\documentclass[12pt]{article} %\pagestyle{empty} \setlength{\topmargin}{-1in} \setlength{\oddsidemargin}{-.3in} \setlength{\textheight}{9in} \setlength{\textwidth}{7.2in} \usepackage{amssymb} \newcommand{\adj}{\mathrm{Adj}} \newcommand{\mbf}{\mathbf} \newcommand{\R}{\mathbb{R}} \newcommand{\ra}{\rightarrow} \newcommand{\dotprod}{\bullet} \newcommand{\dist}{\mathrm{dist}} \newcommand{\answer}{\textbf{Answer}} \begin{document} {\Large Take home I -- Key} \begin{itemize} \item For most problems, there are many methods that can be used. For example, we know several ways to compute curvature. Any correct method you use, as long as your presentation is clear, will be graded as correct. \end{itemize} \begin{enumerate} \item \[ \mbf{r}(t) = (a \cos t, a \sin t, bt) \] I first do some computations: % \begin{eqnarray*} \mbf{v} & = & \mbf{r}' = (-a\sin t, a\cos t, b) \\ \mbf{a} & = & \mbf{r}'' = (-a \cos t, -a \sin t, 0) \\ |\mbf{v}| & = & \sqrt{a^2+b^2} \end{eqnarray*} % \begin{enumerate} \item Is the point moving at constant speed? \answer: we can see that speed ($|\mbf{v}|$) is independent of $t$, so yes, the point is moving at constant speed. (The velocity is non-constant). \item Find $T(t)$, the unit tangent vector. \answer: \[ T(t) = \frac{\mbf{v}}{|\mbf{v}|} = \frac{1}{\sqrt{a^2+b^2}}(-a\sin t, a\cos t, b) \] \item Find $N(t)$, the unit normal vector. \answer: To find $N$, we need to compute $dT/ds$: \begin{eqnarray*} \frac{dT}{ds} & = & \frac{T'}{|\mbf{v}|} = \left(\frac{1}{\sqrt{a^2+b^2}}(-a\cos t, -a\sin t, 0)\right) \left(\frac{1}{\sqrt{a^2+b^2}} \right) \\ & = & \frac{1}{a^2+b^2}(-a\cos t, -a\sin t, 0) \end{eqnarray*} Now, $N$ is the unit vector in this direction, so we need to compute norm of $dT/ds$: \[ \kappa = \left| \frac{dT}{ds} \right| = \frac{a}{a^2+b^2} \] And, now we can find $N$: \[ N = \frac{1}{\kappa}\frac{dT}{ds} = (-\cos t, -\sin t, 0) \] \item Find $B(t)$, the unit bi-normal. \answer: $B=T \times N$, so we just compute this cross product: \[ B = \frac{1}{\sqrt{a^2+b^2}}(b\sin t, -b \cos t, a) \] \item Find $\kappa(t)$, the curvature. \answer: we found this above: \[ \kappa = \frac{a}{a^2+b^2} \] \item Find the acceleration of the curve. Write the acceleration in its tangential and normal components. \answer: $\mbf{a}$ is \[ \mbf{a} = \mbf{r}'' = (-a \cos t, -a \sin t, 0) \] We can find the tangential and normal components of this by computing: \begin{eqnarray*} a_N & = & \mbf{a} \dotprod N = a \\ a_T & = & \mbf{a} \dotprod T = 0 \end{eqnarray*} Therefore we see that \[ \mbf{a} = a N \] This is what we expect because the curve is constant speed. \item Consider the tangent vector and the vector $\mbf{k}=(0,0,1)$. What is the angle between these vectors? Is the angle acute, obtuse or right angled? Does the angle depend on $t$? Explain what all this means geometrically. \answer: We can compute cosine of the angle, notice that $T$ and $\mbf{k}$ are both unit vectors: \[ \cos \theta = \frac{T \dotprod \mbf{k}}{|T||\mbf{k}|} = \frac{b}{\sqrt{a^2+b^2}} \] To get the angle, we take $\arccos$ of the expression. The important thing is that the angle is independent of $t$ and is acute (because $\cos \theta>0$). Thus the angle between $T$ and the $z$-axis is always the same. \item Find the angle between $N(t)$ and $\mbf{k}=(0,0,1)$. Does this angle depend on $t$? Explain what this means geometrically. \answer: We again compute angle: \[ \cos \theta = \frac{N \dotprod \mbf{k}}{|N||\mbf{k}|} = 0 \] This means that the angle is $\pi/2$, and is also independent of $t$. Thus $N$ is always perpendicular with the $z$-axis. \item The plane determined