// Numbas version: finer_feedback_settings {"name": "Maths Support: Implicit differentiation", "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "preventleave": false, "browse": true, "showfrontpage": false, "showresultspage": "never"}, "duration": 0.0, "metadata": {"notes": "", "description": "
Implicit differentiation including finding tangents
", "licence": "Creative Commons Attribution 4.0 International"}, "timing": {"timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "shufflequestions": false, "questions": [], "percentpass": 0.0, "allQuestions": true, "pickQuestions": 0, "type": "exam", "feedback": {"showtotalmark": true, "advicethreshold": 0.0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true, "enterreviewmodeimmediately": false, "showexpectedanswerswhen": "never", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "never"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "SFY0004 Implicit 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "calculus", "derivative", "deriving an implicit relation", "differentiate", "differentiate implicitly", "differentiation", "first derivative using implicit differentiation", "implicit differentiation", "implicit relation"], "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))}\\]
Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\nInput your answer here:
\n$\\displaystyle \\frac{dy}{dx}= $ [[0]]
\n \n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "all,!collectNumbers", "marks": 2.0, "answer": "(({( - a)} + ( - (2 * x))) / ({b} + (2 * y)))", "type": "jme"}], "type": "gapfill", "marks": 0.0}], "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.
20/06/2012:
\n \t\t \t\t \t\tAdded tags.
\n \t\t \t\t \t\tImproved display using \\displaystyle where appropriate.
\n \t\t \t\t \t\tChanged marks to 2.
\n \t\t \t\t \t\t\n \t\t \t\t \t\t
3/07/2012:
Added tags.
\n \t\t \t\t \n \t\t \n \t\t", "description": "\n \t\t \t\t \t\tImplicit differentiation.
\n \t\t \t\t \t\tGiven $x^2+y^2+ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.
\n \t\t \t\t \t\t\n \t\t \t\t \n \t\t \n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SFY0004 Implicit 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "c", "b", "d"], "tags": ["Calculus", "Differentiation", "calculus", "derivative", "deriving an implicit relation", "differentiate", "differentiate implicitly", "differentiation", "equation of tangent", "first derivative using implicit differentiation", "gradient", "implicit differentiation", "implicit relation", "tangent at a point"], "preamble": {"css": "", "js": ""}, "advice": "
On 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)}\\]
\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 $b=1$.
\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\\]
", "rulesets": {"std": ["all", "fractionNumbers"]}, "parts": [{"stepsPenalty": 0, "prompt": "\nUsing 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.
\n ", "marks": 0, "gaps": [{"notallowed": {"message": "Input all numbers as integers or as fractions, not as decimals.
", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "all,!collectNumbers", "scripts": {}, "answer": "(({( - a)} + ( - (2 * x))-{d}y) / ({b} + (2 * y)+{d}x))", "marks": 2, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "steps": [{"type": "information", "showCorrectAnswer": true, "scripts": {}, "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 ", "marks": 0}], "type": "gapfill"}, {"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 ", "marks": 0, "gaps": [{"allowFractions": false, "marks": 1, "maxValue": "-c", "minValue": "-c", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 1, "maxValue": "1", "minValue": "1", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"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]]
\n ", "marks": 0, "gaps": [{"notallowed": {"message": "Input as an integer or as a fraction, not as a decimal.
", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{a-d*c}/{c+1}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"notallowed": {"message": "Input all numbers as integers or as fractions, not as decimals.
", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{a-d*c}/{c+1}*x-{c}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"notallowed": {"message": "Input as an integer or as a fraction, not as a decimal.
", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{-a-d}/{c+1}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}, {"notallowed": {"message": "Input all numbers as integers or as fractions, not as decimals.
", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "{-a-d}/{c+1}*x+{1}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}], "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.
30/04/2013
\n \t\tCreated new question for SFY0004 out of 1041 CBA2_5.
\n \t\tAdded more tags.
\n \t\t", "description": "\n \t\tImplicit differentiation.
\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", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "extensions": [], "custom_part_types": [], "resources": []}