// Numbas version: exam_results_page_options {"name": "MID-TRIMESTER EXAM REVISION 2020 T3", "metadata": {"description": "

Maths 1 B week 1 to week 6 content

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Find the inverse function $f^{-1}(x)$ for $f(x)=\\var{A}e^{\\var{B}x}$.

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If $y=f(x)$ then the inverse function $f^{-1}(y)$ is defined by which of the following statements.

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$f^{-1}(y)=1/f(x)$

", "

$f(y)=x$

", "

$f^{-1}(y)=x$

", "

$y=1/x$

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Which of the following statments follows directly from the original equation?

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$f(x)/\\var{A}=e^{\\var{B}x}$

", "

$f^{-1}(x)=\\frac{1}{\\var{A}}e^{-\\var{B}x}$

", "

$e^{f(x)}=\\var{A}+\\var{B}x$

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If $y/A=e^{Bx}$, what is $\\ln(y/A)$?

", "answer": "B*x", "checkingAccuracy": 0.001, "unitTests": [], "checkingType": "absdiff", "variableReplacements": [], "useCustomName": false, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "marks": 1, "checkVariableNames": false, "customMarkingAlgorithm": "", "vsetRange": [0, 1], "vsetRangePoints": 5, "valuegenerators": [{"name": "b", "value": ""}, {"name": "x", "value": ""}], "failureRate": 1, "type": "jme", "showPreview": true}, {"scripts": {}, "customName": "", "showFeedbackIcon": true, "prompt": "

What is $f^{-1}(x)$? [HINT: use $\\ln$ to indicate a natural logarithm and take care with variable names]

", "answer": "ln(x/{A})/{B}", "checkingAccuracy": 0.001, "unitTests": [], "checkingType": "absdiff", "variableReplacements": [], "useCustomName": false, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "answerSimplification": "All", "marks": "2", "checkVariableNames": false, "customMarkingAlgorithm": "", "vsetRange": [0, 1], "vsetRangePoints": 5, "valuegenerators": [{"name": "x", "value": ""}], "failureRate": 1, "type": "jme", "showPreview": true}], "type": "question"}, {"name": "Finding the formula for a shifted function", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Fig13-exp1.png", "/srv/numbas/media/question-resources/Fig13-exp1.png"], ["question-resources/Fig15-exp3.png", "/srv/numbas/media/question-resources/Fig15-exp3.png"], ["question-resources/Fig20-recip2.png", "/srv/numbas/media/question-resources/Fig20-recip2.png"], ["question-resources/Fig16-sin1.png", "/srv/numbas/media/question-resources/Fig16-sin1.png"], ["question-resources/Fig17-sin2.png", "/srv/numbas/media/question-resources/Fig17-sin2.png"], ["question-resources/Fig18-sin3.png", "/srv/numbas/media/question-resources/Fig18-sin3.png"], ["question-resources/Fig19-recip1.png", "/srv/numbas/media/question-resources/Fig19-recip1.png"], ["question-resources/Fig23-quad2.png", "/srv/numbas/media/question-resources/Fig23-quad2.png"], ["question-resources/Fig21-recip3.png", "/srv/numbas/media/question-resources/Fig21-recip3.png"], ["question-resources/Fig14-exp2.png", "/srv/numbas/media/question-resources/Fig14-exp2.png"], ["question-resources/Fig22-quad1.png", "/srv/numbas/media/question-resources/Fig22-quad1.png"], ["question-resources/Fig24-quad3.png", "/srv/numbas/media/question-resources/Fig24-quad3.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Owen Jepps", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1195/"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "ungrouped_variables": ["originalF", "shiftedF", "Fs", "figList"], "variable_groups": [], "parts": [{"displayType": "radiogroup", "distractors": ["", "", "", ""], "marks": 0, "prompt": "

{image(figlist[Fs[0]])}

If the blue curve represents is {originalF[Fs[0]]}, then the black dotted function is

", "minMarks": 0, "displayColumns": 0, "scripts": {}, "type": "1_n_2", "showCorrectAnswer": true, "choices": ["

{shiftedF[Fs[0]][0]}

", "

{shiftedF[Fs[0]][1]}

", "

{shiftedF[Fs[0]][2]}

", "

{shiftedF[Fs[0]][3]}

{image(figlist[Fs[1]])}

If the blue curve represents is {originalF[Fs[1]]}, then the black dotted function is

", "minMarks": 0, "displayColumns": 0, "scripts": {}, "type": "1_n_2", "showCorrectAnswer": true, "choices": ["

{shiftedF[Fs[1]][0]}

", "

{shiftedF[Fs[1]][1]}

", "

{shiftedF[Fs[1]][2]}

", "

{shiftedF[Fs[1]][3]}

{image(figlist[Fs[2]])}

If the blue curve represents is {originalF[Fs[2]]}, then the black dotted function is

", "minMarks": 0, "displayColumns": 0, "scripts": {}, "type": "1_n_2", "showCorrectAnswer": true, "choices": ["

{shiftedF[Fs[2]][0]}

", "

{shiftedF[Fs[2]][1]}

", "

{shiftedF[Fs[2]][2]}

", "

{shiftedF[Fs[2]][3]}

{image(figlist[Fs[3]])}

If the blue curve represents is {originalF[Fs[3]]}, then the black dotted function is

", "minMarks": 0, "displayColumns": 0, "scripts": {}, "type": "1_n_2", "showCorrectAnswer": true, "choices": ["

{shiftedF[Fs[3]][0]}

", "

{shiftedF[Fs[3]][1]}

", "

{shiftedF[Fs[3]][2]}

", "

{shiftedF[Fs[3]][3]}

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harder functions

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[
'$e^x$','$e^x$','$e^x$',
'$\\\\sin(x)$','$\\\\sin(x)$','$\\\\sin(x)$',
\"$1/x$\", \"$1/x$\", \"$1/x$\",
\"$x^2$\", \"$x^2$\", \"$x^2$\"
]

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For each of the following blue functions $f(x)$, identify the formula for the black dotted funtion.

", "advice": "

Look over the summary and examples in the lecture slides. For a function $f(x)$, and positive constant $c$

--- $f(x\\pm c)$ shifts $f(x)$ to the left or right by $c$

--- $f(x) \\pm c$ shifts $f(x)$ up or down by $c$

--- $f(cx)$ rescales $f(x)$ in the $x$-direction [shrinks if $c>1$, or expands if $c<1$]

--- $c f(x)$ rescales $f(x)$ by $c$ in the $y$-direction

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For each of the following functions $f(x)$, give the inverse function $f^{-1}(x)$.

To write the function, follow the syntax of the examples below:

--- $\\sqrt{x-1}$ is written sqrt(x-1);

--- $x^n$ is written x^n (for any power $n$);

--- $e^{x/2+1}$is written e^(x/2+1), $\\ln (2x)$is written ln(2x);

--- the sinusoids are written using their usual name, e.g. cos(x), sin(2x), tan(x-1)

--- the hyperbolic sinusoids are also written using their usual name, e.g. cosh(x), sinh(x-1), tanh(2x)

--- the inverse (hyperbolic) sinusoids are written using the 'arc-' prefix, e.g. arccos(x), arccosh(x), etc.

", "advice": "

To find the inverse of $f(x)$, write $f(y)=x$, and then keep re-arranging this equation until you reach $y=\\ldots$. During the re-arranging, remember that anything you do to the left-hand side must be done to the right-hand side as well. This process \"undoes\" the operations on $y$, in the reverse order in which they are done when calculating $f(y)$.

So, if $f(x)=\\cos(2x)$, we write

$f(y) = \\cos(2y) = x$

and then re-arrange until we only have $y$ on the left-hand side. To calculate the left-hand side, we double $y$ then find the cosine, so we undo this in the reverse order:

--- first, we take the inverse-cosine -- the arccos -- of both side, to get

$ 2y = \\arccos(x)$

--- second, we halve both side to get

$ y = \\arccos(x)/2$

So if $f(x)=\\cos(2x)$, then the inverse function $f^{-1}(x)=\\arccos(x)/2$

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[
'$\\\\simplify{x^2/{hm}+{hc}}$',
'$\\\\simplify{{hm}*sup(x,1/4)+{hc}}$',
'$\\\\ln(\\\\simplify{x^2/{hm}+{hc}})$',
'$e^{\\\\simplify{{hc}-{hm}x}}$',
'$\\\\sqrt{\\\\simplify{{hc}-{hm}x}}$',
'$\\\\sqrt{\\\\simplify{{hc}+x^2/{hm}}}$',
'$\\\\cos(\\\\simplify{x/{hm}+{hc}*\\pi})$',
'$\\\\arctan(\\\\simplify{{hm}*x-{hc}})$'
]

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harder functions

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[
'$\\\\ln\\\\simplify{{em}*x}$',
'$\\\\ln(\\\\simplify{x-{ec}})$',
'$e^{\\\\simplify{{em}*x}}$',
'$e^{\\\\simplify{x+{ec}}}$',
'$\\\\sqrt{\\\\simplify{x-{ec}}}$',
'$\\\\sqrt{\\\\simplify{x/{em}}}$',
'$\\\\cos(\\\\simplify{x+{ec}*\\pi})$',
'$\\\\arctan(\\\\simplify{{em}*x})$',
'$\\\\sinh(\\\\simplify{x+{ec}})$',
'$\\\\text{arccosh}(\\\\simplify{x/{em}})$'
]

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If {easierF[ef]}, then the inverse function $f^{-1}(x)=$ [[0]]

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If {harderF[hf]}, then the inverse function $f^{-1}(x)=$ [[0]]

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harder functions

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inverses of harder functions

", "definition": "[\n 'sqrt((x+1)/2)',\n '((x-2)/3)^3',\n '3(e^x-1)',\n '(1-ln(x))/2',\n '(x^2+5)/2',\n 'sqrt(2(x^2-5))',\n '(arccos(x)-pi)/2',\n '4(tan(x)-2)'\n]", "group": "Ungrouped variables"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "rulesets": {}, "variable_groups": [], "tags": [], "ungrouped_variables": ["harderF", "harderI", "hn"], "advice": "

So, if $f(x)=\\cos(x/2+1)$, we write

$f(y) = \\cos(y/2+1) = x$

and then re-arrange until we only have $y$ on the left-hand side. To calculate the left-hand side, we halve $y$, add 1, then find the cosine, so we undo this in the reverse order:

--- first, we take the inverse-cosine -- the arccos -- of both side, to get

$ y/2+1 = \\arccos(x)$

--- second, we subtract 1 from both side to get

$ y/2 = \\arccos(x) - 1$

--- finally, we double both sides to get

$ y = 2(\\arccos(x) - 1) = 2\\arccos(x) - 2$

So if $f(x)=\\cos(x/2+1)$, then the inverse function $f^{-1}(x)=2\\arccos(x)-2$

", "metadata": {"licence": "None specified", "description": ""}, "statement": "

For each of the following functions $f(x)$, give the inverse function $f^{-1}(x)$.

