// Numbas version: finer_feedback_settings
{"name": "Implicit Differentiation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"extensions": [], "rulesets": {"std": ["all", "fractionNumbers"]}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "\n    \t\t<p>Implicit differentiation.</p>\n    \t\t            <p>Given $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.</p>\n    \t\t            <p>Also find two points on the curve where $x=0$ and find the equation of the tangent at those points.</p>\n    \t\t            <p>&nbsp;</p>\n    \t\t"}, "variable_groups": [], "variables": {"c": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(2..9 except -a+1)", "name": "c"}, "b": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "c-1", "name": "b"}, "d": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "random(-3..3 except 0)", "name": "d"}, "a": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "-random(1..9)", "name": "a"}}, "parts": [{"type": "gapfill", "marks": 0, "prompt": "\n      <p>Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$&nbsp;in terms of $x$ and $y$.</p>\n      <p>Input your answer here:</p>\n      <p>$\\displaystyle \\frac{dy}{dx}= $ [[0]]</p>\n      <p>Input all numbers as integers not as decimals.</p>\n      <p>If you want more help click on Show steps - you will not lose any marks.</p>\n      ", "gaps": [{"marks": 2, "showPreview": true, "scripts": {}, "answer": "(({( - a)} + ( - (2 * x))-{d}y) / ({b} + (2 * y)+{d}x))", "vsetRange": [0, 1], "answerSimplification": "all,!collectNumbers", "valuegenerators": [{"value": "", "name": "x"}, {"value": "", "name": "y"}], "checkingAccuracy": 0.001, "checkVariableNames": false, "type": "jme", "vsetRangePoints": 5, "unitTests": [], "extendBaseMarkingAlgorithm": true, "customName": "", "variableReplacements": [], "failureRate": 1, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "useCustomName": false, "showCorrectAnswer": true, "checkingType": "absdiff", "notallowed": {"strings": ["."], "showStrings": false, "partialCredit": 0, "message": "<p>Input all numbers as integers or as fractions, not as decimals.</p>"}, "showFeedbackIcon": true}], "scripts": {}, "customName": "", "variableReplacements": [], "stepsPenalty": 0, "unitTests": [], "customMarkingAlgorithm": "", "steps": [{"type": "information", "marks": 0, "prompt": "\n          <p>Note that we regard $y$ as a function of $x$.</p>\n          <p>Hence we have (using the chain rule):</p>\n          <p>$\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$</p>\n          <p>And , using the product rule:</p>\n          <p>$\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.</p>\n          <p>&nbsp;Now differentiate both sides of the relation with respect to $x$.</p>\n          ", "scripts": {}, "customName": "", "variableReplacements": [], "unitTests": [], "customMarkingAlgorithm": "", "showFeedbackIcon": true, "useCustomName": false, "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst"}], "showFeedbackIcon": true, "useCustomName": false, "sortAnswers": false, "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "tags": [], "functions": {}, "name": "Implicit Differentiation", "statement": "<p>Given the following relation between $x$ and $y$<br/>\\[\\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\\]<br/>answer the following questions.</p>", "ungrouped_variables": ["a", "c", "b", "d"], "advice": "<h4>a)</h4>\n<p>On differentiating both sides of the equation implicitly we get<br/>\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0\\]<br/>Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get <br/>\\[( \\simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x-{d}y}\\] and hence on further rearranging:</p>\n<p>\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x-{d}y) / ({b} + (2 * y)+{d}x)}\\]</p>\n<h4>b)</h4>\n<p>On putting $x=0$ in the relation we get:</p>\n<p>\\[\\simplify{y^2+{b}y={c}} \\Rightarrow \\simplify{y^2+{b}y-{c}=0 }\\Rightarrow (y+\\var{c})(y-1)=0\\]</p>\n<p>Hence $a=-\\var{c}$ and $b=1$.</p>\n<h4>c)</h4>\n<p>First we find the tangent at the point $(0,-\\var{c})$.</p>\n<p>We find using the formula we found for $\\frac{dy}{dx}$ in part a) that the gradient at &nbsp;$(0,-\\var{c})$ is:</p>\n<p>\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}+{d*c}}}{\\var{b}-\\var{2*c}}=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}\\]</p>\n<p>As the tangent goes through the point $(0,\\var{-c})$ i.e. at $x=0,\\;\\;y=-\\var{c}$ we see that the equation of the tangent is:</p>\n<p>\\[y=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}x-\\var{c}\\]</p>\n<p>Next we find that the gradient at &nbsp;$(0,1)$ is:</p>\n<p>\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}-{d}}}{\\var{b}+2}=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}\\]</p>\n<p>As the tangent goes through the point $(0,1)$ i.e. at $x=0,\\;\\;y=1$ we see that the equation of the tangent is:</p>\n<p>\\[y=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}x+1\\]</p>", "preamble": {"css": "", "js": ""}, "type": "question", "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Kevin Bohan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3363/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Kevin Bohan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3363/"}]}