Times up!
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", "action": "warnifunattempted"}, "showresultspage": "oncompletion", "reverse": true, "browse": true, "allowregen": true}, "name": "Differentiation Quiz Apprentices", "showQuestionGroupNames": false, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questions": [{"name": "Differentiation 1 General Rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Catherine Palmer", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/423/"}, {"name": "TEAME CIT", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/591/"}, {"name": "Violeta CIT", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1030/"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"ans22": {"description": "", "definition": "pow[1]-1", "group": "Ungrouped variables", "templateType": "anything", "name": "ans22"}, "ans31": {"description": "", "definition": "pow[2]*num[1]", "group": "Ungrouped variables", "templateType": "anything", "name": "ans31"}, "ans12": {"description": "", "definition": "pow[0]-1", "group": "Ungrouped variables", "templateType": "anything", "name": "ans12"}, "ans11": {"description": "", "definition": "pow[0]", "group": "Ungrouped variables", "templateType": "anything", "name": "ans11"}, "pow": {"description": "", "definition": "shuffle(3..7)[0..3]", "group": "Ungrouped variables", "templateType": "anything", "name": "pow"}, "num": {"description": "", "definition": "shuffle(2..5)[0..2]", "group": "Ungrouped variables", "templateType": "anything", "name": "num"}, "ans32": {"description": "", "definition": "pow[2]-1", "group": "Ungrouped variables", "templateType": "anything", "name": "ans32"}, "ans21": {"description": "", "definition": "pow[1]*num[0]", "group": "Ungrouped variables", "templateType": "anything", "name": "ans21"}}, "tags": ["rebelmaths"], "rulesets": {}, "statement": "Remember the rule:
\nIf $y = ax^n$ then $\\frac{dy}{dx} = anx^{n-1}$
\n\nDifferentiate each of the following:
", "variable_groups": [], "preamble": {"js": "", "css": ""}, "functions": {}, "advice": "i)
\n$y = x^{\\var{pow[0]}}$
\n$\\frac{dy}{dx} = \\var{ans11}x^{\\var{pow[0]}-1} = \\var{ans11}x^{\\var{ans12}}$
\nii)
\n$y = \\var{num[0]}x^{\\var{pow[1]}}$
\n$\\frac{dy}{dx} = \\var{ans21}x^{(\\var{pow[1]}-1)} = \\var{ans21}x^{\\var{ans22}}$
\niii)
\n$y = \\var{num[1]}x^{\\var{pow[2]}}$
\n$\\frac{dy}{dx} = \\var{ans31}x^{(\\var{pow[2]}-1)} = \\var{ans31}x^{\\var{ans32}}$
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\n$\\frac{dy}{dx} =$ [[0]]
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\n$\\frac{dy}{dx} =$ [[0]]
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\n$\\frac{dy}{dx} =$ [[0]]
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\nIf $y = ax^n$ then $\\frac{dy}{dx} = anx^{n-1}$
\n\nDifferentiate each of the following:
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\n$y = x^{\\frac{1}{\\var{p1}}}$
\n$\\frac{dy}{dx} = \\var{ans11}x^{(\\frac{1}{\\var{p1}}-1)} = \\var{ans11}x^{\\var{ans12}}$
\nii)
\n$y = x^{\\frac{\\var{p2[0]}}{\\var{p2[1]}}}$
\n$\\frac{dy}{dx} = \\frac{\\var{p2[0]}}{\\var{p2[1]}}x^{(\\frac{\\var{p2[0]}}{\\var{p2[1]}}-1)} = \\var{ans21}x^{\\var{ans22}}$
\niii)
\n$y = \\var{num[0]}x^{\\var{p3}}$
\n$\\frac{dy}{dx} = \\var{ans31}x^{(\\var{p3}-1)} = \\var{ans31}x^{\\var{ans32}}$
\niv)
\n$y = \\var{num[1]}\\sqrt(x) = \\var{num[1]}x^{\\frac{1}{2}}$
\n$\\frac{dy}{dx} = \\frac{\\var{num[1]}}{2}x^{(\\frac{1}{2}-1)} = \\var{ans41}x^{\\var{ans42}}$
\nv)
\n$y = \\frac{\\var{n51}}{x^{\\var{n52}}} = \\var{n51}x^{-\\var{n52}}$
\n$\\frac{dy}{dx} = \\var{ans51}x^{(-\\var{n52}-1)} = \\var{ans51}x^{\\var{ans52}}$
\nvi)
\n$y = \\sqrt(x^\\var{n6}) = x^{\\frac{\\var{n6}}{2}}$
\n$\\frac{dy}{dx} = \\frac{\\var{n6}}{2}x^{(\\frac{\\var{n6}}{2}-1)} = \\var{ans61}x^{\\var{ans62}}$
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\n$\\frac{dy}{dx} =$ [[0]]
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\n$\\frac{dy}{dx} =$ [[0]]
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\nThe above graph shows a graph of a quadratic equation, it is your task to find this equation.
