// Numbas version: finer_feedback_settings {"name": "Denis's copy of Gareth's copy of Differentiation 19 - Finding Tangent and Normal", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"preamble": {"css": "fraction {\n display: inline-block;\n vertical-align: middle;\n}\nfraction > numerator, fraction > denominator {\n float: left;\n width: 100%;\n text-align: center;\n line-height: 2.5em;\n}\nfraction > numerator {\n border-bottom: 1px solid;\n padding-bottom: 5px;\n}\nfraction > denominator {\n padding-top: 5px;\n}\nfraction input {\n line-height: 1em;\n}\n\nfraction .part {\n margin: 0;\n}\n\n.table-responsive, .fractiontable {\n display:inline-block;\n}\n.fractiontable {\n padding: 0; \n border: 0;\n}\n\n.fractiontable .tddenom \n{\n text-align: center;\n}\n\n.fractiontable .tdnum \n{\n border-bottom: 1px solid black; \n text-align: center;\n}\n\n\n.fractiontable tr {\n height: 3em;\n}\n", "js": "document.createElement('fraction');\ndocument.createElement('numerator');\ndocument.createElement('denominator');"}, "name": "Denis's copy of Gareth's copy of Differentiation 19 - Finding Tangent and Normal", "statement": "
Find the following.
", "variables": {"b": {"description": "", "definition": "repeat(random(-8..8 except 0 except 1 except -1),2)", "name": "b", "group": "Ungrouped variables", "templateType": "anything"}, "d": {"description": "", "definition": "random(-2..3 except 0)", "name": "d", "group": "Ungrouped variables", "templateType": "anything"}, "c": {"description": "", "definition": "repeat(random(-10..10 except 0 except 1 except -1),2)", "name": "c", "group": "Ungrouped variables", "templateType": "anything"}, "a": {"description": "", "definition": "repeat(random(2..4),2)", "name": "a", "group": "Ungrouped variables", "templateType": "anything"}}, "parts": [{"type": "gapfill", "gaps": [{"variableReplacements": [], "variableReplacementStrategy": "originalfirst", "checkingtype": "absdiff", "checkvariablenames": false, "showFeedbackIcon": true, "answersimplification": "all", "checkingaccuracy": 0.001, "answer": "2*{a[0]}*x+{b[0]}", "expectedvariablenames": [], "type": "jme", "marks": 1, "vsetrangepoints": 5, "vsetrange": [0, 1], "showCorrectAnswer": true, "showpreview": true, "scripts": {}}, {"variableReplacements": [], "correctAnswerStyle": "plain", "notationStyles": ["plain", "en", "si-en"], "variableReplacementStrategy": "originalfirst", "maxValue": "{b[0]}", "allowFractions": false, "minValue": "{b[0]}", "correctAnswerFraction": false, "type": "numberentry", "marks": 1, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}}, {"variableReplacements": [], "correctAnswerStyle": "plain", "notationStyles": ["plain", "en", "si-en"], "variableReplacementStrategy": "originalfirst", "maxValue": "{c[0]}", "allowFractions": false, "minValue": "{c[0]}", "correctAnswerFraction": false, "type": "numberentry", "marks": 1, "showFeedbackIcon": true, "showCorrectAnswer": true, "scripts": {}}, {"variableReplacements": [], "variableReplacementStrategy": "originalfirst", "checkingtype": "absdiff", "checkvariablenames": false, "showFeedbackIcon": true, "answersimplification": "all", "checkingaccuracy": 0.001, "answer": "{b[0]}*x+{c[0]}", "expectedvariablenames": [], "type": "jme", "marks": "2", "vsetrangepoints": 5, "vsetrange": [0, 1], "showCorrectAnswer": true, "showpreview": true, "scripts": {}}], "variableReplacementStrategy": "originalfirst", "prompt": "Find an equation for the tangent to the curve $\\simplify{y={a[0]}x^2+{b[0]}x+{c[0]}}$ at the point at which it crosses the y-axis.
\n$\\frac{dy}{dx}=$ [[0]]
\nTherefore, the gradient of the tangent at $x=0$ is [[1]]
\nWhen $x=0$, $y=$ [[2]]
\nHence, the equation for the tangent is:
\nSo, $y=$ [[3]]
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\nThe tangent to a line is the gradient of the curve at the point, and the normal is the negative reciprocal of this value (the perpendicular line intersecting the curve).
\nThe equation of a line can be found in many ways, but given a point and gradient for a straight line, two of the simple manipulations are:
\nUsing $y=mx+c$ and substituting in to find $c$
\nOR
\nUsing $\\frac{y_2-y_1}{x_2-x_1}=m$ and rearranging.
", "ungrouped_variables": ["a", "b", "c", "d"], "metadata": {"description": "Using differentiation to find the tangent and normal to a line at a given point
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "contributors": [{"name": "Denis Flynn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1216/"}]}]}], "contributors": [{"name": "Denis Flynn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1216/"}]}