Chapter 5.1 Practice

Fully Worked Out Questions

  1. Take \(D\) to be the disk of radius \(3\) with center at \((2,4)\). Find a path that parameterizes the boundary of \(D\). Be sure that \(D\) moves clockwise.
    Answer Rotation on the unit circle (radius \(1\) and center \((0,0)\)) in a counterclockwise direction is given by \[R_t(1,0)=(\cos(t),\sin(t))\star(1,0)=(\cos(t),\sin(t)),\] where \(t\in[0,2\pi].\) To get a parameterization of the boundary of \(D\), scale symmetrically by \(3\) and then shift the path in the direction \(\langle 2,4\rangle:\) \[\langle 2,4\rangle+\sigma_3(\cos(t),\sin(t))=(3\cos(t)+2,3\sin(t)+4).\] To get a clockwise parameterization, scale \(t\) by \(-1\). The final answer is \[c(t)=(3\cos(-t)+2,3\sin(-t)+4)\] where \(t\in[0,2\pi].\)
  2. Take \(R\) to be the rectangle with vertices \((3,5), (4,11), (-14,14)\) and \((-15,8)\). Find a piecewise linear function \(c\) that parameterizes the boundary of \(R\) and simulate the motion of a particle whose position at time \(t\) is \(c(t)\).
    Answer Start with identifying the vector that moves the vertex \((3,5)\) to the vertex \((4,11).\) The vector is \(V=\langle 1,6\rangle.\) The unit vector in the same direction as \(V\) and its corresponding unit, perpendicular vector are \[\hat{V}=\left\langle\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}} \right\rangle\quad\text{and}\quad\hat{V}_{\perp}=\left\langle -\frac{6}{\sqrt{37}},\frac{1}{\sqrt{37}}\right\rangle.\] The side lengths of the rectangle, \(a\) and \(b\), are \[a=\|V\|=\sqrt{37}\quad\text{and}\quad b=\|(-14,14)-(4,11)\|=3\sqrt{37}.\] The piecewise linear function \(c\) that parameterizes the boundary of \(R\) is \[c(t)=\begin{cases}\sqrt{37}t\left\langle\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}\right\rangle+(3,5)&\text{if }0\leq t<1\\ 3\sqrt{37}(t-1)\left\langle -\frac{6}{\sqrt{37}},\frac{1}{\sqrt{37}}\right\rangle+(4,11)&\text{if }1\leq t<2\\ -\sqrt{37}(t-2)\left\langle\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}\right\rangle+(-14,14)&\text{if }2\leq t<3\\ -3\sqrt{37}(t-3)\left\langle -\frac{6}{\sqrt{37}},\frac{1}{\sqrt{37}}\right\rangle +(-15,8)&\text{if }3\leq t\leq 4.\end{cases}.\] This can be simplified down to look like this \[c(t)=\begin{cases}t\left\langle 1,6\right\rangle+(3,5)&\text{if }0\leq t<1\\ (t-1)\left\langle -18,3\right\rangle+(4,11)&\text{if }1\leq t<2\\ (t-2)\left\langle-1,-6\right\rangle+(-14,14)&\text{if }2\leq t<3\\ (t-3)\left\langle 18,-3\right\rangle +(-15,8)&\text{if }3\leq t\leq 4.\end{cases}.\]
  3. Take \(R\) to be the rectangle with vertices \((3,5), (4,11), (-14,14)\) and \((-15,8)\). Describe \(R\) as the feasible set of inequalities.
    Answer Start with the determining the lines that describe each edge of \(R\). This can be by looking at the piecewise linear path.
    The first piece, \(t\left\langle 1,6\right\rangle+(3,5),\) can be rewritten like this: \(y=6(x-3)+5.\)
    The second piece, \((t-1)\left\langle -18,3\right\rangle+(4,11),\) can be rewritten like this: \(y=-\frac{1}{6}(x-4)+11.\)
    The third piece, \((t-2)\left\langle-1,-6\right\rangle+(-14,14)\), can be rewritten like this: \(y=6(x+14)+14.\)
    The fourth piece, \((t-3)\left\langle 18,-3\right\rangle +(-15,8)\), can be rewritten like this: \(y=-\frac{1}{6}(x+15)+8.\)