\nYou are given the two points of the curve with the x axis, $(\\var{b},0)$ and $(\\var{a},0)$, and the $y$-intercept at $(0,\\var{c})$ as indicated on the diagram.
", "preamble": {"css": "", "js": ""}, "functions": {"eqnline": {"type": "html", "language": "javascript", "definition": "// This function creates the board and sets it up, then returns an\n// HTML div tag containing the board.\n \n// The line is described by the equation \n// y = a*x + b\n\n// This function takes as its parameters the coefficients a and b,\n// and the coordinates (x2,y2) of a point on the line.\n\n// First, make the JSXGraph board.\n// The function provided by the JSXGraph extension wraps the board up in \n// a div tag so that it's easier to embed in the page.\nvar div = Numbas.extensions.jsxgraph.makeBoard('400px','400px',\n{boundingBox: [-13,22,13,-22],\n axis: false,\n showNavigation: false,\n grid: true\n});\n \n// div.board is the object created by JSXGraph, which you use to \n// manipulate elements\nvar board = div.board; \n\n// create the x-axis.\nvar xaxis = board.create('line',[[0,0],[1,0]], { strokeColor: 'black', fixed: true});\nvar xticks = board.create('ticks',[xaxis,2],{\n drawLabels: true,\n label: {offset: [-4, -10]},\n minorTicks: 0\n});\n\n// create the y-axis\nvar yaxis = board.create('line',[[0,0],[0,1]], { strokeColor: 'black', fixed: true });\nvar yticks = board.create('ticks',[yaxis,2],{\ndrawLabels: true,\nlabel: {offset: [-20, 0]},\nminorTicks: 0\n});\n\n// mark the two given points - one on the y-axis, and one at (x2,y2)\n\n\n\n\nboard.create('functiongraph',[function(x){ return (x-a)*(x-b);},-13,13]);\n\nreturn div;", "parameters": [["a", "number"], ["b", "number"], ["x2", "number"], ["y2", "number"]]}}, "rulesets": {}, "parts": [{"type": "gapfill", "gaps": [{"checkingtype": "absdiff", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "vsetrangepoints": 5, "scripts": {}, "marks": 1, "expectedvariablenames": [], "type": "jme", "showpreview": true, "showFeedbackIcon": true, "variableReplacements": [], "answer": "x^2-({a}+{b})x+{a}{b}", "showCorrectAnswer": true, "answersimplification": "all", "checkvariablenames": false, "variableReplacementStrategy": "originalfirst"}], "showFeedbackIcon": true, "variableReplacements": [], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "prompt": "Write the equation of the graph in the diagram.
\n$y=\\;$[[0]]
", "variableReplacementStrategy": "originalfirst"}, {"type": "gapfill", "gaps": [{"checkingtype": "absdiff", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "vsetrangepoints": 5, "scripts": {}, "marks": 1, "expectedvariablenames": [], "type": "jme", "showpreview": true, "showFeedbackIcon": true, "variableReplacements": [], "answer": "({a}+{b})/2", "showCorrectAnswer": true, "checkvariablenames": false, "variableReplacementStrategy": "originalfirst"}, {"checkingtype": "absdiff", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "vsetrangepoints": 5, "scripts": {}, "marks": 1, "expectedvariablenames": [], "type": "jme", "showpreview": true, "showFeedbackIcon": true, "variableReplacements": [], "answer": "(-{b}^2-{a}^2+2{a}{b})/4", "showCorrectAnswer": true, "checkvariablenames": false, "variableReplacementStrategy": "originalfirst"}], "showFeedbackIcon": true, "variableReplacements": [], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "prompt": "Find the coordinates of the turning point of this quadratic
\n$x=$[[0]]
\n$y=$[[1]]
", "variableReplacementStrategy": "originalfirst"}], "variablesTest": {"condition": "", "maxRuns": 100}, "variables": {"b": {"name": "b", "description": "", "templateType": "anything", "definition": "random(-5..5 except [0,a,-a])", "group": "Ungrouped variables"}, "a": {"name": "a", "description": "", "templateType": "anything", "definition": "random(-4..4 except 0)", "group": "Ungrouped variables"}, "y2": {"name": "y2", "description": "", "templateType": "anything", "definition": "x2*a+b", "group": "Ungrouped variables"}, "x2": {"name": "x2", "description": "", "templateType": "anything", "definition": "random(-3..3 except -1..1)", "group": "Ungrouped variables"}, "c": {"name": "c", "description": "", "templateType": "anything", "definition": "a*b", "group": "Ungrouped variables"}}, "metadata": {"description": "Students enter equation and turning point
", "licence": "Creative Commons Attribution 4.0 International"}, "ungrouped_variables": ["a", "x2", "b", "y2", "c"], "advice": "We know that the graph crosses the $x$-axis at both $(\\var{a},0)$ and $(\\var{b},0)$. Since this is a quadratic, we know our equations has two roots, and by the previous observation, they are at $\\var{a}$ and $\\var{b}$. Hence we can write our equation as $\\simplify{y=(x-{a})(x-{b})}$ which simplifies to $\\simplify{y=x^2-({a}+{b})x+({a}*{b})}$.
\n\nTo find the coefficients of the turning point of the quadratic, we know the x-coordinate of the turning point will correspond to the solution to $dy/dx=0$. So we get $\\simplify{2x-({a}+{b})}=0$ hence $\\simplify{x=({a}+{b})/2}$. We substitute this value of x back into the equation of the quadratic to find the corresponding y-coordinate.
", "tags": [], "variable_groups": [], "type": "question"}, {"name": "Applications of differentiation 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "cormac breen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/306/"}], "type": "question", "advice": "On differentiating we get $\\displaystyle \\frac{df}{dx}=\\simplify[std]{{3*a}x^2+{2*b}x+{c}}$.
\nTo find the stationary points we have to solve $\\displaystyle \\frac{df}{dx}=0$ for $x$.
\nSo we have to solve $\\simplify[std]{{3*a}x^2+{2*b}x+{c}=0}$.
\nNote that the quadratic factorises and the equation becomes $\\simplify[std]{({3a}x-{r1})(x-{r2})=0}$.
\nHence we have two stationary points: $x=\\simplify[std]{{r1}/{3a}}$ and $x=\\var{r2}$.
\nTo find out the types of these stationary points we look at the sign of $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6a}*x+{2*b}}$ at the stationary points.
\nIf $\\displaystyle \\frac{d^2f}{dx^2} \\lt 0 $ at a stationary point then it is a MAXIMUM.
\nIf $\\displaystyle \\frac{d^2f}{dx^2} \\gt 0 $ at a stationary point then it is a MINIMUM.
\nIf $\\displaystyle \\frac{d^2f}{dx^2} = 0 $ at a stationary point then we have to do more work!
\nAt $x=\\var{r2}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6*a*r2+2*b}}${lg1}$0$ hence is a {type1}.
\nAt $\\displaystyle x=\\simplify[std]{{r1}/{3a}}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{2*r1+2*b}}${lg2}$0$ hence is a {type2}.
\n", "parts": [{"type": "gapfill", "prompt": "
$f(x)=\\simplify[all,!collectNumbers,!noleadingminus]{{a}x^3+{b}x^2+{c}x+{d}}$
\n$f'(x)=$ [[2]]
\n$f''(x)=$ [[3]]
\n\nFind when $f'(x)=0$, hence find:
\n$x$-coordinate of the stationary point giving a minimum $=$ [[0]]
\n$x$-coordinate of the stationary point giving a maximum $=$ [[1]]
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\n\\(x=\\) [[0]]
\nInput the larger of the two \\(x\\) values.
\n\\(x=\\) [[1]]
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\nTo locate a turning point, differentite the function, set equal to zero and solve.
\n\\(f'(x)=6x^2-\\simplify{6*({a}+{b})x+6*{a}*{b}}=0\\)
\nDivide across by 6 to get the quadratic equation
\n\\(x^2-\\simplify{({a}+{b})x+{a}*{b}}=0\\)
\nThis has factors
\n\\((x-\\var{a})(x-\\var{b})=0\\)
\n\\(x-\\var{a}=0\\) or \\(x-\\var{b}=0\\)
\n\\(x=\\var{a}\\) or \\(x=\\var{b}\\)
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\nHere, the question takes you throught the stages needed to find the solution. The reason we differentiate is that the derivative of a function tells us its gradient at a given point, and we want to find where the function has gradient zero because when the gradient is zero we either have a maximum or a minimum point.
\nPart C
\nThe first part of this question is similar to parts A and B. The tricky bit is the second part! You need to work out the value of $t$ that produces the maximum piont but that is not the final answer - you need to use that value of $t$ to find the maximum height, which you do by substituting $t$ into the original function to find $y$.
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\n\\[ y = \\simplify{ {a}*x^2 + {b}x + {c}} \\]
\nFirstly, differentiate.
\n$\\displaystyle \\frac{dy}{dx}=$ [[1]]
\nGradient at $x=\\var{d}\\;$ is [[0]]
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\n$y=\\simplify {{f}x^2+{g}x+{h}}$
\nFirstly, find the first and second derivatives $y$.
\n$\\displaystyle \\frac{dy}{dx}=$ [[2]]
\n$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]
\n\nSecondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.
\n$x$-coordinate of the turning point $=$ [[0]]
\n$y$-coordinate of the turning point $=$ [[1]]
\nThe turning point is a [[4]]
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\nAt a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula
\n\\[ y=\\var{z}t-\\var{w}t^2. \\]
\nCalculate the maximum height reached by the missile.
\nFirstly, differentiate.
\n$\\displaystyle \\frac{dy}{dt}=$ [[0]]
\nNow use this result and your knowledge of differentiation to find the maximum height of the missile, rounding your answer to $2$ decimal places.
\n$y=$ [[1]]
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\nExam is due to be completed by Sunday 22nd April at 23:00 and is worth 2%.
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