To write the function, you can write, e.g., $\\sqrt{x-1}$ as sqrt(x-1), $x^n$ as x^n, $\\cos(x/2)$ as cos(x/2)

", "preamble": {"js": "", "css": ""}, "parts": [{"prompt": "

If {harderF[hn[0]]}, then the inverse function $f^{-1}(x)=$ [[0]]

If {harderF[hn[1]]}, then the inverse function $f^{-1}(x)=$ [[0]]

If {harderF[hn[2]]}, then the inverse function $f^{-1}(x)=$ [[0]]

If {harderF[hn[3]]}, then the inverse function $f^{-1}(x)=$ [[0]]

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If {easierF[en[0]]}, then the inverse function $f^{-1}(x)=$ [[0]]

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If {easierF[en[1]]}, then the inverse function $f^{-1}(x)=$ [[0]]

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If {easierF[en[2]]}, then the inverse function $f^{-1}(x)=$ [[0]]

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If {easierF[en[3]]}, then the inverse function $f^{-1}(x)=$ [[0]]

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For each of the following functions $f(x)$, give the inverse function $f^{-1}(x)$.

To write the function, you can write, e.g., $\\sqrt{x-1}$ as sqrt(x-1), $x^n$ as x^n, $\\cos(x/2)$ as cos(x/2)

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So, if $f(x)=\\cos(2x)$, we write

$f(y) = \\cos(2y) = x$

and then re-arrange until we only have $y$ on the left-hand side. To calculate the left-hand side, we double $y$ then find the cosine, so we undo this in the reverse order:

--- first, we take the inverse-cosine -- the arccos -- of both side, to get

$ 2y = \\arccos(x)$

--- second, we halve both side to get

$ y = \\arccos(x)/2$

So if $f(x)=\\cos(2x)$, then the inverse function $f^{-1}(x)=\\arccos(x)/2$

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Testing your ability to identify the formulas defining piecewise-defined functions

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Testing your ability to identify the formulas defining piecewise-defined functions

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The graph above is a plot of the formula

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Is not a function

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{image(figList[whichFig[0]])}

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{image(figList[whichFig[1]])}

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For each plot, identify whether the graph corresponds to a function or not (using the appropriate test)

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For each plot, identify whether the graph corresponds to a function or not (using the appropriate test)

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Even functions are symmetric about the $y$-axis ($f(-x)=f(x)$), while odd functions are antisymmetric about the $y$-axis ($f(-x)=-f(x)$). However, functions can be neither symmetric nor antisymmetric.

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Odd

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Even

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Neither

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Odd

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Even

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Neither

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Odd

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Even

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Neither

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For each graph, indicate whether the function is odd, even or neither. Note that the grids have the same units in the $x$ and $y$ directions

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['resources/question-resources/Fig01-even.png','resources/question-resources/Fig02-odd.png','resources/question-resources/Fig03-neither.png','resources/question-resources/Fig04-odd.png','resources/question-resources/Fig05-even.png','resources/question-resources/Fig06-neither.png','resources/question-resources/Fig07-odd.png','resources/question-resources/Fig08-neither.png','resources/question-resources/Fig09-neither.png','resources/question-resources/Fig10-neither.png','resources/question-resources/Fig11-odd.png','resources/question-resources/Fig12-even.png']

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Odd

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Even

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Neither

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Odd

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Even

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Neither

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For each graph, indicate whether the function is odd, even or neither. Note that the grids have the same units in the $x$ and $y$ directions

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", "variablesTest": {"condition": "", "maxRuns": 100}, "metadata": {"licence": "None specified", "description": ""}, "type": "question"}, {"name": "Recognising the graphs of some important functions", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Fig04-x.png", "/srv/numbas/media/question-resources/Fig04-x.png"], ["question-resources/Fig09-sqrt.png", "/srv/numbas/media/question-resources/Fig09-sqrt.png"], ["question-resources/Fig08-x3.png", "/srv/numbas/media/question-resources/Fig08-x3.png"], ["question-resources/Fig06--mxc.png", "/srv/numbas/media/question-resources/Fig06--mxc.png"], ["question-resources/Fig05-mxc.png", "/srv/numbas/media/question-resources/Fig05-mxc.png"], ["question-resources/Fig07-x2.png", "/srv/numbas/media/question-resources/Fig07-x2.png"], ["question-resources/Fig11-cos.png", "/srv/numbas/media/question-resources/Fig11-cos.png"], ["question-resources/Fig13-exp.png", "/srv/numbas/media/question-resources/Fig13-exp.png"], ["question-resources/Fig15-recip.png", "/srv/numbas/media/question-resources/Fig15-recip.png"], ["question-resources/Fig10-sin.png", "/srv/numbas/media/question-resources/Fig10-sin.png"], ["question-resources/Fig14-lnx.png", "/srv/numbas/media/question-resources/Fig14-lnx.png"], ["question-resources/Fig12-tan.png", "/srv/numbas/media/question-resources/Fig12-tan.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Owen Jepps", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1195/"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "variable_groups": [], "metadata": {"licence": "None specified", "description": ""}, "parts": [{"matrix": ["1", 0, 0, 0], "variableReplacementStrategy": "originalfirst", "type": "1_n_2", "scripts": {}, "maxMarks": 0, "minMarks": 0, "displayType": "radiogroup", "displayColumns": 0, "marks": 0, "shuffleChoices": true, "distractors": ["", "", "", ""], "prompt": "

{image(figList[whichFunc[0]])}

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{funcName[options[whichFunc[0]][3]]}

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{funcName[options[whichFunc[1]][0]]}

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{funcName[options[whichFunc[1]][1]]}

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{funcName[options[whichFunc[1]][2]]}

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{funcName[options[whichFunc[1]][3]]}

{image(figList[whichFunc[2]])}

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{funcName[options[whichFunc[2]][0]]}

", "

{funcName[options[whichFunc[2]][1]]}

", "

{funcName[options[whichFunc[2]][2]]}

", "

{funcName[options[whichFunc[2]][3]]}

{image(figList[whichFunc[3]])}

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{funcName[options[whichFunc[3]][3]]}

{image(figList[whichFunc[4]])}

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{funcName[options[whichFunc[4]][1]]}

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{funcName[options[whichFunc[5]][3]]}

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For each graph, identify the plotted function

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0: '$x$',

1: '$\\\\frac{x}{2}+3$',

2: '$4-x$',

3: '$x^2$',

4: '$x^3$',

5: '$\\\\sqrt{x]$',

6: '$\\\\sin x$',

7: '$\\\\cos x$',

8: '$\\\\tan x$',

9: '$e^x$',

10: '$\\\\ln x$',

11: '$\\\\frac{1}{x}$']

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For each graph, identify the plotted function. Note that the grids have the same units in the $x$ and $y$ directions

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2: '$4-x$',

3: '$x^2$',

4: '$x^3$',

5: '$\\\\sqrt{x]$',

6: '$\\\\sin x$',

7: '$\\\\cos x$',

8: '$\\\\tan x$',

9: '$e^x$',

10: '$\\\\ln x$',

11: '$\\\\frac{1}{x}$']

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{image(figList[whichFunc[0]])}

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{funcName[options[whichFunc[0]][3]]}

{image(figList[whichFunc[1]])}

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Consider what the definition of a function says about how many y-values can be obtained for a given x-value.

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The graph of a function, and the graph of its inverse, are reflections of one another across which line?

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The line $y=x$

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The $y$-axis

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The $x$-axis

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The Plimsoll line

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This question is about the relationship between the graph of a function and the graph of its inverse.

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Identify the test used to distinguish whether a graph represents a function.

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The horizontal line test

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The vertical line test

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The $y=x$ test

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The centenary test

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What is the test we use to check whether a function has an inverse, based on its graph?

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Consider what the definition of a function says about how many y-values can be obtained for a given x-value.

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Identify the test used to distinguish whether a graph represents a function.

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This question is about a general, key approach to distinguishing functions that have inverses from functions that do not.

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Consider what the definition of a function says about how many y-values can be obtained for a given x-value.

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This question is about a general, key approach to distinguishing functions from other types of relations, graphically

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What is the test that we use to check that the graph of a relationship represents a function?

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The vertical line test

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The horizontal line test

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The $y=x$ test

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The Myers Briggs test

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Identify the test used to distinguish whether a graph represents a function.

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Evaluate a limit (x-a)/(x^2+(a+b)x+ab)

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Evaluate the following limit

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\$\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{x^2-\\simplify{{a}+{b}}x+\\simplify{{a}*{b}}}\$

We can factorise the denominator

\$\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{x-\\var{a}}{(x-\\var{a})(x-\\var{b})}\$

Cancel out the common factor

\$\\displaystyle{\\lim_{x \\to \\var{a}}}\\frac{1}{x-\\var{b}}\$

Insert the value \$\\var{a}\$ in for \$x\$ to evaluate the limit

\$-\\frac{1}{\\simplify{{b}-{a}}}\$

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Input as a fraction or an integer.

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$\\displaystyle \\simplify{Limit((x + { -a}) / (x ^ 2 + { -a -b} * x + {a * b}),x,{a}) }=\\;$[[0]] (input as a fraction or as an integer).

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Find the limit of the following function.

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Note that on putting $x=\\var{a}$ into $\\displaystyle \\simplify{(x + { -a}) / (x ^ 2 + { -a -b} * x + {a * b}) }$ we get a $0/0$ case and so we have to do more work.

You can factorise $\\simplify{x ^ 2 + { -a -b} * x + {a * b}}$ and then see what happens.

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A graph is drawn. A student is to identify the derivative of this graph from four other graphs.

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This is a non-calculator question

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There are two main ways of thinking about this. Different people have different preferences.

1) This is identical to being asked: \"A displacement-time graph is given, select the corresponding velocity-time graph\". Hence use the same reasoning as in previous questions (when is the velocity positive, negative, zero.)

2) The derivative tells you about the gradient/slope of the original graph. Thus, you want to ask yourself \"when is the gradient positive, when is the gradient negative, when is the gradient zero\".

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Select the graph that shows the derivative of the graph above.

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{plot({n})}

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{plot({n+2})}

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{plot({n+3})}

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For the function below

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Find the value of $f(\\var{Input1})$ [[0]]

Find the value of $f(\\var{Input2})$ [[1]]

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1. Find \\[\\lim_{x \\to \\var{a}}(\\simplify[std]{{b}x+{c}})\\]

Limit = [[0]].

2.Find \\[\\lim_{x \\to \\var{a1}}(\\simplify[std]{{b1}x^2+{c1}x+{d1}})\\]

Limit = [[1]].

3. Find \\[\\lim_{x \\to \\var{a2}}\\left(\\simplify[std]{({b2}x+{c2})/({b3}x+{c3})}\\right)\\]

Limit = [[2]]

Enter all numbers as either integers or fractions but not as decimals.

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Enter all numbers as either integers or fractions but not as decimals.

