{formcorrectchoice}, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$
", "{formincorrectchoice1}, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$
", "{formincorrectchoice2}, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$
", "{formincorrectchoice3}, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$
"], "prompt": "What is the correct parametric representation of the curve?
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", "{dirincorrectchoice}
"], "prompt": "In which direction is the curve traversed?
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index=15, 'From $\\\\pmatrix{1,0}$ to $\\\\pmatrix{1,2}$',\n index=16, 'From $\\\\pmatrix{1,1}$ to $\\\\pmatrix{0,0}$ and then back to $\\\\pmatrix{1,1}$',\n index=17, 'Clockwise',\n index=18, 'From $\\\\pmatrix{1,-1}$ to $\\\\pmatrix{1,1}$',\n index=19, 'From $\\\\pmatrix{4.63,-3.53}$ to $\\\\pmatrix{4.63,3.53}$',\n false\n )", "name": "dircorrectchoice", "description": ""}, "formcorrectchoice": {"group": "Ungrouped variables", "templateType": "anything", "definition": "switch (\n index=0, '$t\\\\rightarrow\\\\pmatrix{t^3,t}$',\n index=1, '$t\\\\rightarrow\\\\pmatrix{t,\\\\frac{5}{t}}$',\n index=2, '$t\\\\rightarrow\\\\pmatrix{t,\\\\arctan(t)}$',\n index=3, '$t\\\\rightarrow\\\\pmatrix{t,\\\\tan(t)}$',\n index=4, '$t\\\\rightarrow\\\\pmatrix{t,\\\\cot(\\\\frac{t}{2}+5)}$',\n index=5, '$t\\\\rightarrow\\\\pmatrix{3t,-7t}$',\n index=6, '$t\\\\rightarrow\\\\pmatrix{-t^3,t}$',\n index=7, '$t\\\\rightarrow\\\\pmatrix{t^3,\\\\sin(-5t)}$',\n index=8, 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"definition": "random(0..19)", "name": "index", "description": ""}, "image": {"group": "Ungrouped variables", "templateType": "anything", "definition": "'parametric-'+index+'.png'", "name": "image", "description": ""}, "f1": {"group": "Ungrouped variables", "templateType": "anything", "definition": "switch (\n index=0 or index=11, random(1..9),\n index=1 or index=2 or index=3 or index=4 or index=19, 3*random(1..3),\n index=5 or index=6 or index=7 or index=13 or index=14 or index=15 or index=16 or index=17 or index=18, random(1..4),\n false\n )", "name": "f1", "description": ""}}, "ungrouped_variables": ["index", "formcorrectchoice", "g1", "f1", "image", "range", "a1", "dirincorrectchoice", "formincorrectchoice3", "formincorrectchoice1", "formincorrectchoice2", "b1", "c1", "e1", "dircorrectchoice", "d1"], "name": "Identify correct parametric representation of a curve", "functions": {"show": {"type": "html", "language": "javascript", "definition": "return $('');", "parameters": [["n", "number"]]}}, "variable_groups": [], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "{show(index)}
\nYou are given the following parametric representation of a curve.
\n{image('resources/images/'+image)}
", "tags": ["checked2015"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Determine the correct parametric representation of a given curve. Curve is randomly chosen from a set of 20.
\nThe graph of the curve was not displayed on my machine.
"}, "extensions": [], "advice": "a)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t^3,t}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t^3$ and $y=t$, therefore $x=y^3$ or $y=x^\\frac{1}{3}$. Substituting the values of $t=-1$ and $t=1$ for $x$ and $y$ reveals the end-points of the curve.
\nAlso, $\\frac{\\mathrm{d}y}{\\mathrm{d}x}=\\frac{1}{3}x^{-\\frac{2}{3}}$, which diverges at the origin, implying that the tangent to the curve is vertical there.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{-1,-1}$ to $\\pmatrix{1,1}$, because $t=-1$ implies $x=-1$ and $y=-1$, and $t=1$ implies $x=1$ and $y=1$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t,\\frac{5}{t}}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t$ and $y=\\frac{5}{t}$, therefore $y=\\frac{5}{x}$ or $xy=5$. This is the equation of a rectangular hyperbola in the right half-plane, where the asymptotes coincide with the $x$- and $y$-axes. You can check this by taking the limit of $y=\\frac{5}{x}$ as $x\\rightarrow 0$. This limit diverges. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{0,\\infty}$ to $\\pmatrix{1,5}$, because $t=0$ implies $x=0$ and $y\\rightarrow\\infty$, and $t=1$ implies $x=1$ and $y=5$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t,\\arctan(t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t$ and $y=\\arctan(t)$, therefore $y=\\arctan(x)$. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\nSince $x=t$, the range of $x$ coincides with the range of $t$, and so $-5\\leqslant t\\leqslant 2$.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{-5,-1.37}$ to $\\pmatrix{2,1.11}$, because $t=-5$ implies $x=-5$ and $y\\approx -1.37$, and $t=2$ implies $x=2$ and $y\\approx 1.11$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t,\\tan(t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t$ and $y=\\tan(t)$, therefore $y=\\tan(x)$. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\nThe lines $x=-\\frac{\\pi}{2}$ and $x=\\frac{\\pi}{2}$ are asymptotes for $y=\\tan(x)$.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{-\\frac{\\pi}{2},-\\infty}$ to $\\pmatrix{\\frac{\\pi}{2},\\infty}$, because $t=-\\frac{\\pi}{2}$ implies $x=-\\frac{\\pi}{2}$ and $y\\rightarrow -\\infty$, and $t=\\frac{\\pi}{2}$ implies $x=\\frac{\\pi}{2}$ and $y\\rightarrow \\infty$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t,\\cot(\\frac{t}{2}+5)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t$ and $y=\\cot(\\frac{t}{2}+5)$, therefore $y=\\cot(\\frac{x}{2}+5)$. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\nSince $x=t$, the range of $t$ coincides with the range of $x$, and so $-2\\leqslant x\\leqslant 1$.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{-2,0.86}$ to $\\pmatrix{1,-1}$, because $t=-2$ implies $x=-2$ and $y\\approx 0.86$, and $t=1$ implies $x=1$ and $y\\approx -1$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{3t,-7t}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=3t$ and $y=7t$, therefore $y=-\\frac{7}{3}x$, and this is the equation of a straight line with gradient $-7/3$, passing through the origin. In addition, substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{0,0}$ to $\\pmatrix{3,-7}$, because $t=0$ implies $x=0$ and $y=0$, and $t=1$ implies $x=3$ and $y=-7$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{-t^3,t}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=-t^3$ and $y=t$, therefore $x=-y^3$ or $y=-x^\\frac{1}{3}$. Also note that $\\frac{\\mathrm{d}y}{\\mathrm{d}x}=-\\frac{1}{3}x^{-\\frac{2}{3}}$, which diverges at the origin, so the tangent to the curve is vertical there. In addition, substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{1,-1}$ to $\\pmatrix{-1,1}$, because $t=-1$ implies $x=1$ and $y=-1$, and $t=1$ implies $x=-1$ and $y=1$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t^3,\\sin(-5t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t^3$ and $y=\\sin(-5t)$, therefore $y=\\sin\\left(-5x^\\frac{1}{3}\\right)$. The curve is a non-uniformly stretched sinusoid, due to the term $x^\\frac{1}{3}$, and the stretching is stronger at larger $x$, because $x=t^3$ grows faster as $t$ increases.
\nThe values of $x$ range from $0$ to $27$ because $t$ ranges from $0$ to $3$ and $x=t^3$. In addition, substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{0,0}$ to $\\pmatrix{27,-0.65}$, because $t=0$ implies $x=0$ and $y=0$, and $t=3$ implies $x=27$ and $y\\approx -0.65$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{2\\sin(t),-2\\cos(t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=2\\sin(t)$ and $y=-2\\cos(t)$, therefore $x^2+y^2=4$, and this is the equation for a circle of radius $2$, centred at the origin. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\nIn addition, since $-\\frac{\\pi}{2}\\leqslant t\\leqslant\\frac{\\pi}{2}$, the curve is only defined in the lower half of the $\\pmatrix{x,y}$-plane.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{-2,0}$ to $\\pmatrix{2,0}$, because $t=-\\frac{\\pi}{2}$ implies $x=-2$ and $y=0$, and $t=\\frac{\\pi}{2}$ implies $x=2$ and $y=0$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{3t,t}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=3t$ and $y=t$, therefore $y=\\frac{1}{3}x$, which is the equation of a straight line, with gradient $1/3$, passing through the origin. Since the range of $t$ given does not allow $x$ and $y$ to be equal to zero, the segment of the line shown does not include the origin. In addition, substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{3,1}$ to $\\pmatrix{15,1}$, because $t=1$ implies $x=3$ and $y=1$, and $t=5$ implies $x=15$ and $y=5$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{-t^2,t}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=-t^2$ and $y=t$, therefore $x=-y^2$, which is only defined in the left half-plane, and the derivative at the origin is divergent, so the tangent to the curve is vertical there. In addition, substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{-1,-1}$ to $\\pmatrix{-1,1}$, because $t=-1$ implies $x=-1$ and $y=-1$, and $t=1$ implies $x=-1$ and $y=1$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t,t^\\frac{3}{2}}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t$ and $y=t^\\frac{3}{2}$, therefore $y=x^\\frac{3}{2}$. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{1,1}$ to $\\pmatrix{2,2.83}$, because $t=1$ implies $x=1$ and $y=1$, and $t=2$ implies $x=2$ and $y\\approx 2.83$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t,-t^2}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t$ and $y=-t^2$, therefore $y=-x^2$, which is only defined in the lower half-plane, and the derivative at the origin is zero, corresponding to a maximum.
\nSince $x=t$, the range of $x$ coincides with the range of $t$.
\nIn addition, substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n \nb)
\nThe curve is traversed from $\\pmatrix{-1,-1}$ to $\\pmatrix{1,-1}$, because $t=-1$ implies $x=-1$ and $y=-1$, and $t=1$ implies $x=1$ and $y=-1$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t^2,\\sin(t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t^2$ and $y=\\sin(t)$, therefore $y=\\sin(\\sqrt{x})$. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\nThe curve is part of a sinusoid that has been stretched in the $x$-direction due to the term $\\sqrt{x}$, with the stretching stronger at larger $x$, because $x=t^2$ grows faster as $t$ increases.
\nAlso note that $\\frac{\\mathrm{d}y}{\\mathrm{d}x}=\\frac{\\cos(\\sqrt{x})}{2\\sqrt{x}}$, so the curve has a maximum where $\\sqrt{x}=\\frac{1}{2}(2n+1)\\pi$, or $x=\\frac{1}{4}(2n+1)^2\\pi^2$, for $n=0,1,2,\\ldots$. Given the range of $t$, the only valid value of $x$ is when $n=0$, so the curve has a maximum at $x=\\frac{\\pi^2}{4}$.
\nAlso note that $\\frac{\\mathrm{d}y}{\\mathrm{d}x}$ diverges at the origin, so the curve is vertical there.
\n \nb)
\nThe curve is traversed from $\\pmatrix{0,0}$ to $\\pmatrix{9,0.14}$, because $t=0$ implies $x=0$ and $y=0$, and $t=3$ implies $x=9$ and $y\\approx 0.14$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{3\\cos(t),-3\\sin(t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=3\\cos(t)$ and $y=-3\\sin(t)$, therefore $x^2+y^2=9$, which is the equation of a circle, centred at the origin, having radius $3$. From the range of $t$ we can determine that the range of valid $x$ and $y$ values corresponds to the right half-plane.
\n \nb)
\nThe curve is traversed from $\\pmatrix{0,3}$ to $\\pmatrix{0,-3}$, because $t=-\\frac{\\pi}{2}$ implies $x=0$ and $y=3$, and $t=\\frac{\\pi}{2}$ implies $x=0$ and $y=-3$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{1,2t}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=1$ and $y=2t$ therefore, for all values of $t$, $x=1$ and $y$ increases with $t$. The graph is therefore a segment of a straight line.
\n \nb)
\nThe curve is traversed from $\\pmatrix{1,0}$ to $\\pmatrix{1,2}$, because $t=0$ implies $x=1$ and $y=0$, and $t=1$ implies $x=1$ and $y=2$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t^2,t^4}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t^2$ and $y=t^4$, therefore $y=x^2$. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\nAlso, note that $\\frac{\\mathrm{d}y}{\\mathrm{d}x}=2x$, which is zero at the origin, so the tangent to the curve is horizontal there.
\n \nb)
\nThe curve is traversed from $\\pmatrix{1,1}$ to $\\pmatrix{0,0}$, then back to $\\pmatrix{1,1}$, because $t=-1$ implies $x=1$ and $y=1$, and $t=1$ also implies $x=1$ and $y=1$, but $t=0$ implies $x=0$ and $y=0$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{5\\sin(t),3\\cos(t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=5\\sin(t)$ and $y=3\\cos(t)$, therefore $\\left(\\frac{x}{5}\\right)^2+\\left(\\frac{y}{3}\\right)^2=1$, which is the equation of an ellipse, with semi-major axis equal to $5$, and semi-minor axis equal to $3$.
\nSince $t$ takes values from $-\\pi$ to $\\pi$, the curve is defined over the entire $\\pmatrix{x,y}$-plane.
\n \nb)
\nThe curve is traversed in a clockwise direction. Substituting $t=-\\pi$ for $x$ and $y$ gives $x=0$ and $y=-3$. Then choosing $t=-\\frac{\\pi}{2}$, say, gives $x=-5$ and $y=0$, which is $90^\\circ$ clockwise from $\\pmatrix{0,-3}$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{t^2,t^3}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=t^2$ and $y=t^3$, therefore $y=x^\\frac{3}{2}$. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\nAlso, note that $\\frac{\\mathrm{d}y}{\\mathrm{d}x}=\\frac{3}{2}x^\\frac{1}{2}$, which is zero at the origin, so the tangent to the curve is horizontal there.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{1,-1}$ to $\\pmatrix{1,1}$, because $t=-1$ implies $x=1$ and $y=-1$, and $t=1$ implies $x=1$ and $y=1$.
\na)
\nThe correct answer in this case is $t\\rightarrow\\pmatrix{3\\cosh(t),3\\sinh(t)}$, $\\var{range[0]}\\leqslant t\\leqslant\\var{range[1]}$.
\nTo see this note that $x=3\\cosh(t)$ and $y=3\\sinh(t)$, therefore $x^2-y^2=9$, which is the equation of a rectangular hyperbola in the right half-plane. Substituting values of $t$ for $x$ and $y$ in the given range determines points along the curve.
\n\n
b)
\nThe curve is traversed from $\\pmatrix{4.63,-3.53}$ to $\\pmatrix{4.63,3.53}$, because $t=-1$ implies $x\\approx 4.63$ and $y\\approx -3.53$, and $t=1$ implies $x\\approx 4.63$ and $y\\approx 3.53$.
\n