// Numbas version: finer_feedback_settings {"name": "Simon's copy of Laplace of trig functions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"condition": "a<>b and a<>c", "maxRuns": 100}, "extensions": [], "parts": [{"marks": 1, "checkVariableNames": false, "vsetRange": [0, 1], "scripts": {}, "showCorrectAnswer": true, "variableReplacements": [], "answer": "s/(s^2+{a}^2)", "checkingType": "absdiff", "extendBaseMarkingAlgorithm": true, "vsetRangePoints": 5, "unitTests": [], "failureRate": 1, "checkingAccuracy": 0.001, "type": "jme", "prompt": "

Find $L[\\cos(\\var{a}t)]$

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Find $L[\\sin(\\var{b}t)]$

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Find $L[\\cos(\\frac{t}{\\var{a}})+\\sin(\\frac{t}{\\var{c}})]$

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", "expectedVariableNames": [], "showPreview": true, "customMarkingAlgorithm": "", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst"}], "name": "Simon's copy of Laplace of trig functions", "statement": "

Note that

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$L[\\cos(bt)]=\\frac{s}{s^2+b^2}$

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$L[\\sin(bt)]=\\frac{b}{s^2+b^2}$

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Use these results to carry out the Laplace transforms below:

", "functions": {}, "ungrouped_variables": ["a", "b", "c"], "variables": {"a": {"definition": "random(2,4,5,8,10,20)", "description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "a"}, "b": {"definition": "random(2,4,5,8,10,20)", "description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "b"}, "c": {"definition": "random(2,4,5,8,10,20)", "description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "c"}}, "advice": "

(a)

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$L[\\cos(bt)]=\\frac{s}{s^2+b^2}$

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In this example $b=\\var{a}$

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Hence $L[\\cos(\\var{a}t)]=\\frac{s}{s^2+(\\var{a})^2}$

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(b)

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Note: $L[\\sin(bt)]=\\frac{b}{s^2+b^2}$

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In this example $b=\\var{b}$

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Hence  $L[\\sin(\\var{b}t)]=\\frac{(\\var{b})}{s^2+(\\var{b})^2}$

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(c)

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Note: $L[\\cos(bt)]=\\frac{s}{s^2+b^2}$ and  $L[\\sin(bt)]=\\frac{b}{s^2+b^2}$

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In the first part $b=\\frac{1}{\\var{a}}$ and in the second part $b=\\frac{1}{\\var{c}}$

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Hence $L[\\cos(\\frac{t}{\\var{a}})+\\sin(\\frac{t}{\\var{b}})]=\\frac{s}{s^2+(\\frac{1}{\\var{a}})^2}+\\frac{\\frac{1}{\\var{c}}}{s^2+(\\frac{1}{\\var{c}})^2}=\\simplify{(s/(s^2+1/{a}^2)+(1/{c})/(s^2+1/{c}^2))}$

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", "tags": [], "metadata": {"description": "

rebelmaths

", "licence": "Creative Commons Attribution 4.0 International"}, "preamble": {"js": "", "css": ""}, "rulesets": {}, "variable_groups": [], "type": "question", "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}