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1. To find this limit we simply substitute $x=\\var{a}$ into $\\simplify[std]{{b}x+{c}}$ to get \\[\\lim_{x \\to \\var{a}}(\\simplify[std]{{b}x+{c}})=\\simplify[]{{b}*{a}+{c}={ans1}}\\]

2. Similarly to find this limit we simply substitute $x=\\var{a1}$ into $\\simplify[std]{{b1}x^2+{c1}x+{d1}}$ to get \\[\\lim_{x \\to \\var{a1}}(\\simplify[std]{{b1}x^2+{c1}x+{d1}}) =\\simplify[]{{b1}*{a1}^2+{c1}*{a1}+{d1}={ans2}}\\]

3. Once again we could simply substitute $x=\\var{a2}$ into $\\displaystyle \\simplify[std]{({b2}x+{c2})/({b3}x+{c3})}$. However before doing this we must make sure that the denominator is not $0$ as otherwise we have a problem and the limit may not exist.

But $\\simplify[]{{b3}*{a2}+{c3}={b3*a2+c3} }\\neq 0$ and so we can make the substitution safely.

So \\[\\lim_{x \\to \\var{a2}}\\left(\\simplify[std]{({b2}x+{c2})/({b3}x+{c3})}\\right)=\\simplify[]{({b2}*{a2}+{c2})/({b3}*{a2}+{c3})}=\\simplify[all,fractionNumbers]{{b2*a2+c2}/{b3*a2+c3}}\\]

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Find the following limits.

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Using simple substitution to find $\\lim_{x \\to a} bx+c$, $\\lim_{x \\to a} bx^2+cx+d$ and $\\displaystyle \\lim_{x \\to a} \\frac{bx+c}{dx+f}$ where $d\\times a+f \\neq 0$.

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Solve an exponential equation

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Determine the value for \$x\$ that satisfies the equation: \$\\var{a}e^{\\var{m}x+\\var{c}}=\\var{k}\$

", "advice": "

\$\\var{k}=\\var{a}e^{\\var{m}x+{\\var{c}}}\$

\$\\frac{\\var{k}}{\\var{a}}=e^{\\var{m}x+\\var{c}}\$

\$ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)=\\var{m}x+\\var{c}\$

\$ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}=\\var{m}x\$

\$\\frac{ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}}{\\var{m}}=x\$

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Input your answer correct to three decimal places.

\$x = \$ [[0]]

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JSXGraph code based on original by Christian Lawson-Perfect

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{eqnline(a,b,x2,y2)}

The graph above shows a line cutting the curve at $x=4$ and $x=4+h$.

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Use the slider to show that, as $h$ tends to $0$ the line becomes the tangent to the curve at $x=4$

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$g: \\mathbb{R} \\rightarrow \\mathbb{R}, g(x)=\\frac{ax}{x^2+b^2}$. Find stationary points and local maxima, minima. Using limits, has $g$ a global max, min?

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Let $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ be the function given by:
\\[g(x)=\\simplify{{a}*x/(x^2+{b}^2)}\\]

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The function $g(x)$ is continuous and differentiable at all points in $\\mathbb{R}$.

Using the quotient rule for differentiation we see that
\\[\\begin{eqnarray*}g'(x)&=&\\simplify{({a}*(x^2+{b^2})-{2*a}*x^2)/(x^2+{b^2})^2}\\\\ &=&\\simplify{({-a}*(x-{b})(x+{b}))/(x^2+{b^2})^2} \\end{eqnarray*} \\]

Stationary Points.

The stationary points are given by solving $g'(x)=0$.

$g'(x)=0 \\Rightarrow \\simplify{{-a}*(x-{b})(x+{b})=0} \\Rightarrow x=\\var{b} \\mbox{ or } x=\\var{-b}$

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

Using the quotient rule for differentiation we see that
\\[\\begin{eqnarray*}g''(x)&=&\\simplify[std]{({-2*a}*x*(x^2+{b^2})^2+{4*a}*x*(x^2-{b^2})(x^2+{b^2}))/(x^2+{b^2})^4}\\\\ &=&\\simplify[std]{({2*a}*x*(x^2-{3*b^2}))/(x^2+{b^2})^3} \\end{eqnarray*} \\]

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

For $x= \\var{lma}$ we have: \\[g''(\\var{lma})= \\simplify[std]{-{abs(a)}/{2*b^3}} \\lt 0\\]

Hence is a local maximum.

Evaluating the function at $x=\\var{lma}$ gives $g(\\var{lma})=\\var{valmax}$ to 3 decimal places.

For $x= \\var{lmi}$ we have: \\[g''(\\var{lmi})= \\simplify[std]{{abs(a)}/{2*b^3}} \\gt 0\\]

Hence is a local minimum.

Evaluating the function at $x=\\var{lmi}$ gives $g(\\var{lmi})=\\var{valmin}$ to 3 decimal places.

The Limits.

If we divide $g(x)$ top and bottom by $x^2$ (OK as $x \\neq 0$ at any time) we obtain: \\[g(x)=\\simplify[std]{({a}/x)/(1+{b^2}/x^2)}\\]

Then using the fact that $\\displaystyle \\frac{1}{x}$ and $\\displaystyle \\frac{1}{x^2}$ both tend to $0$ as $ x \\rightarrow \\pm\\infty$ we see that

$\\displaystyle \\lim_{x \\to \\infty}g(x)=\\frac{0}{1}=0$ and similarly

$\\lim_{x \\to -\\infty}g(x)=0$

Global Maximum and Minimum

Since $g$ has a finite limit of $0$ as $x \\rightarrow \\pm\\infty$ and we have that $0$ lies between the local minimum $\\var{valmin}$ and the local maximum $\\var{valmax}$

Then:

Global Maximum: The local maximum of $g$ we have found at $x=\\var{lma}$ must be a global maximum and similarly,

Global Minimum: The local minimum of $g$ we have found at $x=\\var{lmi}$ must be a global minimum.

So we have shown \\[\\forall x \\in \\mathbb{R},\\;\\;\\var{valmin} \\le g(x) \\le \\var{valmax}\\]

\n ", "parts": [{"matrix": [1, 0], "showCorrectAnswer": true, "marks": 0, "useCustomName": false, "choices": ["

Yes

", "

"], "scripts": {}, "customName": "", "type": "1_n_2", "displayColumns": 0, "showCellAnswerState": true, "minMarks": 0, "showFeedbackIcon": true, "displayType": "radiogroup", "prompt": "\n

Is $g(x)$ continuous at all points of $\\mathbb{R}$?

\n \n

Choose Yes or No.

\n \n ", "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "shuffleChoices": false, "extendBaseMarkingAlgorithm": true, "maxMarks": 0, "customMarkingAlgorithm": "", "unitTests": []}, {"gaps": [{"showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "valuegenerators": [{"value": "", "name": "x"}], "notallowed": {"partialCredit": 0, "strings": ["^", "x*x", "xx", "x x"], "showStrings": false, "message": "

Factorise the expression

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Factorise the expression

The first derivative of $g$ can be written in the form $\\displaystyle \\frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)=(x^2+\\var{b^2})^2$ are polynomials.

Input the numerator $p(x)$ of the first derivative of $g$ here, factorised into a product of two linear factors in the form
\\[p(x)=c(x-a)(x-b)\\]for suitable integers $a$, $b$ and $c$:

$p(x)\\;=\\;$[[0]]

Yes

", "

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Is $g(x)$ differentiable at all points of $\\mathbb{R}$?

\n \n

Choose Yes or No.

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Find the stationary points of $g$.

\n \n

Least stationary point: [[0]]

\n \n

Greatest stationary point: [[1]]

\n \n ", "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "customName": "", "variableReplacements": [], "useCustomName": false, "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "unitTests": []}, {"gaps": [{"showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "valuegenerators": [{"value": "", "name": "x"}], "notallowed": {"partialCredit": 0, "strings": ["x^3"], "showStrings": false, "message": "

Factorise the expression as asked in the question.

"}, "useCustomName": false, "scripts": {}, "customName": "", "type": "jme", "checkingType": "absdiff", "musthave": {"partialCredit": 0, "strings": ["(", ")"], "showStrings": false, "message": "

Factorise the expression

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The second derivative of $g$ can be written in the form $\\displaystyle \\frac{r(x)}{s(x)}$ where $r(x)$ and $s(x)=(x^2+\\var{b^2})^3$ are polynomials.

Input the numerator $r(x)$ of the second derivative of $g$ here, factorised into a product of a linear factor and a quadratic factor in the form
\\[r(x)=a_1x(x^2-a_2)\\] for suitable integers $a_1$, $a_2$

$r(x)=\\;\\;$ [[0]]

Hence find all local maxima and minima given by the stationary points

Local maximum is at $x=\\;\\;$ [[1]] and the value of the function at the local maximum (to 3 decimal places)= [[2]]

Local minimum is at $x=\\;\\;$ [[3]] and the value of the function at the local minimum (to 3 decimal places) = [[4]]

\n ", "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "customName": "", "variableReplacements": [], "useCustomName": false, "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "unitTests": []}, {"gaps": [{"displayColumns": 0, "matrix": [0, 0, 0, 0, 1], "showFeedbackIcon": true, "showCorrectAnswer": true, "marks": 0, "displayType": "radiogroup", "distractors": ["", "", "", "", ""], "variableReplacementStrategy": "originalfirst", "type": "1_n_2", "choices": ["

$-\\infty$

", "

$\\infty$

", "

$\\var{b}$

", "

$\\var{valmax}$

", "

$0$

"], "scripts": {}, "customName": "", "variableReplacements": [], "useCustomName": false, "shuffleChoices": true, "extendBaseMarkingAlgorithm": true, "minMarks": 0, "maxMarks": 0, "customMarkingAlgorithm": "", "showCellAnswerState": true, "unitTests": []}, {"displayColumns": 0, "matrix": [0, 0, 0, 0, 1], "showFeedbackIcon": true, "showCorrectAnswer": true, "marks": 0, "displayType": "radiogroup", "distractors": ["", "", "", "", ""], "variableReplacementStrategy": "originalfirst", "type": "1_n_2", "choices": ["

$-\\infty$

", "

$\\infty$

", "

$\\var{a}$

", "

$\\var{valmin}$

", "

$0$

"], "scripts": {}, "customName": "", "variableReplacements": [], "useCustomName": false, "shuffleChoices": true, "extendBaseMarkingAlgorithm": true, "minMarks": 0, "maxMarks": 0, "customMarkingAlgorithm": "", "showCellAnswerState": true, "unitTests": []}, {"displayColumns": 0, "matrix": [1, 0], "showFeedbackIcon": true, "showCorrectAnswer": true, "marks": 0, "displayType": "radiogroup", "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "type": "1_n_2", "choices": ["

Yes

", "

"], "scripts": {}, "customName": "", "variableReplacements": [], "useCustomName": false, "shuffleChoices": true, "extendBaseMarkingAlgorithm": true, "minMarks": 0, "maxMarks": 0, "customMarkingAlgorithm": "", "showCellAnswerState": true, "unitTests": []}, {"displayColumns": 0, "matrix": [1, 0], "showFeedbackIcon": true, "showCorrectAnswer": true, "marks": 0, "displayType": "radiogroup", "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "type": "1_n_2", "choices": ["

Yes

", "

What are the following limits?