by the point $\mbf{r}(t)$ and spanned by the two vectors $T(t)$ and $N(t)$ is called the \emph{osculating plane} (see 14.5 in book). Notice that it is a different plane for every value of $t$. Find an equation for the osculating plane (this equation will necessarily have a $t$ in it since the plane depends on $t$). \answer: The vector $B$ is normal to the plane, so our data for our plane is the vector $B$ and the point $\mbf{r}(t)$. Since changing $B$ by multiplying by a scalar, won't change the direction (and hence we will still have the same plan) we are free to do this. So let $B'=(b \sin t, -b \cos t, a)$ represent the normal to the osculating plane. So, our data is: \begin{eqnarray*} \mbox{normal vector: } & & B' = (b \sin t, -b \cos t, a) \\ \mbox{point: } && \mbf{r} = (a \cos t, a \sin t, bt) \end{eqnarray*} This gives the equation: \begin{eqnarray*} B' \dotprod ((x,y,z)-\mbf{r}) & = & 0 \\ (b \sin t, -b \cos t, a) \dotprod (x- a \cos t , x- a \sin t , z-bt) & = & 0 \\ (b \sin t)x - (b \cos t)y + az & = & abt \end{eqnarray*} And this is the equation of the plane. In other words, if you plug in something for $t$, you get an equation of a plane in $x,y,z$. \item For curves in $3$-space, the osculating circle is the circle in the osculating plane, tangent to the curve with the same curvature as the curve at the given point. This is analogous to 13.5 in the book, except the circle lives in the osculating plane. Find the radius of the osculating circle at time $t$, and the center of the osculating circle at time $t$. \answer: We know the curvature and hence the radius of the osculating circle: \[ R = \frac{1}{\kappa} = \frac{a^2+b^2}{a} \] We also know that the center of the osculating circle in on the line determined by $N$. Thus, the center of the osculating circle must be a distance of $R$ away from the point $\mbf{r}(t)$. So the center of the osculating circle is \begin{eqnarray*} C & = & \mbf{r}(t)+RN = (a \cos t, a \sin t, bt) + \left(\frac{a^2+b^2}{a}\right)(-\cos t, -\sin t, 0) \\ & = & \left( -\frac{b^2\cos t}{a}, -\frac{b^2\sin t}{a}, bt \right) \end{eqnarray*} \end{enumerate} \item (15 points) For the three points \[ P_1=(1,-2,-1) \qquad P_2=(-1,-8,13) \qquad P_3=(2,1,-8) \] I begin by finding the following vectors: \begin{eqnarray*} v & = & \stackrel{\longrightarrow}{P_1P_2} = (-2, -6, 14) \\ u & = & \stackrel{\longrightarrow}{P_1P_2} = (1,3,-7) \end{eqnarray*} \begin{enumerate} \item Do these three points span a line or a plane in $3$-space? \answer: Here I notice that $v=2u$. This means that the points cannot span a plane, they must all line on the same line. Another way to see this is to compute $u \times v = (0,0,0)$. \item If the points lie in a line, find the equation of this line. Otherwise, find the equation of the plane they lie in. \answer: The equation of the line is \begin{eqnarray*} (x,y,z) & = & P_1+tu \\ & = & (1+t,-2+3t,-1-7t) \end{eqnarray*} \item Consider the point $Q=(9,3,1)$. Find the distance from $Q$ to either the line or plane determined by $\{P_1, P_2, P_3\}$ (depending on if they span a plane or a line). \answer: Let \[ w=\stackrel{\longrightarrow}{P_1Q} = (8,5,2) \] Now, we don't know exactly which vector goes through the point $Q$ and the line. This is different from the case when we compute distance to a plane because there is really only one choice for a normal direction. But, we know that we can project the vector $w$ to the vector $u$. This projection will be an orthogonal projection. Lets compute this projection: \begin{eqnarray*} \mbf{Proj}_u w & = & \frac{u \dotprod w}{|u|^2} u \\ & = & \frac{9}{59} (1,3,-7) = \left( \frac{9}{59}, \frac{27}{59}, -\frac{63}{59} \right) \end{eqnarray*} Now, we can let $a=w-\mbf{Proj}_u w$. Notice that $a+\mbf{Proj}_u w = w$, and the vector $a$ is orthogonal to the line (you can check this by doing a dot product). Therefore the distance between the line and the point $Q$ is $|a|$: \begin{eqnarray*} \mathrm{Distance} & = & |a| = |w-\mbf{Proj}_u w| \\ & = & \left| (8,5,2) - \left( \frac{9}{59}, \frac{27}{59}, -\frac{63}{59} \right) \right| \\ & = & \left| \left( \frac{463}{59}, \frac{268}{59}, \frac{181}{59} \right) \right| \\ & = & \sqrt{\frac{5406}{59}} \cong 9.5722 \end{eqnarray*} \end{enumerate} \item (20 points) Consider a point moving in the cycloid in the plane: \begin{eqnarray*} \mbf{r}(t) & = & a(t-\sin t)\mbf{i} + a(1-\cos t)\mbf{j} \qquad (a>0) \\ & = & (a(t-\sin t), a(1-\cos t)) \end{eqnarray*} I will apply Theorem~A of section 13.5. To do this we need some derivatives: \begin{eqnarray*} x = a(t-\sin t) && y = a(1-\cos t) \\ x' = a(1-\cos t) && y' = a\sin t \\ x'' = a \sin t && y'' = a\cos t \end{eqnarray*} \begin{enumerate} \item Compute the curvature, at time $t$, of this plane curve. \answer: We now apply theorem A: \begin{eqnarray*} \kappa & = & \frac{|x'y'' - x''y'|}{(x'^2+y'^2)^{3/2}} \\ & = & \frac{|a^2(1-\cos t)\cos t - a^2 \sin^2 t|}{(a^2(1-\cos t)^2 + a^2\sin^2t)^{3/2}} \\ & = & \frac{|\cos t -1|}{a(2-2\cos t)^{3/2}} \\ & = & \frac{1}{a\sqrt{8}\sqrt{1-\cos t}} \\ \end{eqnarray*} \item For what values of $t$, if any, is the curvature function undefined? What is happening at these points? \answer: clearly $\kappa$ is undefined when $\cos t=1$, so when $t=2n\pi$. This corresponds to the ``corners'' in the graph of the cycloid. \item Are there any points on the curve with a minimal or maximal curvature? Explain why or why not. \answer: As $t$ approaches $2n\pi$ (for any $n$), we see that curvature approaches $\infty$. Thus, there will be no points with maximum curvature. But, the points where $1- \cos t$ is maximum, we will have minimum curvature. Evaluating where this happens, we see that it happens when $t=2n\pi +\pi$. (You should work out all the details of why these are minimum curvature points). Thus, the minimum curvature is $1/(a\sqrt{16})$. \item If there are points on the curve with minimum or maximum curvature, find them. \answer: see above. You should really find the $x$ and $y$ coordinates of these points as well. \end{enumerate} \item (15 points) Consider the sphere in $\R^3$ defined by the equation \[ x^2+y^2+z^2-4x+6y+2z-11 = 0 \] \begin{enumerate} \item Find the center and radius of the sphere. (Remember to show your work.) \answer: We just complete the square, getting: \[ (x-2)^2 + (y+3)^2 + (z+1)^2 = 25 \] Thus the center is $C=(2,-3,-1)$ and radius is $5$. \item Find the distance from the sphere to the point $P_1=(2,-2,-2)$. (Explain why what you found is the distance). \answer: we compute the distance to the center \[ \dist(P_1,C) = \sqrt{ (2-2)^2 + (-2-(-3))^2 + (-2-(-1))^2 } = \sqrt{2} \] Notice that this is closer to the center of the sphere then the radius of the sphere. Thus, the point $P_1$ lies inside the sphere and the distance to the sphere itself must be \[ 5-\sqrt{2} \] \item Find the distance from the sphere to the point $P_2=(3,0,5)$. (Explain why what you found is the distance). \answer: Here we do the same as above and get: \[ \dist(P_1,C) = \sqrt{ (3-2)^2 + (0-(-3))^2 + (5-(-1))^2 } = \sqrt{46} \] This point must be outside the sphere, and the distance from $P_2$ to the sphere must be: \[ \sqrt{46}-5 \] \end{enumerate} \end{enumerate} \end{document}