// Numbas version: exam_results_page_options {"name": "Hyperbolic #1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "variable_groups": [], "functions": {}, "ungrouped_variables": ["k", "r", "s"], "variables": {"k": {"group": "Ungrouped variables", "definition": "random(2..9#1)", "name": "k", "templateType": "randrange", "description": ""}, "r": {"group": "Ungrouped variables", "definition": "random(2..18#1)", "name": "r", "templateType": "randrange", "description": ""}, "s": {"group": "Ungrouped variables", "definition": "random(2..21#1)", "name": "s", "templateType": "randrange", "description": ""}}, "name": "Hyperbolic #1", "statement": "

Solve the following equation for \\(a\\) and \\(b\\):

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\\(aCosh(\\var{k}t)+bSinh(\\var{k}t)=\\var{r}e^{\\var{k}t}+\\var{s}e^{-\\var{k}t}\\)

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\\(a=\\) [[0]]

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\\(b=\\) [[1]]

"}], "advice": "

\\(aCosh(\\var{k}t)+bSinh(\\var{k}t)=\\var{r}e^{\\var{k}t}+\\var{s}e^{-\\var{k}t}\\)

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\\(a.\\frac{1}{2}\\left(e^{\\var{k}t}+e^{-\\var{k}t}\\right)+b.\\frac{1}{2}\\left(e^{\\var{k}t}-e^{-\\var{k}t}\\right)=\\var{r}e^{\\var{k}t}+\\var{s}e^{-\\var{k}t}\\)

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\\(\\frac{a}{2}e^{\\var{k}t}+\\frac{a}{2}e^{-\\var{k}t}+\\frac{b}{2}e^{\\var{k}t}-\\frac{b}{2}e^{-\\var{k}t}=\\var{r}e^{\\var{k}t}+\\var{s}e^{-\\var{k}t}\\)

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\\(\\left(\\frac{a}{2}+\\frac{b}{2}\\right)e^{\\var{k}t}+\\left(\\frac{a}{2}-\\frac{b}{2}\\right)e^{-\\var{k}t}=\\var{r}e^{\\var{k}t}+\\var{s}e^{-\\var{k}t}\\)

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Comparing like with like leads to two linear equations:

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\\(\\left(\\frac{a}{2}+\\frac{b}{2}\\right)e^{\\var{k}t}=\\var{r}e^{\\var{k}t}\\)  \\(\\implies \\frac{a}{2}+\\frac{b}{2}=\\var{r}\\) equation (i)

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and

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\\(\\left(\\frac{a}{2}-\\frac{b}{2}\\right)e^{-\\var{k}t}=\\var{s}e^{-\\var{k}t}\\)  \\(\\implies \\frac{a}{2}-\\frac{b}{2}=\\var{s}\\)  equation (ii)

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Adding equation (i) and equation (ii) gives

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\\(a=\\simplify{{r}+{s}}\\)

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Subtracting would give

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\\(b=\\simplify{{r}-{s}}\\)

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