1) $\\lim_{x \\to \\infty}g(x)\\;\\;$

Choose one of the following [[0]]

2) $\\lim_{x \\to -\\infty}g(x)$

Choose one of the following [[1]]

Does $g$ have a finite global maximum? Click on Yes or No
[[2]]

\n ", "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "customName": "", "variableReplacements": [], "useCustomName": false, "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "unitTests": []}]}, {"name": "Find the limit of an algebraic fraction", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}], "advice": "

Note that on putting $x=\\var{a}$ into $\\displaystyle \\simplify{(x + { -a}) / (x ^ 2 + { -a -b} * x + {a * b}) }$ we get a $0/0$ case and so we have to do more work.

You can factorise $\\simplify{x ^ 2 + { -a -b} * x + {a * b}}$ and then see what happens.

", "variable_groups": [], "showQuestionGroupNames": false, "ungrouped_variables": ["a", "b"], "tags": ["checked2015", "MAS2224"], "variablesTest": {"condition": "", "maxRuns": 100}, "question_groups": [{"pickQuestions": 0, "pickingStrategy": "all-ordered", "questions": [], "name": ""}], "type": "question", "statement": "

Find the limit of the following function.

", "preamble": {"js": "", "css": ""}, "functions": {}, "parts": [{"marks": 0, "scripts": {}, "prompt": "

$\\displaystyle \\simplify{Limit((x + { -a}) / (x ^ 2 + { -a -b} * x + {a * b}),x,{a}) }=\\;$[[0]] (input as a fraction or as an integer).

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Input as a fraction or an integer.

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Let the function $f$ be given by $\\displaystyle f(t)=\\simplify{({a3} * t + {b3}) / ({a2} * t ^ 2 + {2 * b2 * a2} * t + {c2 + a2 * b2 ^ 2}) }$

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Find the following limit:

$\\displaystyle \\simplify{Limit(f(t),t,{d2}) }= \\;$[[0]].

Enter your answer as a fraction or an integer and not as a decimal.

Enter as a fraction or an integer and not as a decimal.

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Also find:

$\\displaystyle \\simplify{Limit(f(t),t,a) }= \\;$[[0]], where $a \\in \\mathbb{R}$ is any point.

Why can we evaluate this limit? [[1]]

", "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "showFeedbackIcon": true, "marks": 0, "unitTests": [], "useCustomName": false, "gaps": [{"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": true, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "({a3}*a+{b3})/({a2}*a^2+{2*a2*b2}*a+{c2+a2*b2^2})", "valuegenerators": [{"value": "", "name": "a"}], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": 1, "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}, {"customName": "", "customMarkingAlgorithm": "", "showCellAnswerState": true, "maxMarks": 0, "scripts": {}, "minMarks": 0, "matrix": [0, 1, 0, 0, 0], "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "type": "1_n_2", "shuffleChoices": true, "showFeedbackIcon": true, "marks": 0, "displayType": "radiogroup", "unitTests": [], "distractors": ["Not true as $f(t)=0$ for a value of $t$.", "", "", "", ""], "displayColumns": 0, "useCustomName": false, "variableReplacements": [], "choices": ["

Because $f(t) \\neq 0, \\; \\forall t \\in \\mathbb{R}$.

", "

Because $f$ is continuous at all points in $\\mathbb{R}$.

", "

Because $f$ is a function defined in terms of polynomials.

", "

Because all ratios of polynomials are continuous.

", "

Because $f$ is differentiable at all points.

"]}], "variableReplacements": []}], "rulesets": {}, "advice": "

Graph of $f$.

{plotf(a2,b2,c2,a3,b3,stat1,stat2)}

Note that $\\simplify{{a2} * t ^ 2 + {2 * b2 * a2} * t + {c2 + a2 * b2 ^ 2} ={a2}*(t+{b2})^2+{c2}} \\gt 0$.

Hence the denominator of $f(t) \\neq 0,\\;\\forall t \\in \\mathbb{R}$ and so $f$ is continuous at all points in $\\mathbb{R}$.

This means that in part a) we can take the limit by simply subsituting $t=\\var{d2}$ into the expression for $f(t)$ and we get:

\\[\\lim_{x \\to \\var{d2}}f(t)=\\simplify[all,!otherNumbers,fractionNumbers,!collectNumbers]{({a3} * {d2} + {b3}) / ({a2} * {d2}^ 2 + {2 * b2 * a2} * {d2}+ {c2 + a2 * b2 ^ 2})={v}/{w} }\\]

Similarly in part b) we have :

\\[\\lim_{x \\to a}f(t)=\\simplify[all,!collectNumbers,!otherNumbers,fractionNumbers]{({a3} * a + {b3}) / ({a2} * a^ 2 + {2 * b2 * a2} * a+ {c2 + a2 * b2 ^ 2})}\\]

As noted above we can find this limit by simply putting $t=a$ into the formula for the function as $f$ is continuous at all points in $\\mathbb{R}$.

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A basic introduction to differentiation

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Differentiate the following polynomials.

Note: some questions may not include all the possible terms.

", "advice": "

If $y=ax^n$,

$\\frac{dy}{dx}=anx^{n-1}$ for all rational $n$.

We'll take one of the terms from Part a as an example:

$\\var{cc[0]}x^\\var{cp}$

All we have to do to terms where $x$ is to a power of anything is times the coefficient of $x$ by the original power, and then take one away from the original power.

If you are not familiar with this kind of work, these instructions may sound confusing, but it is much easier once you have seen it in practice.

We take

$\\var{cc[0]}x^\\var{cp}$

and times $\\var{cc[0]}$ by $\\var{cp}$, to get

$(\\var{cc[0]}*\\var{cp})x^\\var{cp}=\\simplify{{cc[0]}*{cp}x^{cp}}$.

We then subtract one from the original power, $\\var{cp}$.

This gives us the final answer of

$\\simplify{{cc[0]}*{cp}x^{cp-1}}$.

Remember, don't be confused if there is no coefficient. The fact the term is there means the coefficient must be $1$, but we don't tend to write it out as, for example $1x$, we just say $x$.

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$\\simplify[basic,zeroterm,zerofactor,unitfactor]{{ac[1]}x^{ap}+{bc[1]}x^{bp}+{cc[1]}x^{cp}+{dc[1]}x^{dp}+{ec[1]}x+{fc[1]}}$

$f'(x)=$ [[0]]

$f''(x)=$ [[1]]

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$\\simplify[basic,zeroterm,zerofactor,unitfactor]{{ac[2]}x^{ap}+{bc[2]}x^{bp}+{cc[2]}x^{cp}+{dc[2]}x^{dp}+{ec[2]}x+{fc[2]}}$

$f'(x)=$ [[0]]

$f''(x)=$ [[1]]

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Standard derivatives asked for (e.g. $x^n$, $1/x^n$, $\\sqrt(x)$, $\\ln(x)$, $\\sin(x)$, etc.) .

", "licence": "None specified"}, "statement": "", "advice": "

See 10.1, 10.2, 10.4 and 10.5.

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[
0[\"(\"+b0[0]+\"*\",\")\"],
1[\"(\",\")^\"+n1],
2[\"(\",\")^\"+n2],
3[\"1/(\",\")\"],
4[\"1/((\",\")^\"+n3+\")\"],
5[\"sqrt(\",\")\"],
6[\"sin(\",\")\"],
7[\"cos(\",\")\"],
8[\"e^(\",\")\"],
9[\"ln(\",\")\"]
]

don't use 0 for product rule

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Differentiate the following with respect to $\\simplify{{expression(var)}}$.

$\\simplify[fractionNumbers, all]{{f0[0]}+{f0[1]}}$. [[0]]

$\\simplify[fractionNumbers, all]{{f0[2]}+{f0[3]}+{f0[4]}}$. [[1]]

$\\simplify[fractionNumbers, all]{{f0[5]}+{f0[6]}+{f0[7]}+{f0[8]}}$. [[2]]

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Q1 is true/false question covering some core facts, notation and basic examples. Q2 has two functions for which second derivative needs to be determined.

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Correct answers are not given for the first question, because you should read the information provided to determine what is correct.

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Which of the following are true and which are false?

In the following, $f(x) = \\sin(x)$ and $g(t) = \\cos(t)$.

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True

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False

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$f(x) = \\simplify{{a1}*x^{n1} + {b1}sin(x)}$. What is $\\frac{d^2 f}{dx^2}$?

[[0]]

$g(t) = \\simplify{{a2}ln(t) + {b2} cos(t)}$. What is $g''(t)$?

[[1]]

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Given a set of curves on axes, generated from a function and its first two derivatives, identify which curve corresponds to which derivative.

"}, "statement": "\n

The following graph (which may take a little while to load) shows three curves: a solid line, a dashed line and a dotted line. These curves represent a function and its derivatives; if we call the function $f$, then one curve represents $f$ and the other two curves represent $f'$ and $f''$.

{geogebra_applet('BsRYG6PV',defs)}

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Here are some questions to consider for this problem.

What happens to even functions when you differentiate them? What about odd functions?
Can you identify any maxima and minima and line them up with zeroes in other functions?

", "parts": [{"choices": ["

Solid line

", "

Dashed line

", "

Dotted line

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For each curve, select the corresponding derivative. You will score $2$ points for each curve correctly identified, and $-1$ point for each curve incorrectly identified.

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$f$

", "

$f'$

", "

$f''$

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$f(x) = \\var{a}$

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$f(x) = \\var{b}x^2 -\\var{c}x+\\var{d}$

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$f(x) = x^2(1-\\var{f}x)$

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$f(x) = -\\frac{\\var{g}}{x^\\var{h}}$

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$f(x) = \\sqrt{x}$

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$f(x) = x^2 + e^x$

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Some basic derivatives.

rebelmaths

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Differentiate the following with respect to $x$.

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(i) {plotgraph(1,x11,x12,a1,b1,c1,{x11-2},{x12+2},-3,15)}

This is the graph of the function $f(x) = \\simplify{{a1}*(x+{b1})+{c1}}$.

Determine the minimum and maximum $x$-values of the region. Enter the minimum first.

[[0]],[[1]]

(ii) {plotgraph(2,x21,x22,a2,0,0,-0.5,10,-1.2,1.2)}

This is the graph of the function $f(x) = \\sin(x-\\frac{\\var{a2*4}}{4}\\pi)$.

Determine the minimum and maximum $x$-values of the region.

Enter the minimum first. Enter the values exactly in terms of $\\pi$. To enter $\\pi$, type `pi'.

[[2]], [[3]]

(iii) {plotgraph(3,x31,x32,a3,b3,0,-1,x32*1.6,-1.2*a3,1.2*a3)}

This curve has equation $y = \\simplify{{a3}*sin(x/{b3})}$.

Determine the minimum and maximum $x$-values of the region.

Enter the minimum first. Enter the values exactly in terms of $\\pi$. To enter $\\pi$, type `pi'.

[[4]], [[5]]

"}], "statement": "", "metadata": {"description": "

Three graphs are given with areas underneath them shaded. Student is asked to determine the minimum and maximum $x$-values of the regions. This will involve solving a linear equation and two trigonmetric equations.

