// Numbas version: finer_feedback_settings {"name": "Simon's copy of SFY0004 Implicit 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"parts": [{"extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "sortAnswers": false, "stepsPenalty": 0, "marks": 0, "unitTests": [], "prompt": "
Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\nInput your answer here:
\n$\\displaystyle \\frac{dy}{dx}= $ [[0]]
\nInput all numbers as integers not as decimals.
\nIf you want more help click on Show steps - you will not lose any marks.
", "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "\nNote that we regard $y$ as a function of $x$.
\nHence we have (using the chain rule):
\n$\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$
\nAnd , using the product rule:
\n$\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.
\nNow differentiate both sides of the relation with respect to $x$.
\n ", "type": "information", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "variableReplacements": [], "unitTests": [], "customMarkingAlgorithm": "", "marks": 0, "extendBaseMarkingAlgorithm": true}], "gaps": [{"notallowed": {"strings": ["."], "message": "Input all numbers as integers or as fractions, not as decimals.
", "partialCredit": 0, "showStrings": false}, "checkingType": "absdiff", "showPreview": true, "failureRate": 1, "answerSimplification": "all,!collectNumbers", "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "vsetRangePoints": 5, "vsetRange": [0, 1], "marks": 2, "unitTests": [], "checkingAccuracy": 0.001, "checkVariableNames": false, "type": "jme", "variableReplacementStrategy": "originalfirst", "answer": "(({( - a)} + ( - (2 * x))-{d}y) / ({b} + (2 * y)+{d}x))", "variableReplacements": [], "expectedVariableNames": [], "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true}], "variableReplacements": [], "customMarkingAlgorithm": "", "type": "gapfill"}, {"extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "sortAnswers": false, "marks": 0, "unitTests": [], "prompt": "\nFind the two points $(0,a),\\;\\;(0,b),\\;\\;a \\lt b$ which lie on the curve given by the relation.
\n$a=\\;$[[0]]
\n$b=\\;$[[1]]
\n(remember that $a \\lt b$).
\n ", "variableReplacementStrategy": "originalfirst", "gaps": [{"minValue": "-c", "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "marks": 1, "unitTests": [], "mustBeReduced": false, "mustBeReducedPC": 0, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "maxValue": "-c", "correctAnswerFraction": false, "variableReplacements": [], "customMarkingAlgorithm": "", "allowFractions": false, "extendBaseMarkingAlgorithm": true}, {"minValue": "1", "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "marks": 1, "unitTests": [], "mustBeReduced": false, "mustBeReducedPC": 0, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "maxValue": "1", "correctAnswerFraction": false, "variableReplacements": [], "customMarkingAlgorithm": "", "allowFractions": false, "extendBaseMarkingAlgorithm": true}], "variableReplacements": [], "customMarkingAlgorithm": "", "type": "gapfill"}, {"extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}, "sortAnswers": false, "marks": 0, "unitTests": [], "prompt": "\nHence find the equations of the tangents at the points $(0,a)$ and $(0,b)$.
\nInput all numbers as integers or as fractions, not as decimals.
\nEquation of tangent at $(0,a)$:
\nFind the gradient of the tangent at $(0,a)$.
\nGradient=[[0]].
\nHence the equation of the tangent at $(0,a)$ is:
\n$y = \\;$[[1]]
\nEquation of tangent at $(0,b)$:
\nFind the gradient of the tangent at $(0,b)$.
\nGradient=[[2]].
\nHence the equation of the tangent at $(0,b)$ is:
\n$y = \\;$[[3]]
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\n \t\tGiven $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\n \t\tAlso find two points on the curve where $x=0$ and find the equation of the tangent at those points.
\n \t\t\n \t\t"}, "name": "Simon's copy of SFY0004 Implicit 2", "statement": "
Given the following relation between $x$ and $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\\]
answer the following questions.
Note from the chain rule that $\\frac{d}{dx}(xy) = x\\frac{dy}{dx}+y\\frac{d}{dx}(x)= x\\frac{dy}{dx}+y$
\n\n
(a)
\nOn 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)}\\]
\n\nOn putting $x=0$ in the relation we get:
\n\\[\\simplify{y^2+{b}y={c}} \\Rightarrow \\simplify{y^2+{b}y-{c}=0 }\\Rightarrow (y+\\var{c})(y-1)=0\\]
\nHence $a=-\\var{c}$ and $b=1$.
\n\nFirst we find the tangent at the point $(0,-\\var{c})$.
\nWe find using the formula we found for $\\frac{dy}{dx}$ in part (a) that the gradient at $(0,-\\var{c})$ is:
\n\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}+{d*c}}}{\\var{b}-\\var{2*c}}=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}\\]
\nAs 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:
\n\\[y=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}x-\\var{c}\\]
\nNext we find that the gradient at $(0,1)$ is:
\n\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}-{d}}}{\\var{b}+2}=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}\\]
\nAs the tangent goes through the point $(0,1)$ i.e. at $x=0,\\;\\;y=1$ we see that the equation of the tangent is:
\n\\[y=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}x+1\\]
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