# solution_pdf-21 - gallegos(rg32245 Postclass 13.4...

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gallegos (rg32245) – Postclass 13.4 – sadun – (56620) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find the curvature, κ ( t ), of the curve parametrized by r ( t ) = 6 t i + 3 sin t j + 3 cos t k 1. κ ( t ) = 1 15 2. κ ( t ) = 45 3. κ ( t ) = 1 15 correct 4. κ ( t ) = 45 5. κ ( t ) = 3 Explanation: Now r ( t ) = 6 i + 3 cos t j 3 sin t k while r ′′ ( t ) = 3 sin t j 3 cos t k . But κ ( t ) = bardbl r ( t ) × r ′′ ( t ) bardbl bardbl r ( t ) bardbl 3 . In this case r ( t ) × r ′′ ( t ) = 9 i + 18 cos t j 18 sin t k and bardbl r ( t ) × r ′′ ( t ) bardbl = radicalBig 9 2 + 18 2 (cos 2 t + sin 2 t ) = 405 = 1 15 . Since bardbl r ( t ) bardbl = 36 + 9 = 15 , it follows that κ ( t ) = 1 15 (15) (3 / 2) = 1 15 . keywords: 002 10.0points For what value of x does the graph of f ( x ) = 1 5 e x have maximum curvature? 1. x = ln 5 2. x = ln parenleftbigg 5 2 2 parenrightbigg correct 3. x = 5 2 2 4. x = 5 5. x = 5 2 6. x = ln parenleftBig 5 2 parenrightBig Explanation: The curvature of the graph of y = f ( x ) is given by κ ( x ) = | f ′′ ( x ) | (1 + | f ( x ) | 2 ) 3 / 2 . But when f ( x ) = 1 5 e x , f ( x ) = f ′′ ( x ) = 1 5 e x . Thus κ ( x ) = 25 e x (25 + e 2 x ) 3 / 2 . Now the maximum value of κ ( x ) occurs at a critical value of κ . By the Quotient Rule, however, κ ( x ) = 25 parenleftBig e x (25 + e 2 x ) 3 / 2 3 e 3 x (25 + e 2 x ) 1 / 2 (25 + e 2 x ) 3 parenrightBig = 25 e x braceleftBig 25 2 e 2 x (25 + e 2 x ) 5 / 2 bracerightBig ,
gallegos (rg32245) – Postclass 13.4 – sadun – (56620) 2 so the only critical value of κ occurs when 2 e 2 x = 25, i.e. , when x