", "licence": "Creative Commons Attribution 4.0 International"}, "tags": [], "rulesets": {}, "preamble": {"js": "", "css": ""}, "functions": {"plotgraph": {"definition": "// Shading under a graph! This function plots one of three graphs\n// depending on the value of q. It shades the area between the\n// graph and x-axis depending between x1 and x2\n\n\n\n// First, make the JSXGraph board.\nvar div = Numbas.extensions.jsxgraph.makeBoard(\n '300px',\n '250px',\n {\n boundingBox: [xmin,ymax,xmax,ymin],\n axis: false,\n showNavigation: false,\n grid: false\n }\n);\n\n\n\n// div.board is the object created by JSXGraph, which you use to \n// manipulate elements\nvar brd = div.board; \n\n// create the x-axis.\nvar xaxis = brd.create('axis',\t[ [0,0],[1,0] ]);\nxaxis.removeAllTicks();\nbrd.create('ticks',[xaxis,1],{\n strokeColor:'#ccc',\n majorHeight:-1,\n drawLabels: true,\n label: {offset: [-4, -10]},\n minorTicks: 0\n});\n\n\n// create the y-axis\nvar yaxis = brd.create('axis',\t[ [0,0],[0,1] ]);\nyaxis.removeAllTicks();\nbrd.create('ticks', [yaxis, 1], {\n strokeColor:'#ccc',\n majorHeight:-1, // Need this because the JXG.Options one doesn't apply\n drawLabels:true, // Only works for equidistant ticks\n label: {offset: [7, -2]},\n minorTicks:1, // The NUMBER of small ticks between each Major tick\n drawZero:false\n }\n);\n\n\n\n\n\n// This function shades in the area below the graph of f\n// between the x values x1 and x2\n\nvar shade = function(f,x1,x2,colour) {\n var dataX1 = [x1,x1];\n var dataY1 = [0,f(x1)];\n\n var dataX2 = [];\n var dataY2 = [];\n for (var i = x1; i <= x2; i = i+0.01) {\n dataX2.push(i);\n dataY2.push(f(i));\n }\n\n var dataX3 = [x2,x2];\n var dataY3 = [f(x2),0];\n\n dataX = dataX1.concat(dataX2).concat(dataX3);\n dataY = dataY1.concat(dataY2).concat(dataY3);\n\nvar shading = brd.create('curve', [dataX,dataY],{strokeWidth:0, fillColor:colour, fillOpacity:0.2});\n\nreturn shading;\n}\n\n\n//Define your functions\nvar f1 = function(x) {\n return a*(x+b)+c;\n}\n\nvar f2 = function(x) {\n return Math.sin(x-a*3.141);\n}\n\nvar f3 = function(x) {\n return a*Math.sin(x/b);\n}\n\n\n//Plot the graph and do shading\nswitch(q) {\n case 1:\n brd.create('functiongraph', [f1]);\n shade(f1,x1,x2, 'red');\n break;\n case 2:\n brd.create('functiongraph', [f2]);\n shade(f2,x1,x2,'red');\n break;\n case 3:\n brd.create('functiongraph', [f3]);\n shade(f3,x1,x2,'red');\n break;\n}\n\n\n\nreturn div;", "parameters": [["q", "number"], ["x1", "number"], ["x2", "number"], ["a", "number"], ["b", "number"], ["c", "number"], ["xmin", "number"], ["xmax", "number"], ["ymin", "number"], ["ymax", "number"]], "type": "html", "language": "javascript"}}, "advice": "

(i) The maximum can be read from the graph, and it is $\\var{x12}$.

To determine the minimum, we need to solve $f(x)=0$. So,

$\\simplify{{a1}(x+{b1})+{c1} = 0}$

$\\simplify{{a1}(x+{b1}) = -{c1}}$

$\\simplify{x+{b1} = -{c1}/{a1}}$

$\\simplify{x = -{c1}/{a1} - {b1} = -{c1+b1*a1}/{a1}}$.

So the minimum $x$-value is $\\simplify{-{c1+b1*a1}/{a1}}$.

(ii) By looking at the graph, the minimum and maximum $x$-values correspond to solutions of $f(x)=0$. In particular, they are the 2nd and 3rd positive solutions. Solving $f(x)=0$:

$\\sin(x-\\frac{\\var{a2*4}}{4}\\pi) =0$

$x-\\frac{\\var{a2*4}}{4}\\pi = \\ldots,-2\\pi, -\\pi,0,\\pi,2\\pi,3\\pi,\\ldots$

$x = \\ldots, \\frac{\\var{a2*4-8}}{4}\\pi, \\frac{\\var{a2*4-4}}{4}\\pi, \\frac{\\var{a2*4}}{4}\\pi,\\frac{\\var{a2*4+4}}{4}\\pi,\\frac{\\var{a2*4+8}}{4}\\pi, \\frac{\\var{a2*4+12}}{4}\\pi,\\ldots$. (These values were obtained by adding $\\frac{\\var{a2*4}}{4}\\pi$ to the previous line.)

The 2nd and 3rd positive solutions are $\\frac{\\var{a2*4}}{4}\\pi$ and $\\frac{\\var{a2*4+4}}{4}\\pi$, which are the numbers we want.

(iii) The minimum $x$-value can be read off the graph, and it is $0$. The maximum $x$-value is the smallest positive solution of $\\simplify{{a3}*sin(x/{b3})} =0$. Solving this:

$\\simplify{{a3}*sin(x/{b3})} =0$,

$\\simplify{sin(x/{b3})} =0$,

$\\simplify{x/{b3}} =\\ldots,-2\\pi, -\\pi,0,\\pi,2\\pi,3\\pi,\\ldots$,

$ x = \\ldots, -\\var{2*b3} \\pi, -\\var{b3}\\pi, 0 , \\var{b3}\\pi,\\var{2*b3}\\pi,\\var{2*b3}\\pi, \\ldots$. (These values were obtained by multiplying the previous line by $\\var{b3}$.)

The smallest positive solution is $\\var{b3}\\pi$, so this is the maximum $x$-value of the shaded region.

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A cubic with a maximum and minimum point is given. Question is to determine coordinates of the minimum and maximum point. Non-calculator. Advice is given.

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{plotgraph(a,b,c,d)}

Above is the graph of some function $f$.

What are the coordinates of its maximum point? ([[0]],[[1]])

What are the coordinates of its minimum point? ([[2]],[[3]])

", "showFeedbackIcon": true, "scripts": {}, "showCorrectAnswer": true, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "variableReplacementStrategy": "originalfirst"}], "tags": [], "advice": "

(i) A maximum point is a point where regardless if you move right or left, the height will decrease. A visual analogy would be a hill: if you're at the top of a hill, no matter which direction you go your height will decrease. So you're looking for a part of the graph which is 'like a hill', and in this graph the point is at $(\\var{xmax}, \\var{ymax})$.

(ii) A minimum point is the opposite of a maximum point (or an upside-down version of a maximum point, if you like). The analogy in this case would be a valley: no matter which direction you go your height will increase. In this graph, the minimum point is at $(\\var{xmin}, \\var{ymin})$.

", "statement": "

Finding stationary points on a graph.

", "type": "question"}, {"name": "Implicit differentiation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "name": "b"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "-random(1..9)", "description": "", "name": "a"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "name": "c"}}, "ungrouped_variables": ["a", "c", "b"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"scripts": {}, "gaps": [{"answer": "(({( - a)} + ( - (2 * x))) / ({b} + (2 * y)))", "showCorrectAnswer": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "answersimplification": "all,!collectNumbers", "marks": 2, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n

Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

Input your answer here:

$\\displaystyle \\frac{dy}{dx}= $ [[0]]

\n ", "showCorrectAnswer": true, "marks": 0}], "statement": "

Given the following relation between $x$ and $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{a}x+{b}y}=\\var{c}\\]
answer the following question.

", "tags": ["calculus", "Calculus", "checked2015", "derivative", "derivative ", "deriving an implicit relation", "differentiate", "differentiate implicitly", "differentiation", "first derivative using implicit differentiation", "implicit differentiation", "implicit relation", "mas1601", "MAS1601"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \t\t

20/06/2012:

\n \t\t

Added tags.

\n \t\t

Improved display using \\displaystyle where appropriate.

\n \t\t

Changed marks to 2.

\n \t\t

3/07/2012:

\n \t\t

Added tags.

\n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "\n \t\t

Implicit differentiation.

\n \t\t

Given $x^2+y^2+ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

\n \t\t

\n \t\t"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "

On differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) + {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[(\\var{b} + 2y) \\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x}\\] and hence on further rearranging:
\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x) / ({b} + (2 * y))}\\]

"}, {"name": "Implicit Differentiation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Gareth Woods", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/978/"}, {"name": "Aoife O'Brien", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1693/"}, {"name": "Antonia Wilmot-Smith", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2410/"}, {"name": "Leticija Dubickaite", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2461/"}, {"name": "Kevin Bohan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3363/"}], "advice": "

Hint:

Note that we regard $y$ as a function of $x$. Hence we have (using the Chain Rule): $\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$. And, using the Product Rule: $\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.

Now differentiate both sides of the relation with respect to $x$. Below is a worked solution to the problem.

a) By differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[( \\simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x-{d}y}\\] and hence on further rearranging:

\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x-{d}y) / ({b} + (2 * y)+{d}x)}\\]

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Implicit differentiation.

\n \t\t

Given $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

\n \t\t

Also find two points on the curve where $x=0$ and find the equation of the tangent at those points.

\n \t\t

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Input all numbers as integers or as fractions, not as decimals.

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Input your answer here:

$\\displaystyle \\frac{dy}{dx}= $ [[0]]

Input all numbers as integers not as decimals.

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You are given the following relation between $x$ and $y$
\\[\\simplify{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\\]
where $y=y(x)$. Find $\\dfrac{dy}{dx}$.

", "tags": [], "rulesets": {"std": ["all", "fractionNumbers"]}, "variable_groups": [], "type": "question"}, {"name": "Implicit differentiation 1 (Basic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}, {"name": "JPO AddMath", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2346/"}], "variables": {"a": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "Random(-6..6 except 0)", "name": "a"}, "c": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "Random(-3..3)", "name": "c"}, "constant": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "{a}*{c}-{b}*{d}", "name": "constant"}, "d": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "if(c=0, random(-3..3),0)", "name": "d"}, "b": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "Random(-6..6 except 0 except a except -a)", "name": "b"}}, "preamble": {"css": "", "js": ""}, "functions": {}, "rulesets": {}, "parts": [{"useCustomName": false, "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "sortAnswers": false, "scripts": {}, "variableReplacements": [], "gaps": [{"checkVariableNames": false, "useCustomName": false, "showPreview": true, "showCorrectAnswer": false, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "checkingType": "absdiff", "answer": "-({a}+2x*y^2)/(2x^2*y-{b})", "scripts": {}, "checkingAccuracy": 0.001, "variableReplacements": [], "valuegenerators": [{"value": "", "name": "x"}, {"value": "", "name": "y"}], "vsetRange": [0, 1], "showFeedbackIcon": false, "unitTests": [], "failureRate": 1, "marks": 1, "type": "jme", "vsetRangePoints": 5, "customName": "", "customMarkingAlgorithm": "malrules:\n [\n [\"3/y+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1+3/y)\", \"There are two main errors here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"],\n [\"-y/(x+1/y^3)\", \"Be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/y^3+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1/y^3+1)\", \"There are three things to watch here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$. Finally, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))Hint: Type $2x^2y$ as 2x^2*y

$\\frac{dy}{dx}=$ [[0]]

", "marks": 0, "type": "gapfill", "customName": "", "customMarkingAlgorithm": ""}, {"useCustomName": false, "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "sortAnswers": false, "scripts": {}, "variableReplacements": [], "gaps": [{"checkVariableNames": false, "useCustomName": false, "showPreview": true, "showCorrectAnswer": false, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "checkingType": "absdiff", "answer": "{a}/{b}", "scripts": {}, "checkingAccuracy": 0.001, "variableReplacements": [], "valuegenerators": [], "vsetRange": [0, 1], "showFeedbackIcon": false, "unitTests": [], "failureRate": 1, "marks": 1, "type": "jme", "vsetRangePoints": 5, "customName": "", "customMarkingAlgorithm": ""}], "showFeedbackIcon": true, "unitTests": [], "prompt": "

$\\frac{dy}{dx} \\bigg|_{(\\var{c},\\var{d})}=$[[0]]

(Answer in fraction form if necessary)

", "marks": 0, "type": "gapfill", "customName": "", "customMarkingAlgorithm": ""}], "variable_groups": [], "ungrouped_variables": ["a", "b", "c", "d", "constant"], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Implicit differentiation question with customised feedback to catch some common errors.