    The feasible set of inequalities looks like this: \[\begin{cases} y\geq 6(x-3)+5\\ y\leq-\frac{1}{6}(x-4)+11\\ y\leq 6(x+14)+14\\ y\geq -\frac{1}{6}(x+15)+8 \end{cases}\]
  4. Take \(R\) to be the rectangle with ordered vertex set \(((3,5), (4,11), (-14,14), (-15,8))\). Is \(R\) positively oriented?
    Answer Take \(p_1=(3,5)\), \(p_2=(4,11)\) and \(p_3=(-14,14).\) To determine the orientation, check whether \[p_3-p_2=\|p_3-p_2\|\hat{V}_{\perp},\] where \(V=p_2-p_1.\) For this problem,\(V=\langle 1,6\rangle\) and so \(\hat{V}_{\perp}=\left\langle -\frac{6}{\sqrt{37}},\frac{1}{\sqrt{37}} \right\rangle.\) Also \(p_3-p_2=\langle-18,3 \rangle\) and \(\|p_3-p_2\|=3\sqrt{37}.\) Calculate \(\|p_3-p_2\|\hat{V}_{\perp}\) to get that \[\begin{align*}\|p_3-p_2\|\hat{V}_{\perp}&=3\sqrt{37}\left\langle -\frac{6}{\sqrt{37}},\frac{1}{\sqrt{37}} \right\rangle\\ &=\langle -18,3\rangle.\end{align*}\] This is equal to \(p_3-p_2\), which means that \(R\) is positively oriented.
  5. Take \(R\) to be the rectangle with ordered vertex set \(((3,5), (4,11), (-14,14), (-15,8))\). Find a translation and rotation so that \((3,5)\) is moved to \((0,0)\) and one of the edges of \(R\) is on the \(x\)-axis.
    Answer First translate all points by the vector that moves \((3,5)\) to \((0,0),\) which is the vector \(V=\langle -3,-5\rangle.\) The original vertices of \(R\) \[(3,5),(4,11),(-14,14),(-15,8)\] translated by \(V=\langle -3,-5\rangle\) are now the vertices \[(0,0),(1,6),(-17,9),(-18,3).\] This new rectangle is \(V+R.\) To get one of the edges of \(V+R\) in the \(x\)-axis, rotate \(V+R\) by the right amount. To find this focus on the vertex \(p=(1,6).\) Its projection onto the unit circle, \(q=\left(\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}\right)\), is an angle that takes \((1,0)\), a point on the \(x\)-axis, to \(q=\left(\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}\right)\). To go the other way around, calculate \(q^{-1}\), which is \(q^{-1}=\left(\frac{1}{\sqrt{37}},-\frac{6}{\sqrt{37}}\right).\) Therefore, rotate \(V+R\) by \(q^{-1}\) to achieve our goal. The vertices of \(q^{-1}\star(V+R)\) are \[q^{-1}\star (0,0)=(0,0),\] \[q^{-1}\star (1,6)=\left(\frac{37}{\sqrt{37}},0\right),\] \[q^{-1}\star (-17,9)=\left(\frac{37}{\sqrt{37}},\frac{111}{\sqrt{37}}\right),\] \[q^{-1}\star (-18,3)=\left(0,\frac{111}{\sqrt{37}}\right).\] So the final answer is that a translation by \(V=\langle -3,-5\rangle\) and a rotation by \(q^{-1}=\left(\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}\right)\) will produce the desired results.
  6. Take \(R\) to be the oriented triangle with ordered vertex set equal to \(((2,3),(8,4),(6,9))\). Find a continuous, piecewise linear parameterization for the boundary of \(R\), where the domain of the path is equal to \([0,5].\)
    Answer The are many different parameterizations of the triangle \(R\) with a domain of \([0,5]\). The example below breaks the path to time intervals \([0,2]\), \((2,4]\) and \((4,5]:\) \[c(t)=\begin{cases}\frac{t}{2}\langle 6,1 \rangle +(2,3)&\text{if }0\leq t\leq 2\\ \frac{t-2}{2}\langle -2,5\rangle +(8,4)&\text{if }2<t\leq 4\\ (t-4)\langle-4 ,-6\rangle+(6,9)&\text{if }4<t\leq 5\end{cases}.\]

