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Practice Exam 2 Solved

# Practice Exam 2 Solved - Create assignment 57950 Exam 2 Oct...

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Create assignment, 57950, Exam 2, Oct 25 at 1:12 pm 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC3g06a 49:05, calculus3, multiple choice, > 1 min, wording-variable. 001 Determine d 2 y/dx 2 when x 2 + y 2 = 2 . 1. d 2 y dx 2 = 2 y 3 correct 2. d 2 y dx 2 = 2 y 3 3. d 2 y dx 2 = 2 y 2 4. d 2 y dx 2 = 2 y 2 5. d 2 y dx 2 = 1 y 3 Explanation: Differentiating implicitly with respect to x we see that 2 x + 2 y dy dx = 0 , so dy dx = x y . But then d 2 y dx 2 = d dx x y = y x dy dx y 2 = 1 y 2 y + x 2 y . Thus d 2 y dx 2 = 1 y 3 x 2 + y 2 = 2 y 3 . CalC3g16c 49:05, calculus3, multiple choice, < 1 min, wording-variable. 002 If x = x ( t ) is defined implicitly by 5 e 3 x = 3 tx + 5 t, find the value of x ( t ) at (1 , 0) . 1. x = 5 12 2. x = 1 3 3. x = 5 18 4. x = 1 2 5. x = 5 12 correct Explanation: Differentiating 5 e 3 x 3 tx 5 t = 0 implicitly with respect to t we see that 15 e 3 x x 3( t x + x ) 5 = 0 . Thus x = 3 x + 5 15 e 3 x 3 t , so at (1 , 0) x = 5 12 . CalC3g52a 49:05, calculus3, multiple choice, > 1 min, wording-variable. 003

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Create assignment, 57950, Exam 2, Oct 25 at 1:12 pm 2 The points P and Q on the graph of y 2 xy + 6 = 0 have the same x -coordinate x = 5. Find the point of intersection of the tangents to the graph at P and Q . 1. intersect at = 12 5 , 24 5 2. intersect at = 12 5 , 12 5 3. intersect at = 24 5 , 12 5 correct 4. intersect at = 24 5 , 24 5 5. intersect at = 12 5 , 4 5 Explanation: The y -coordinate at P, Q will be the solu- tions of ( ) y 2 xy + 6 = 0 at x = 5, i.e. , the solutions of y 2 5 y + 6 = ( y 3)( y 2) = 0 . Thus P = (5 , 3) , Q = (5 , 2) . To determine the tangent lines we need also the value of the derivative at P and Q . But by implicit differentiation, 2 y dy dx x dy dx y = 0 . so dy dx = y 2 y x . Thus dy dx P = 3 , dy dx Q = 2 . By the point-slope formula, therefore, the equation of the tangent line at P is y 3 = 3( x 5) , while that at Q is y 2 = 2( x 5) . Consequently, the tangent lines at P and Q are y 3 x = 12 and y + 2 x = 12 respectively. These two tangent lines intersect at = 24 5 , 12 5 . StewartC5 03 07 37 49:05, calculus3, multiple choice, > 1 min, wording-variable. 004 Find an equation of the tangent line to the ellipse x 2 a 2 + y 2 b 2 = 1 at the point ( x o , y o ). 1. x o x a 2 + y o y b 2 = 1 correct 2. x o x a 2 y o y b 2 = 1 3. x a 2 y b 2 = x o y o 4. x a 2 + y b 2 = 1 5. x o x a 2 + y o y b 2 = x o y o Explanation: x 2 a 2 + y 2 b 2 = 1 2 x a 2 + 2 yy b 2 = 0 y = b 2 x a 2 y
Create assignment, 57950, Exam 2, Oct 25 at 1:12 pm 3 An equation of the tangent line at ( x o , y o ) is y y o = b 2 x o a 2 y o ( x x o ) Multiplying both sides by y o b 2 gives y o y b 2 y 2 o b 2 = x o x a 2 + x 2 o a 2 x o x a 2 + y o y b 2 = x 2 o a 2 + y 2 o b 2 Since ( x o , y o ) lies on the ellipse, we have x 2 o a 2 + y 2 o b 2 = 1 x o x a 2 + y o y b 2 = 1 CalC3h01b 49:05, calculus3, multiple choice, > 1 min, wording-variable.

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Practice Exam 2 Solved - Create assignment 57950 Exam 2 Oct...

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