"}, "tags": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "advice": "

$\\frac{d}{dx}(\\var{a}x+x^2y^2)=\\frac{d}{dx}(\\simplify{{constant}+{b}y})$

$\\var{a}+2x.y^2+x^2.2y.\\frac{dy}{dx}=\\var{b}.\\frac{dy}{dx}$

$(\\simplify{2x^2y-{b}}) \\frac{dy}{dx}=\\simplify{-{a}-2x*y^2}$

$\\frac{dy}{dx}=\\frac{\\simplify{-{a}-2x*y^2}}{\\simplify{2x^2*y-{b}}} =\\simplify{-(2x*y^2+{a})/(2x^2*y-{b})}$

$\\ $

$\\frac{dy}{dx} \\bigg|_{(\\var{c},\\var{d})}=\\simplify{{a}/{b}}$

", "statement": "

Find the gradient of the curve $\\simplify{{a}x}+x^2y^2=\\simplify{{constant}+{b}y}$ at the point $(\\var{c},\\var{d})$.

", "type": "question"}, {"name": "Implicit differentiation 2 (Basic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}, {"name": "JPO AddMath", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2346/"}], "preamble": {"js": "", "css": ""}, "parts": [{"useCustomName": false, "prompt": "

Hint: Type $2xy$ as 2x*y

$\\frac{dy}{dx}=$ [[0]]

", "extendBaseMarkingAlgorithm": true, "gaps": [{"vsetRangePoints": 5, "failureRate": 1, "useCustomName": false, "vsetRange": [0, 1], "extendBaseMarkingAlgorithm": true, "unitTests": [], "scripts": {}, "showFeedbackIcon": false, "answer": "-{b}y^2/({a}+{b}x*y)", "checkingType": "absdiff", "checkVariableNames": false, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": false, "checkingAccuracy": 0.001, "variableReplacements": [], "type": "jme", "showPreview": true, "customName": "", "valuegenerators": [{"value": "", "name": "x"}, {"value": "", "name": "y"}], "marks": 1, "customMarkingAlgorithm": "malrules:\n [\n [\"3/y+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1+3/y)\", \"There are two main errors here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"],\n [\"-y/(x+1/y^3)\", \"Be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/y^3+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1/y^3+1)\", \"There are three things to watch here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$. Finally, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))$\\frac{dy}{dx} \\bigg|_{(\\var{c},1)}=$[[0]]

(Answer in fraction form if necessary)

Implicit differentiation question with customised feedback to catch some common errors.

"}, "rulesets": {}, "statement": "

Find the gradient of the curve $\\ln(y^\\var{a})+\\var{b}xy=\\simplify{{c}*{b}}$ at the point $(\\var{c},1)$.

", "variable_groups": [], "advice": "

$\\frac{d}{dx}(\\ln(y^\\var{a})+\\var{b}xy)=\\frac{d}{dx}(\\simplify{{c}*{b}})$

$(\\frac{1}{y^\\var{a}}.\\simplify{{a}y^{a-1}}).\\frac{dy}{dx}+(\\var{b}.1.y+\\var{b}x.\\frac{dy}{dx})=0$

$(\\frac{\\var{a}}{y}+\\var{b}x)\\frac{dy}{dx}=-\\var{b}y$

$\\frac{dy}{dx}=\\frac{-\\var{b}y}{\\frac{\\var{a}}y+\\var{b}x}$

$\\frac{dy}{dx}=\\frac{-\\var{b}y}{\\frac{\\var{a}}y+\\var{b}x} \\times \\frac{y}{y}$

$\\frac{dy}{dx}=\\simplify{(-{b}y^2)/({a}+{b}x*y)}$

$\\ $

$\\frac{dy}{dx} \\bigg|_{(\\var{c},1)}=\\simplify{(-{b})/({a}+{b}*{c})}$

", "functions": {}, "variables": {"b": {"group": "Ungrouped variables", "name": "b", "definition": "Random(3..7 except 6 except a)", "templateType": "anything", "description": ""}, "a": {"group": "Ungrouped variables", "name": "a", "definition": "Random(3..7 except 6)", "templateType": "anything", "description": ""}, "c": {"group": "Ungrouped variables", "name": "c", "definition": "Random(-4,-3,-2)", "templateType": "anything", "description": ""}}, "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question"}, {"name": "Maria's copy of Maximum/minimum", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Untitled.jpg", "/srv/numbas/media/question-resources/Untitled.jpg"], ["question-resources/Untitled_qCawkyB.jpg", "/srv/numbas/media/question-resources/Untitled_qCawkyB.jpg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "ungrouped_variables": ["l", "w"], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "rulesets": {}, "parts": [{"marks": 0, "prompt": "

Input the value for \$x\$ correct to one decimal place.

\$x = \$ [[0]]

", "scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "gaps": [{"correctAnswerFraction": false, "precisionPartialCredit": 0, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "precisionType": "dp", "minValue": "({w}+{l}-sqrt({w}^2+{l}^2-{w}*{l}))/6", "showPrecisionHint": false, "precision": "1", "marks": 1, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "maxValue": "({w}+{l}-sqrt({w}^2+{l}^2-{w}*{l}))/6"}], "showCorrectAnswer": true}], "advice": "

The length of the box is \$\\var{l}-2x\$, the width is \$\\var{w}-2x\$ and the height is \$x\$.

The volume is then given by

\$V=(\\var{l}-2x).(\\var{w}-2x).x\$

\$V=\\simplify{4x^3-2*({w}+{l})x^2+{l}*{w}x}\$

\$\\frac{dV}{dx}=\\simplify{12x^2-4*({w}+{l})x+{l}*{w}}\$

This is a quadratic equation.

\$x=\\frac{\\simplify{4*({w}+{l})}\\pm\\sqrt(\\simplify{16*({w}+{l})^2-48*{w}*{l}})}{24}\$

\$x=\\frac{\\simplify{{w}+{l}}\\pm\\sqrt(\\simplify{{w}^2-{w}*{l}+{l}^2})}{6}\$

\$\\frac{d^2V}{dx^2}=\\simplify{24x-4*({w}+{l})}\$

when \$x=\\simplify{({w}+{l}-sqrt({w}^2-{w}*{l}+{l}^2))/6}\$ \$\\frac{d^2V}{dx^2}<0\$ and therefore is the value that gives a maximum.

", "variable_groups": [], "tags": [], "functions": {}, "metadata": {"description": "

Maximising the volume of a rectangular box

", "licence": "Creative Commons Attribution 4.0 International"}, "variables": {"w": {"description": "", "name": "w", "templateType": "randrange", "definition": "random(10..28#2)", "group": "Ungrouped variables"}, "l": {"description": "", "name": "l", "templateType": "randrange", "definition": "random(30..100#5)", "group": "Ungrouped variables"}}, "statement": "

A rectangular sheet of metal of length = \$\\var{l}cm\$ and width = \$\\var{w}cm\$ has a square of side \$x\\,cm\$ cut from each corner. The ends and sides will be folded upwards to form an open box.

Determine the value of \$x\$ that will maximise the volume of this box.

", "type": "question"}, {"name": "Maximum/minimum Cylinder", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Untitled2.jpg", "/srv/numbas/media/question-resources/Untitled2.jpg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "

Problem on a closed cylindrical tank having minimum surface area

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A closed cylindrical tank is to be built having a volume of \$\\var{v}\$ cm³.

Determine the required height, \$h\$, and radius, \$r\$, if the total surface area is to be a minimum.

", "advice": "

\$\\pi r^2h=\\var{v}\$

\$h=\\frac{\\var{v}}{\\pi r^2}\$

The total surface area is to be a minimum.

Lid + curved surface area + base

\$A=\\pi r^2+2\\pi rh+\\pi r^2\$

\$A=2\\pi r^2+2\\pi r\\left(\\frac{\\var{v}}{\\pi r^2}\\right)\$

\$A=2\\pi r^2+\\simplify{2*{v}}r^{-1}\$

\$\\frac{dA}{dr}=4\\pi r-\\simplify{2{v}}r^{-2}=0\$

\$4\\pi r=\\simplify{2*{v}}/{r^2}\$

\$r^3=\\frac{\\var{v}}{2\\pi}\$

\$r=\\simplify{({v}/(2*pi))^(1/3)}\$

From the second line we have the relation \$h=\\frac{\\var{v}}{\\pi r^2}\$ to get

\$h=2*\\simplify{({v}/(2*pi))^(1/3)}\$

", "rulesets": {}, "variables": {"v": {"name": "v", "group": "Ungrouped variables", "definition": "random(50 .. 300#5)", "description": "", "templateType": "randrange"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["v"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Input the cyinder height, correct to two decimal places.

\$h = \$ [[0]]

Input the required cylinder radius, correct to two decimal places.

\$r = \$ [[1]]

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Given a graph of some function $f(x)$ (a cubic), the student is asked to write the coordinates of the maximum and minimum points. The student then finds the maximum and minimum points of a second cubic function without using a graph, by finding the derivative, solving the quadratic equation that results from setting the derivative equal to zero, and finally testing the value of the second derivative.

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Remember that the coordinates of a point are given in the order $(x,y)$.

The local maximum is the 'top of the hill' and the local minimum is the 'bottom of the valley'.

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Shown below is the plot of the function $f(x)=\\simplify[all,!noLeadingMinus]{{a}x^3+{b}x^2+{c}x+{d}}$.

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What are the coordinates of its local maximum point? ([[0]],[[1]])

What are the coordinates of its local minimum point? ([[2]],[[3]])

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There are parts of the graph where the function is seen to be increasing ('going up', from left to right) and parts where it is decreasing ('going down'). The exact rate at which the function is increasing or decreasing - its rate of change - at any given point is given by the slope of the tangent to the curve at that point. This is also known as the derivative of the function.

To find the derivative, we differentiate. For a function $f(x)$, we call the derivative $\\frac{\\mathrm{d}f}{\\mathrm{d}x}$.

Any time we meet a polynomial (like a quadratic or cubic) function, we can differentiate it piece by piece like this: \\[f(x)=a\\cdot x^n\\rightarrow\\frac{\\mathrm{d}f}{\\mathrm{d}x}=n\\cdot a\\cdot x^{n-1}\\]

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Differentiate the function $f(x)=\\simplify[all,!noLeadingMinus]{{a}x^3+{b}x^2+{c}x+{d}}$:

$\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\;$[[0]]$x^2+$[[1]]$x+$[[2]].

(Be careful to enter the negative sign where needed.)

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Note that at the turning points, the curve must go between 'going up' and 'going down', which means that the slopes of the tangents lines will go between being positive and being negative. This means the tangents must have a slope of zero at some point - a slope which is neither positive nor negative. And of course, to get the slope of the tangent at any point, we differentiate.

So, we find the derivative $\\frac{\\mathrm{d}f}{\\mathrm{d}x}$ and put it equal to zero.

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Now let us find the coordinates of the stationary points (also called turning points, extreme points, or local maximum and local minimum) of a different function using algebra.

We're going to use the function: \\[f(x)=\\simplify[all,!noLeadingMinus]{{a1}x^3+{b1}x^2+{c1}x+{d1}}\\]

To find the turning points, we first differentiate and then set the answer equal to zero:

$f(x)=\\simplify[all,!noLeadingMinus]{{a1}x^3+{b1}x^2+{c1}x+{d1}}$

$\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\;$[[0]]$\\;=0$

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We should be able to find these points without drawing a graph, but you can of course use Geogebra, Google, WolframAlpha or just pen and paper to plot the graph and verify what it looks like.

Finding the answer to the previous part should leave us with a quadratic equation. (The derivative of a cubic function will be a quadratic function, and we have made the function equal to zero, so now we have an equation.)

You should be able to use either the quadratic formula $(x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a})$ or factorise the quadratic equation in order to find the two solutions.

These solutions are the $x$-coordinates of the two turning points. For each $x$ value, you can substitute back into the original cubic equation to find the corresponding $y$-coordinates.

You should be able to see from the $y$ values which point is higher or lower, and so tell which one is the local maximum and which one is the local minimum.

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Now solve the previous quadratic equation and then enter the $x$ and $y$ values of the turning points below:

Local maximum: ([[0]],[[1]])

Local minimum: ([[2]],[[3]])

Note: We can use the second derivative to show whether a turning point is a local maximum or a local minimum. Find the second derivative and substitute the $x$ values that you have identified above. A negative value for the second derivative means that we have found a local maximum; a positive value means we have found a local minimum.

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Finding stationary points on a graph.

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Part (a)

A local maximum point is a point where regardless if you move right or left, the height will decrease, like the top of a hill. A minimum point is the opposite: no matter which direction you go your height will increase, like the bottom of a valley.

In this graph, the local minimum point is at $(\\var{xmin}, \\var{ymin})$ and the local maximum is at $(\\var{xmax}, \\var{ymax})$

Part (b)

In full, the function is: \\[f(x)=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor]{{a}x^3+{b}x^2+{c}x^1+{d}x^0}\\]

And so following the rules for differentiating polynomials we have:

\\[f(x)=a\\cdot x^n\\rightarrow\\frac{\\mathrm{d}f}{\\mathrm{d}x}=n\\cdot a\\cdot x^{n-1}\\]

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}= (3\\times\\var{a})x^2+(2\\times\\var{b})x^1+(1\\times\\var{c})x^0+(0\\times\\var{d})x^{-1}\\]

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor,!zeroTerm]{{3*a}x^2+{2*b}x^1+{c}x^0+0}\\]

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroTerm]{{3*a}x^2+{2*b}x^1+{c}x^0}\\]

Part (c)

We are now using a different function, but differentiating in the same way as before:

\\[f(x)=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor,!zeroTerm]{{a1}x^3+{b1}x^2+{c1}x^1+{d1}x^0}\\]

\\[f(x)=a\\cdot x^n\\rightarrow\\frac{\\mathrm{d}f}{\\mathrm{d}x}=n\\cdot a\\cdot x^{n-1}\\]

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}= (3\\times\\var{a1})x^2+(2\\times\\var{b1})x^1+(1\\times\\var{c1})x^0+(0\\times\\var{d1})x^{-1}\\]

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor,!zeroTerm]{{3*a1}x^2+{2*b1}x^1+{c1}x^0+0}\\]

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroTerm]{{3*a1}x^2+{2*b1}x^1+{c1}x^0}\\]

Part (d)

Now let's solve the quadratic equation $\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroTerm]{{3*a1}x^2+{2*b1}x^1+{c1}}=0$ using the quadratic formula $x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$. We have $a=\\var{3*a1}$, $b=\\var{2*b1}$, and $c=\\var{c1}$.

\\[x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\]

\\[x=\\frac{-(\\var{2*b1})\\pm\\sqrt{(\\var{2*b1})^2-4(\\var{3*a1})(\\var{c1})}}{2(\\var{3*a1})}\\]

\\[x=\\frac{\\var{-1*2*b1}\\pm\\sqrt{\\var{(2*b1)^2}-\\var{4*3*a1*c1}}}{\\var{2*3*a1}}\\]

\\[x=\\frac{\\var{-1*2*b1}\\pm\\sqrt{\\var{(2*b1)^2-4*3*a1*c1}}}{\\var{2*3*a1}}\\]

\\[x=\\var{x1min} \\mathrm{\\ \\ or }\\ \\ \\ x=\\var{x1max}\\]

Substituting these back in to the original cubic equation:

\\[f(x)=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor,!zeroTerm]{{a1}x^3+{b1}x^2+{c1}x^1+{d1}}\\]

\\[f(\\var{x1min})=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor,!zeroTerm,!otherNumbers,!collectNumbers]{{a1}({x1min})^3+{b1}({x1min})^2+{c1}({x1min})^1+{d1}}\\]

\\[f(\\var{x1min})=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor,!zeroTerm,!otherNumbers,!collectNumbers]{{a1*x1min^3}+{b1*x1min^2}+{c1*x1min}+{d1}}\\]

\\[f(\\var{x1min})=\\var{y1min}\\]

Similarly, we find $f(\\var{x1max})=\\var{y1max}$.

So we have two points and by comparing the $x$ and $y$ coordinates we can tell which one is the local minimum and which one is the local maximum:

Local minimum: ($\\var{x1min}$,$\\var{y1min}$). Local maximum: ($\\var{x1max}$,$\\var{y1max}$)

To really prove that these points are local maximum and minimum points, we look at the second deriviative. We find this by differentiating the first derivative one more time:

\\[f(x)=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroPower,!unitPower,!unitFactor]{{a1}x^3+{b1}x^2+{c1}x^1+{d1}x^0}\\]

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroTerm]{{3*a1}x^2+{2*b1}x^1+{c1}x^0}\\]

\\[\\frac{\\mathrm{d}^2f}{\\mathrm{d}x^2}=\\simplify[all,!noLeadingMinus,!zeroFactor,!zeroTerm]{{6*a1}x+{2*b1}}\\]

We now test the value of $\\frac{\\mathrm{d}^2f}{\\mathrm{d}x^2}$ at the two $x$-values we have found so far.

When $x=\\var{x1min}$, we have $\\frac{\\mathrm{d}^2f}{\\mathrm{d}x^2}=\\simplify[all,!noLeadingMinus,!zeroTerm]{{6*a1}{x1min}+{2*b1}}=\\var{6*a1*x1min+2*b1}$

When $x=\\var{x1max}$, we have $\\frac{\\mathrm{d}^2f}{\\mathrm{d}x^2}=\\simplify[all,!noLeadingMinus,!zeroTerm]{{6*a1}{x1max}+{2*b1}}=\\var{6*a1*x1max+2*b1}$

When the value of $\\frac{\\mathrm{d}^2f}{\\mathrm{d}x^2}$ is positive, the slopes are increasing in value and so we are looking at a local minimum. So we know ($\\var{x1min}$,$\\var{y1min}$) is a local minimum.

When the value of $\\frac{\\mathrm{d}^2f}{\\mathrm{d}x^2}$ is negative, we are looking at a local maximum. So we know ($\\var{x1max}$,$\\var{y1max}$) is a local maximum.

Bonus graph:

In fact, the function $f(x)=\\simplify[all,!noLeadingMinus]{{a1}x^3+{b1}x^2+{c1}x+{d1}}$ that we have been looking at looks like this:

{plotgraph(a1,b1,c1,d1)}

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(i) {plotgraph(1,x11,x12,a1,b1,c1,{x11-2},{x12+2},-3,15)}

This is the graph of the function $f(x) = \\simplify{{a1}*(x+{b1})+{c1}}$.

Determine the minimum and maximum $x$-values of the region. Enter the minimum first.

[[0]],[[1]]

(ii) {plotgraph(2,x21,x22,a2,0,0,-0.5,10,-1.2,1.2)}

This is the graph of the function $f(x) = \\sin(x-\\frac{\\var{a2*4}}{4}\\pi)$.

Determine the minimum and maximum $x$-values of the region.

Enter the minimum first. Enter the values exactly in terms of $\\pi$. To enter $\\pi$, type `pi'.

[[2]], [[3]]

(iii) {plotgraph(3,x31,x32,a3,b3,0,-1,x32*1.6,-1.2*a3,1.2*a3)}

This curve has equation $y = \\simplify{{a3}*sin(x/{b3})}$.

Determine the minimum and maximum $x$-values of the region.

Enter the minimum first. Enter the values exactly in terms of $\\pi$. To enter $\\pi$, type `pi'.

[[4]], [[5]]

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(i) The maximum can be read from the graph, and it is $\\var{x12}$.

To determine the minimum, we need to solve $f(x)=0$. So,

$\\simplify{{a1}(x+{b1})+{c1} = 0}$

$\\simplify{{a1}(x+{b1}) = -{c1}}$

$\\simplify{x+{b1} = -{c1}/{a1}}$

$\\simplify{x = -{c1}/{a1} - {b1} = -{c1+b1*a1}/{a1}}$.

So the minimum $x$-value is $\\simplify{-{c1+b1*a1}/{a1}}$.

(ii) By looking at the graph, the minimum and maximum $x$-values correspond to solutions of $f(x)=0$. In particular, they are the 2nd and 3rd positive solutions. Solving $f(x)=0$:

$\\sin(x-\\frac{\\var{a2*4}}{4}\\pi) =0$

$x-\\frac{\\var{a2*4}}{4}\\pi = \\ldots,-2\\pi, -\\pi,0,\\pi,2\\pi,3\\pi,\\ldots$

The 2nd and 3rd positive solutions are $\\frac{\\var{a2*4}}{4}\\pi$ and $\\frac{\\var{a2*4+4}}{4}\\pi$, which are the numbers we want.