Questions

  1. Take \(f\) to be the function from \([-2,1]\) to \(\mathbb R\) that is given by \[f(x)=2x^2+1.\] Find parameterizations for \(f\) (viewed as a subset of \(\mathbb R^2\)) that have domains equal to the following intervals:

    1. \([-2,1]\)
    2. \([0,3]\)
    3. \([-3,3]\)

  2. Take \(D\) to be the disk of radius \(5\) with center at \((-1,5).\) Find a path that parameterizes the boundary of \(D\) clockwise.

  3. Take \(D\) to be the disk of radius \(2\) with center at \((-2,-2).\) Find a path that parameterizes the boundary of \(D\) counter-clockwise.

  4. Take \(D\) to be the disk of radius \(6\) with center \((2,-6)\).

    1. Determine a parameterization of \(\partial D\), the boundary of \(D\).
    2. Determine an inequality for which \(D\) is a feasible set.
    3. Identify a point in \(D\), a point on \(\partial D\), and a point outside of \(D.\)

  5. Take \(E\) to be the solid ellipse that is given by the equation \[\frac{(x-3)^2}{9}+(y+5)^2 \leq 9.\]

    1. Find a point that is inside \(E.\)
    2. Find a path \(c\) that parameterizes the ellipse \(\partial E.\)

  6. Take \(R\) to be the rectangle with vertices \((1,3)\), \((4,4)\), \((3,7)\), and \((0,6).\)

    1. Find a piecewise linear function \(c\) that parameterizes \(\partial R\) and simulate the motion of a particle whose position at time \(t\) is \(c(t).\)
    2. Describe \(R\) as the feasible set for a system of inequalities.
    3. Determine whether \(p=(2,3.33332)\) is inside or outside \(R.\)

Answers

    1. \(c(t)=(t,2t^2+1)\) where \(t\) is in \([-2,1]\)
    2. \((t-2,2(t-2)^2+1)\) where \(t\) is in \([0,3]\)
    3. \(\left(\frac{1}{2}(t-1),2(\frac{1}{2}(t-1))^2+1\right)\) where \(t\) is in \([-3,3]\)

  2. \(c(t)=(5\cos(-t)-1,5\sin(-t)+5)\) where \(t\) is in \([0,2\pi]\)

  3. \(c(t)=(2\cos(t)-2,2\sin(t)-2)\) where \(t\) is in \([0,2\pi]\)

    1. \(c\colon [0,2\pi]\to\mathbb{R}^2\) with \(c(t)=(6\cos(t)+2,6\sin(t)-6)\)
    2. \((x-2)^2+(y+6)^2\leq 36\)
    3. \(p=(2,-5)\) is in \(D\), \(q=(2,-12)\) is in \(\partial D\), and \(u=(2,6)\) is not in \(D\).
    1. Any point in the disk centered at \((0,0)\) with radius \(r=3\) that is transformed by \(\langle 3, -5\rangle+X_3.\) For example, \((1,1)\) is in the disk centered at \((0,0)\) with radius \(r=3\). So the point \(p=\langle 3,-5\rangle +X_3(1,1)\) will be in \(E\). The point is \(p=(6,-4).\)
    2. \((9\cos(t)+3,3\sin(t)-5)\) where \(t\) is in \([0,2\pi]\)
    1. \[c(t)=\begin{cases}\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle t+(1,3) &\text{ if }0\leq t< 1\\ \sqrt{10}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle (t-1)+\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle +(1,3) &\text{ if }1\leq t< 2\\ -\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle (t-2)+\sqrt{10}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle +\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle +(1,3) &\text{ if }2\leq t< 3\\ -\sqrt{10}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}} \right\rangle (t-3)+\sqrt{10}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}} \right\rangle+(1,3) &\text{ if }3\leq t< 4 \end{cases}\] or \[c(t)=\begin{cases}\left\langle 3,1\right\rangle t+(1,3) &\text{ if }0\leq t< 1\\ \left\langle -1,3\right\rangle (t-1)+(4,7) &\text{ if }1\leq t< 2\\ \left\langle- 3,-1\right\rangle (t-2)+(3,7) &\text{ if }2\leq t< 3\\ \left\langle 1,-3 \right\rangle (t-3)+(0,6) &\text{ if }3\leq t< 4 \end{cases}\]
    2. \[\begin{cases}y\geq\frac{1}{3}(x-1)+3\\y\leq-3(x-4)+4\\ y\leq\frac{1}{3}x+6\\ y\geq-3(x-1)+3\end{cases}\]
    3. Check if it satisfies the above inequalities or translate and rotate the rectangle so one of the edges is on the \(x\)-axis. If you do it the second way, the rotated rectangle has vertices at \((0,0),\left(\frac{10}{\sqrt{10}},0\right),\left(\frac{10}{\sqrt{10}},\frac{10}{\sqrt{10}}\right),\left(0,\frac{10}{\sqrt{10}}\right).\) You can also rewrite these vertices as \((0,0),\left(\sqrt{10},0\right),\left(\sqrt{10},\sqrt{10}\right),\left(0,\sqrt{10}\right).\) Take the point \(p\), and do the same translation and rotation you did to rotate the rectangle. The new point is \(\left(\frac{3.33332}{\sqrt{10}},-\frac{0.00004}{\sqrt{10}}\right).\) For this translated and rotated point to be in the translated and rotated rectangle, its \(x\) and \(y\) coordinates must satisfy \(0\leq x\leq\frac{10}{\sqrt{10}}\) and \(0\leq y\leq \frac{10}{\sqrt{10}}\). The \(y\)-coordinate requirement is not satisfied. So the original \(p\) is not in the rectangle \(R.\)

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