$ would be equal to the identity matrix $I_2$. It has a convergent subsequence for all $\theta$ because it is a sequence on the compact set of the unit cube in $\R^3$.

Problem 1.29

Problem 1.35

2015 Fall Final

2018 Fall Take Home Prelim

• My Solution

  Let $f$ be the reflection in the $xw$-plane and $g$ be the reflection in the $yw$-plane.

  \[f=\begin{bmatrix}1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{bmatrix} g=\begin{bmatrix} -1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{bmatrix}\] \[g\circ f = \begin{bmatrix} -1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{bmatrix}\]

  To project this into $\R^3$, we multiply it by a $4\times 5$ identity matrix.

  \[\begin{bmatrix} 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0 \end{bmatrix} \begin{bmatrix} -1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{bmatrix} = \begin{bmatrix} -1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ \end{bmatrix}\]

  By multiplying it by a $5\times 4$ identity matrix, we get the final projection.

  \[[T]=\begin{bmatrix} -1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ \end{bmatrix} \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0 \end{bmatrix}\] \[[T] = \begin{bmatrix} -1&0&0&0\\ 0&-1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix}\]

• My Solution

  \[[Df(t)] = \begin{bmatrix}-\sin t\\\cos t\\1\end{bmatrix}\] \[L(t) =[Df\begin{pmatrix}1/\sqrt{2}\\ 1/\sqrt{2}\\ \pi / 4\end{pmatrix}] = \begin{bmatrix} -\sin (\pi/4)t\\ \cos (\pi/4)t\\ t \end{bmatrix}+ \begin{pmatrix}1/\sqrt{2}\\ 1/\sqrt{2}\\ \pi / 4\end{pmatrix}\] \[L(t)= \begin{bmatrix} -(1/\sqrt{2})t\\ (1/\sqrt{2})t\\ t \end{bmatrix}+ \begin{pmatrix}1/\sqrt{2}\\ 1/\sqrt{2}\\ \pi / 4\end{pmatrix}\] \[L(t) = \begin{pmatrix} -\frac{1}{\sqrt{2}}t+ \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}t+ \frac{1}{\sqrt{2}}\\ t+\frac{\pi}{4} \end{pmatrix}\]

• My Solution

  The directional derivative is given by the equation

  \[\lim_{t\to 0}\frac{1}{t}f\begin{pmatrix}\begin{bmatrix}2&-1\\3&5\end{bmatrix} +t\begin{bmatrix}a&b\\c&d\end{bmatrix}\end{pmatrix} - f\begin{pmatrix}\begin{bmatrix}2&-1\\3&5\end{bmatrix}\end{pmatrix}\] \[=\lim_{t\to0 }\frac{1}{t}\begin{pmatrix} \begin{bmatrix} 2+at&-1+bt\\ 3+ct&5+dt \end{bmatrix} \begin{bmatrix} 2+at&3+ct\\ -1+bt&5+dt \end{bmatrix}- \begin{bmatrix}5&1\\1&34\end{bmatrix} \end{pmatrix}\] \[= \lim_{t\to 0}\frac{1}{t} \begin{bmatrix} 4at-2bt+a^2t^2+b^2t^2& ct+3at+5bt-dt+act^2+dbt^2\\ ct+3at+5bt-dt+act^2+dbt^2& 10dt+6ct+d^2t^2+c^2t^2\end{bmatrix}\] \[=\begin{bmatrix} 4a-2b&3a+5b+c-d\\ 3a+5b+c-d&6c+10d \end{bmatrix}\]

  The directional derivative is linear because each variable of $B$ only appears once for each entry in the matrix.

Worked Solutions

http://pi.math.cornell.edu/~sjamaar/classes/2230/problems/2015-final-solution.pdf

For 1.6.10, from Romin: