// Numbas version: exam_results_page_options {"name": "Parameterisation of a curve - find tangent and length", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "variables": {"num3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "4*9*(a^2+t1*b^2)", "description": "", "name": "num3"}, "num4": {"templateType": "anything", "group": "Ungrouped variables", "definition": "3*a", "description": "", "name": "num4"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..4)", "description": "", "name": "b"}, "s": {"templateType": "anything", "group": "Ungrouped variables", "definition": "precround((8/(27*4*b^2))*((9*a^2+((9*t2*4*b^2)/4))^(3/2)-(9*a^2+((9*t1*4*b^2)/4))^(3/2)),2)", "description": "", "name": "s"}, "t1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..2)", "description": "", "name": "t1"}, "t2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..2)+t1", "description": "", "name": "t2"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..2)", "description": "", "name": "a"}, "num1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "9*4*b^2", "description": "", "name": "num1"}, "num2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "27*4*b^2", "description": "", "name": "num2"}}, "ungrouped_variables": ["a", "num4", "b", "num1", "num2", "num3", "t2", "t1", "s"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "name": "Parameterisation of a curve - find tangent and length", "functions": {}, "showQuestionGroupNames": false, "parts": [{"prompt": "

Find the tangent vector $\\boldsymbol{u}$ to the curve.

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$\\boldsymbol{u}=($[[0]]$,$[[1]]$)$.  (Do not enter decimals in your answers.)

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The arc-length along the curve can be written in the form $s(t)=f(t)-f(t_1)$.  Find $f(t)$.

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$f(t)=$ [[0]].  (Do not enter decimals in your answers.)

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Find the total length of the curve, $S$, given $t_1=\\var{t1}$ and $t_2=\\var{t2}$.

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$S=$ [[0]].  (Enter your answer to 2d.p.)

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$s\\longmapsto\\pmatrix{\\simplify{{3a}/{b^2}}\\left\\{\\left[\\simplify{{b^2}s/2}+\\left(\\var{a^2+t1*b^2}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\},\\simplify{2/{b^2}}\\left\\{\\left[\\left(\\simplify{{b^2}s/2}+\\left(\\var{a^2+t1*b^2}\\right)^\\frac{3}{2}\\right)^\\frac{2}{3}-\\var{a^2}\\right]\\right\\}^\\frac{3}{2}}$

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$s\\longmapsto\\pmatrix{\\simplify{{a}/{3*b^2}}\\left\\{\\left[\\simplify{{a^2}s/2}+\\left(\\var{a^2+t1*b^2}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\},\\simplify{2/{27*b^2}}\\left\\{\\left[\\left(\\simplify{{a^2}s/2}+\\left(\\var{a^2+t1*b^2}\\right)^\\frac{3}{2}\\right)^\\frac{2}{3}-\\var{a^2}\\right]\\right\\}^\\frac{3}{2}}$

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$s\\longmapsto\\pmatrix{\\simplify{{3*a}}\\left\\{\\left[\\simplify{{a^2+t1*b^2}s/2}+\\left(\\var{b^2}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\},\\simplify{{2*b}}\\left\\{\\left[\\left(\\simplify{{a^2+t1*b^2}s/2}+\\left(\\var{b^2}\\right)^\\frac{3}{2}\\right)^\\frac{2}{3}-\\var{a^2}\\right]\\right\\}^\\frac{3}{2}}$

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$s\\longmapsto\\pmatrix{\\simplify{{3}/{a*b^2}}\\left\\{\\left[\\simplify{{b^2}s/2}+\\left(\\var{a^2}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2+t1*b^2}\\right\\},\\simplify{2/{a^3*b^2}}\\left\\{\\left[\\left(\\simplify{{b^2}s/2}+\\left(\\var{a^2}\\right)^\\frac{3}{2}\\right)^\\frac{2}{3}-\\var{a^2+t1*b^2}\\right]\\right\\}^\\frac{3}{2}}$

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Which of the following corresponds to an alternative parametric representation of the curve, again with $t_1=\\var{t1}$ and $t_2=\\var{t2}$, using the arc-length $s$ as the curve parameter?

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You are given the curve $t\\longmapsto\\pmatrix{\\var{3*a}t,\\var{2*b}t^\\frac{3}{2}}$, where $t_1\\leqslant t\\leqslant t_2$.

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Calculation of the length and alternative form of the parameteric representation of a curve.

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The tangent vector to the curve $t\\longmapsto\\pmatrix{x,y}$ is given by $\\boldsymbol{u}\\equiv\\pmatrix{\\frac{\\mathrm{d}x}{\\mathrm{d}t},\\frac{\\mathrm{d}y}{\\mathrm{d}t}}$.

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The length $s$ of the curve in the range $t_1\\leqslant t\\leqslant t_2$ is given by

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\\[s=\\int_{t_1}^{t_2}{u\\,\\mathrm{d}t},\\]

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where $u^2=\\lvert\\boldsymbol{u}\\rvert^2=\\boldsymbol{u\\cdot u}$.

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In this question, therefore, $\\boldsymbol{u}=\\pmatrix{\\frac{\\mathrm{d}}{\\mathrm{d}t}(\\var{3*a}t),\\frac{\\mathrm{d}}{\\mathrm{d}t}\\left(\\var{2*b}t^\\frac{3}{2}\\right)}=\\pmatrix{\\var{3*a},\\simplify{{3*b}*t^(1/2)}}$, and $u^2=9\\left(\\var{a^2}+\\simplify{{b^2}*t}\\right)$.

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Then

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\\[\\begin{align}s=\\int_{t_1}^{t_2}{u\\,\\mathrm{d}t}&=3\\int_{t_1}^{t_2}{\\sqrt{\\var{a^2}+\\simplify{{b^2}*t}}\\,\\mathrm{d}t}\\\\&=\\simplify{2/{b^2}}\\left[\\left(\\var{a^2}+\\simplify{{b^2}*t}\\right)^\\frac{3}{2}\\right]_{t_1}^{t_2},\\end{align}\\]

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so $f(t)=\\simplify{2/{b^2}}\\left(\\var{a^2}+\\simplify{{b^2}*t}\\right)^\\frac{3}{2}$.

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Finally, substitute $t_1=\\var{t1}$ and $t_2=\\var{t2}$ into the expression for $s$ to find the length of the curve over the given range of $t$.

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Hence $s=\\simplify{2/{b^2}}\\left\\{\\left(\\var{a^2+t2*b^2}\\right)^\\frac{3}{2}-\\left(\\var{a^2+t1*b^2}\\right)^\\frac{3}{2}\\right\\}=\\var{s}$ to 2d.p.

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An alternative parametric representation, using $s$ as the curve parameter is given by

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\\[\\begin{align}s(t)=\\int_{t_1}^t{u\\,\\mathrm{d}\\tau}&=3\\int_{t_1}^t{\\sqrt{\\var{a^2}+\\simplify{{b^2}*tau}}\\,\\mathrm{d}\\tau}\\\\&=\\simplify{2/{b^2}}\\left[\\left(\\var{a^2}+\\simplify{{b^2}*tau}\\right)^\\frac{3}{2}\\right]_{t_1}^t\\\\&=\\simplify{2/{b^2}}\\left\\{\\left(\\var{a^2}+\\simplify{{b^2}*t}\\right)^\\frac{3}{2}-\\left(\\var{a^2}+\\simplify{{b^2}*t_1}\\right)^\\frac{3}{2}\\right\\}.\\end{align}\\]

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Now rearrange this expression for $t(s)$, so

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\\[t(s)=\\simplify{1/{b^2}}\\left\\{\\left[\\simplify{{b^2}/2}s+\\left(\\var{a^2}+\\simplify{{b^2}*t_1}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\},\\]

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and substitute into the original representation of the curve $t\\longmapsto\\pmatrix{\\var{3*a}t,\\var{2*b}t^\\frac{3}{2}}$ with $t_1\\leqslant t\\leqslant t_2$.  Hence

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\\[s\\longmapsto\\pmatrix{\\simplify{{3*a}/{b^2}}\\left\\{\\left[\\simplify{{b^2}/2}s+\\left(\\var{a^2}+\\simplify{{b^2}*t_1}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\},\\simplify{2/{b^2}}\\left\\{\\left[\\simplify{{b^2}/2}s+\\left(\\var{a^2}+\\simplify{{b^2}*t_1}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\}^\\frac{3}{2}}\\]

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with $0\\leqslant s\\leqslant\\simplify{2/{b^2}}\\left(\\left(\\var{a^2}+\\simplify{{b^2}*t_2}\\right)^\\frac{3}{2}-\\left(\\var{a^2}+\\simplify{{b^2}*t_1}\\right)^\\frac{3}{2}\\right)$.

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Finally, substitute $t_1=\\var{t1}$ and $t_2=\\var{t2}$ into the above expressions, to find the specific parametric representation corresponding to the given range of t:

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\\[s\\longmapsto\\pmatrix{\\simplify{{3*a}/{b^2}}\\left\\{\\left[\\simplify{{b^2}/2}s+\\left(\\var{a^2+b^2*t1}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\},\\simplify{2/{b^2}}\\left\\{\\left[\\simplify{{b^2}/2}s+\\left(\\var{a^2+b^2*t1}\\right)^\\frac{3}{2}\\right]^\\frac{2}{3}-\\var{a^2}\\right\\}^\\frac{3}{2}}\\]

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with $0\\leqslant s\\leqslant\\simplify{2/{b^2}}\\left(\\left(\\var{a^2+b^2*t2}\\right)^\\frac{3}{2}-\\left(\\var{a^2+b^2*t1}\\right)^\\frac{3}{2}\\right)$.

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