(iii) The minimum $x$-value can be read off the graph, and it is $0$. The maximum $x$-value is the smallest positive solution of $\\simplify{{a3}*sin(x/{b3})} =0$. Solving this:

$\\simplify{{a3}*sin(x/{b3})} =0$,

$\\simplify{sin(x/{b3})} =0$,

$\\simplify{x/{b3}} =\\ldots,-2\\pi, -\\pi,0,\\pi,2\\pi,3\\pi,\\ldots$,

$ x = \\ldots, -\\var{2*b3} \\pi, -\\var{b3}\\pi, 0 , \\var{b3}\\pi,\\var{2*b3}\\pi,\\var{2*b3}\\pi, \\ldots$. (These values were obtained by multiplying the previous line by $\\var{b3}$.)

The smallest positive solution is $\\var{b3}\\pi$, so this is the maximum $x$-value of the shaded region.

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Find the gradient of the curve $y$ at the point $x=\\var{d}$, giving your answer to $2$ decimal places if necessary.

\\[ y = \\simplify{ {a}*x^2 + {b}x + {c}} \\]

Firstly, differentiate.

$\\displaystyle \\frac{dy}{dx}=$ [[1]]

Gradient at $x=\\var{d}\\;$ is [[0]]

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You have not given your answer to the correct precision.

Find the coordinates of the turning point of the function below and state whether it is a maximum or a minimum point. Give your answers to $2$ decimal places where necessary.

$y=\\simplify {{f}x^2+{g}x+{h}}$

Firstly, find the first and second derivatives $y$.

$\\displaystyle \\frac{dy}{dx}=$ [[2]]

$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]

Secondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.

$x$-coordinate of the turning point $=$ [[0]]

$y$-coordinate of the turning point $=$ [[1]]

The turning point is a [[4]]

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You have not given your answer to the correct precision.

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maximum

", "

minimum

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An unpowered missile is launched vertically from the ground.

At a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula

\\[ y=\\var{z}t-\\var{w}t^2. \\]

Calculate the maximum height reached by the missile.

Firstly, differentiate.

$\\displaystyle \\frac{dy}{dt}=$ [[0]]

Now use this result and your knowledge of differentiation to find the maximum height of the missile, rounding your answer to $2$ decimal places.

$y=$ [[1]]

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Parts A and B

Here, 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.

Part C

The 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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Is the stationary point a maximum?

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The square-based box will have volume

\\[V=x^2h,\\]

where $x$ is the base side length and $h$ is the box height. Further, the surface area $S$, of the box is

\\[S=2x^2+4xh.\\]

For this particular problem, $V={volume}$, so we have

\\[\\var{volume}=x^2h \\Rightarrow h=\\frac{\\var{volume}}{x^2},\\]

providing a function for $h$ in terms of $x$. Substituting this into the surface area function we then have

\\[S(x)=2x^2+4x\\left(\\frac{\\var{volume}}{x^2}\\right)=2x^2+4\\frac{\\var{volume}}{x}.\\]

Minima and maxima of $S(x)$ occur when $dS/dx=0$. We have

\\[\\frac{dS}{dx}=4x-4\\frac{\\var{volume}}{x^2},\\]

and so we must solve

\\[0=4x-4\\frac{\\var{volume}}{x^2}.\\]

Rearranging, we find that

\\[4x=4\\frac{\\var{volume}}{x^2}\\]

\\[\\Rightarrow x^3=\\var{volume}\\]

\\[\\Rightarrow x=\\var{volume}^{1/3}=\\var{xsol}.\\]

Finally, substituting this value back into the $h$ equation, we see that

\\[h=\\frac{\\var{volume}}{\\var{volume}^{2/3}}=\\var{hsol}.\\]

We should confirm that this set of dimensions does in fact produce a minimum surface area (rather than a maximum). This can be carried out by noting that the 2nd derivative of $S$ with respect to $x$ is

\\[\\frac{d^2S}{dx^2}=4+8\\frac{\\var{volume}}{x^3},\\]

which is positive for all positive $x$, and hence our proposed side length will produce a minimised box surface area.

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The equation for box volume can be used to express box height in terms of the side length of the base of the box. If the side length is denoted $x$, enter the function here

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Use your result from part (a) to write a function for the surface area of cardboard used in the construction of the box. Your function should be entirely in terms of the side length of the base of the box, $x$.

To minimise the surface area of cardboard, we need the derivative of the surface area function with respect to $x$. This derivative is $\\displaystyle \\frac{dS}{dx}=$

The minimum surface area will occur when the length of the sides of the box base is $x=$

Finally, the height of the box must then be $h=$

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The box volume is fixed for each student. Here it lies between 0.3 and 3$m^3$.

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This question guides students through the process of determining the dimensions of a box to minimise its surface area whilst meeting a specified volume.

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A square based box is to be constructed such that its volume is exactly {volume}m$^3$. The following questions guide you through the process of determining the base side length and box height such that the minimum amount of cardboard is used to construct the box.

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Another practical application of differentiation

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An open metal tank of square base has a volume of $\\var{v}\\text{ m}^3$

Given that the square base has sides of length $x$ metres, find expressions, in terms of $x$, for the following.

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This section draws on the skills learnt the previous parts of the 'Differentiation' series of questions, and some geometry knowledge.

The hint under the steps should be all the extra information you need.

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Describe the height of the tank in terms of $x$

$h=$ [[0]] m

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Describe the surface area of the tank in terms of $x$

$S=$ [[0]] $\\text{m}^2$

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Find the first derivative

$S'(x)=$ [[1]]

and given that the surface area is a minimum, find the value of $x$. Therefore,

$x=$ [[0]] (give your answer to 2 decimal places)

Find the second derivative

$S''(x)=$ [[2]]

Check this is a minimum.

Substitute your value for $x$ into $S''(x)$ and determine whether is it a minimum.

Type '$Y$' for yes, '$N$' for no, or '$U$' for undefined.

[[3]]

Hence, calculate the minimum area of metal used

$A_{min}=$ [[4]] (give your answer to 2 decimal places)

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Hint: find $x$ at stationary point

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Solve a trigonometric equation

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Evaluate for all angles of $\\theta$ in the interval $0^\\circ \\leq \\theta < 360^\\circ $ satisfying:

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$\\simplify{{a}{function}(theta)+{b} = 0}$

In which quadrants do the solutions exist (select two)?

[[0]]

Give all angles below in degrees.

1st solution (lowest angle)

[[1]]$^\\circ$

2nd solution (highest angle)

[[2]]$^\\circ$

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Quadrant 1

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Quadrant 2

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Quadrant 3

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Quadrant 4

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What is the equation of this curve?

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$y=\\simplify[unitFactor,fractionNumbers]{{a} sin({b}x+{c})+{d}}$

", "

$y=\\simplify[unitFactor,fractionNumbers]{{a} sin({1/b}x+{c})+{d}}$

", "

$y=\\simplify[unitFactor,fractionNumbers]{-{a} sin({b}x+{c})+{d}}$

", "

$y=\\simplify[unitFactor,fractionNumbers]{{2a} sin({b}x-{c})+{d}}$

", "

$y=\\simplify[unitFactor,fractionNumbers]{{a+1} sin({b}x-{c})+{2*d}}$

", "

$y=\\simplify[unitFactor,fractionNumbers]{{a} sin({2*b}x+{c})+{d}}$

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This uses an embedded Geogebra graph of a sine curve $y=a\\sin (bx+c)+d$ with random coefficients set by NUMBAS.

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The graph shows a sine curve.

{geogebra_applet('https://ggbm.at/GqxVC76x', defs, [])}

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Simplifying expressions such as $b^{\\log_b(x)}$ and $b^{\\log_b(x)}$.

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", "templateType": "anything"}, "q": {"name": "q", "group": "Ungrouped variables", "definition": "list[1]", "description": "

", "templateType": "anything"}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "random(-12..12 except b)", "description": "", "templateType": "anything"}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(2..7)", "description": "

", "templateType": "anything"}, "p": {"name": "p", "group": "Ungrouped variables", "definition": "list[0]", "description": "

", "templateType": "anything"}, "list": {"name": "list", "group": "Ungrouped variables", "definition": "shuffle([0,0,random(-12..12 except 0),random(-12..12 except 0),random(-12..12 except 0)])", "description": "", "templateType": "anything"}}, "statement": "

Simplify the following expressions:

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You might recall that $b^{\\log_{b}(a)}=a$ (for $a>0$), and therefore

\\[\\var{b}^{\\log_{\\var{b}}(\\simplify{x+{d}})}=\\simplify{x+{d}}, \\quad\\text{ for }x>\\var{-d}.\\]

We can think of the logarithm and the exponential (of the same base) cancelling each other out, or undoing each other since they are the inverse of each other. Some people might prefer to see more evidence of this, so here is a longer explanation:

Let $\\simplify{log(x+{d},{b})}=n$. Recall the definition of log says that this is equivalent to \\[\\simplify{{b}^n=x+{d}}.\\] Now substitute $\\simplify{log(x+{d},{b})}$ for $n$ into this equation (since we defined them to be equal) and we have

\\[\\simplify{{b}^(log(x+{d},{b}))=x+{d}}\\]

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This expression can be simplified to not contain logs or exponentials.

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$\\var{b}^{\\log_{\\var{b}}(\\simplify{x+{d}})}=$ [[0]], for $x>\\var{-d}$.

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Recall that $\\ln$ denotes $\\log_e$.

Here is one way to do this simplification (use the log laws $b\\log(a)=\\log(a^b)$ and $b^{\\log_b(a)}=a$):

$\\begin{align}e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}&=e^{\\ln\\left(\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\right)}\\\\&=\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\end{align}$

Another way (use the index law $a^{bc}=(a^b)^c$ and $b^{\\log_b(a)}=a$):

$\\begin{align}e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}&=\\left(e^{\\ln(\\simplify{x^{a}+{d}})}\\right)^\\var[fractionNumbers]{c}\\\\&=\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\end{align}$

", "scripts": {}, "useCustomName": false, "showCorrectAnswer": true, "unitTests": [], "variableReplacements": [], "extendBaseMarkingAlgorithm": true}], "variableReplacementStrategy": "originalfirst", "marks": 0, "showFeedbackIcon": true, "gaps": [{"customName": "", "type": "jme", "vsetRangePoints": 5, "answer": "(x^{a}+{d})^{c}", "showFeedbackIcon": true, "failureRate": 1, "checkingType": "absdiff", "checkVariableNames": false, "notallowed": {"showStrings": false, "message": "

This expression can be simplified to not contain logs or exponentials.

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$e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}=$ [[0]], for $\\simplify{x^{a}+{d}}>0$.

Recall that $\\ln$ denotes $\\log_e$.

To simplify $\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)$ we might think of the logarithm and the exponential (of the same base) cancelling each other out, or undoing each other since they are the inverse of each other. Some people might prefer to see more evidence of this, so here is a longer explanation:

Let $\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=n$. Recall the definition of log says that this is equivalent to \\[e^n=e^\\simplify{{p}x^2+{q}x+{r}}\\]

Therefore $n=\\simplify{{p}x^2+{q}x+{r}}$ and so we can conclude

\\[\\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=\\simplify{{p}x^2+{q}x+{r}}.\\]

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This expression can be simplified to not contain logs or exponentials.

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$\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=$ [[0]]

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