17a360…: “Until a spring tetrahedron”

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\,d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}\hspace {0in}} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \DeclareMathOperator {\proj }{\mathbf {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Select all of the following statements that are true.

If $\vec {p}(t)$ uses arc length as a parameter and $\uvec {t}(t)$ denotes the unit tangent vector, then $\uvec {t}(t) = \vec {p}'(t)$ . If $\vecl (t) = \vec {v}t+\vec {P}_0$ is an arc length parameterization of a line, then $\vec {v}$ must be a unit vector. If the curve $C$ is traced out by $\vec {r}(t) = \vector {t^2,\frac {1}{2}t}, -\infty < t < \infty$ , then the curve is parameterized by arc length since $|\vec {r}'(t)| = \sqrt {4t^2+\frac {1}{4}} = 1 \textrm { when } t = \frac {\sqrt {3}}{4}.$ If the curve $C$ is parameterized by $\vec {p}(t)$ , and it is known that $\vec {p}(0) = \vector {0,0}$ and $\vec {p}(3) = \vector {2,4}$ , it is possible that $\vec {p}(t)$ is an arc length parameterization. If the curve $C$ is parameterized by $\vec {p}(t)$ , and it is known that $\vec {p}(0) = \vector {0,0}$ and $\vec {p}(8) = \vector {2,4}$ , it is possible that $\vec {p}(t)$ is an arc length parameterization.

(Use the hint to see the logic behind each choice)

For the statement

True or False: If $\vec {p}(t)$ uses arc length as a parameter and $\uvec {t}(t)$ denotes the unit tangent vector, then $\uvec {t}(t) = \vec {p}'(t)$

do you want to see an explanation?

Yes. No

In general, the unit tangent vector is given by

$\uvec {t}(t) = \frac {\vec {p}'(t)}{\left |\vec {p}'(t)\right |}.$ Since $\vec {p}(t)$ uses arc length as a parameter, we have that $\left |\vec {p}'(t)\right | = \answer {1}$ .

For the statement

True or False: If $\vecl (t) = \vec {v}t+\vec {P}_0$ is an arc length parameterization of a line, then $\vec {v}$ must be a unit vector.

do you want to see an explanation?

Yes. No

If $\vecl (t)$ is an arc length parameterization, $\left |\vecl '(t)\right | = \answer {1}$ . Since the $\vec {v}$ and $\vec {P}_0$ are constant vectors,

$\vecl '(t) = \vec {v}.$

Hence, $\left |\vecl '(t)\right | = \left |\vec {v}\right |$ for all $t$ .

For the statement

True or False: If the curve $C$ is traced out by $\vec {r}(t) = \vector {t^2,\frac {1}{2}t}, -\infty < t < \infty$ , then the curve is parameterized by arc length since

$|\vec {r}'(t)| = \sqrt {4t^2+\frac {1}{4}} = 1 \textrm { when } t = \frac {\sqrt {3}}{4}.$

do you want to see an explanation?

Yes. No

The calculation is correct; since $\vec {r}(t) = \vector {t^2,\frac {1}{2}t}$ , we have

$\vec {r}'(t) = \vector {2t,\frac {1}{2}}.$ Thus, $|\vec {r}'(t)| = \sqrt {(2t)^2+\left (\frac {1}{2}\right )^2} = \sqrt {4t^2+\frac {1}{4}}$ .

However, the curve is parameterized by arc length if and only if $|\vec {r}'(t)| =1$ for all $t$ , not just a specific $t$ -value.

For the statement

True or False: If the curve $C$ is parameterized by $\vec {p}(t)$ , and it is known that $\vec {p}(0) = \vector {0,0}$ and $\vec {p}(3) = \vector {2,4}$ , it is possible that $\vec {p}(t)$ is an arc length parameterization.

do you want to see an explanation?

Yes. No

If the curve uses arc length as a parameter, $\vec {p}(3)$ should give us the position $(x,y)$ after we have travelled $3$ units along the curve. The question is, can we travel from the points associated to $\vec {p}(0)$ and $\vec {p}(3)$ by going exactly $3$ units?

Note, the shortest distance between the points associated to $\vec {p}(0)$ and $\vec {p}(3)$ is the length of the line segment between them, and this line segment has length

$\sqrt {(2-0)^2+(4-0)^2} = \sqrt {20} >3.$

For the statement

True or False: If the curve $C$ is parameterized by $\vec {p}(t)$ , and it is known that $\vec {p}(0) = \vector {0,0}$ and $\vec {p}(8) = \vector {2,4}$ , it is possible that $\vec {p}(t)$ is an arc length parameterization.

do you want to see an explanation?

Yes. No

If the curve uses arc length as a parameter, $\vec {p}(8)$ should give us the position $(x,y)$ after we have travelled $8$ units along the curve. The question is, can we travel from the points associated to $\vec {p}(0)$ and $\vec {p}(8)$ by going exactly $8$ units?

Note, the shortest distance between the points associated to $\vec {p}(0)$ and $\vec {p}(8)$ is the length of the line segment between them, and this line segment has length

$\sqrt {(2-0)^2+(4-0)^2} = \sqrt {20} <8.$

It thus should be possible.

In fact, an explicit example is obtained geometrically by tracing along the segment below at a rate of one unit of distance per unit of time, so we trace out a border of each block each unit of time.

It is of course possible to construct other examples as well.