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midterm 2 solutions

# midterm 2 solutions - Version 089 Exam 2 Schultz(56395 This...

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Version 089 – Exam 2 – Schultz – (56395) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine dy/dx when 2 cos x sin y = 7 . 1. dy dx = tan xy 2. dy dx = tan x tan y correct 3. dy dx = cot x tan y 4. dy dx = cot x cot y 5. dy dx = tan x Explanation: Differentiating implicitly with respect to x we see that 2 braceleftBig cos x cos y dy dx sin y sin x bracerightBig = 0 . Thus dy dx cos x cos y = sin x sin y . Consequently, dy dx = sin x sin y cos x cos y = tan x tan y . 002 10.0 points Find the slope of the tangent line to the graph of x 3 2 y 3 xy = 0 at the point P ( 1 , 1). 1. slope = 4 5 2. slope = 5 4 3. slope = 4 5 correct 4. slope = 5 4 5. slope = 2 3 6. slope = 3 2 Explanation: Differentiating implicitly with respect to x we see that 3 x 2 6 y 2 dy dx y x dy dx = 0 . Consequently, dy dx = 3 x 2 y 6 y 2 + x . Hence at P ( 1 , 1) slope = dy dx vextendsingle vextendsingle vextendsingle P = 4 5 . 003 10.0 points Find the differential dy when y = 3 + sin x 1 sin x . 1. dy = 3 cos x 1 sin x dx 2. dy = 3 sin x (1 sin x ) 2 dx 3. dy = sin x (1 sin x ) 2 dx 4. dy = 4 cos x (1 sin x ) 2 dx correct 5. dy = 4 cos x 1 sin x dx 6. dy = 4 cos x (1 sin x ) 2 dx

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Version 089 – Exam 2 – Schultz – (56395) 2 Explanation: After differentiation of y = 3 + sin x 1 sin x using the quotient rule we see that dy = (1 sin x ) cos x + cos x (3 + sin x ) (1 sin x ) 2 dx . Consequently, dy = 4 cos x (1 sin x ) 2 dx . 004 10.0 points If f is the function whose graph is given by 2 4 6 2 4 6 which of the following properties does f have? A. local maximum at x = 4 , B. critical point at x = 2 , C. f ( x ) > 0 on ( 1 , 2) . 1. B only 2. A and B only correct 3. all of them 4. B and C only 5. C only 6. A only 7. none of them 8. A and C only Explanation: The given graph has a removable disconti- nuity at x = 4 and a critical point at x = 2. On the other hand, recall that f has a local maximum at a point c when f ( x ) f ( c ) for all x near c . Thus f could have a local max- imum even if the graph of f has a removable discontinuity at c ; similarly, the definition of local minimum allows the graph of f to have a local minimum at a removable disconitu- ity. So it makes sense to ask if f has a local extremum at x = 4. Inspection of the graph now shows of the three properties A. f has , B. f has , C. f does not have . 005 10.0 points Find the absolute minimum value of f ( x ) = 4 + 2 sin 2 x on [ π, π ]. 1. abs minimum value = 4 correct 2. abs minimum value = 5 3. abs minimum value = 1 4. abs minimum value = 7 5. abs minimum value = 2 6. abs minimum value = 6 Explanation: The absolute minimum value of f on [ π, π ] occurs (a) either at an endpoint x = π or x = π , (b) or at a critical point of f in ( π , π ).
Version 089 – Exam 2 – Schultz – (56395) 3 To determine the critical points note first that f is differentiable everywhere in ( π, π ), so its critical points are the solutions of f ( x ) = 4 sin x cos x = 0 in ( π, π ). These are x = π 2 , 0 , π 2 .

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midterm 2 solutions - Version 089 Exam 2 Schultz(56395 This...

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