YellowFinalSolutions

# YellowFinalSolutions - yellow MATH 32A Final Exam LAST NAME...

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yellow MATH 32A Final Exam March 17, 2008 LAST NAME FIRST NAME ID NO. Your TA: To receive credit, you must write your answer in the space provided . DO NOT WRITE BELOW THIS LINE 1 (20 pts) 5 (20 pts) 2 (20 pts) 6 (20 pts) 3 (20 pts) 7 (20 pts) 4 (20 pts) TOTAL

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2 PROBLEM 1 (20 Points) Let P be the plane through the points P 1 = (1 , 2 , 0) , P 2 = (2 , 0 , 1) , P 3 = (0 , 0 , 2) (A) Find an equation for the plane P . Answer : (B) Find the coordinates of the point Q where the plane P intersects the line through the two points A = (1 , 0 , 0) , B = (0 , 2 , 1) Answer :
3 Solution: (A) We have P 1 P 2 = h 2 , 0 , 1 i - h 1 , 2 , 0 i = h 1 , - 2 , 1 i P 1 P 3 = h 0 , 0 , 2 i - h 1 , 2 , 0 i = h- 1 , - 2 , 2 i We ﬁnd a normal to the plane by computing the cross prod- uct P 1 P 2 × P 1 P 3 = h 1 , - 2 , 1 i × h- 1 , - 2 , 2 i = h- 2 , - 3 , - 4 i Since P 1 = (1 , 2 , 0) lies on the plane, the equation of the plane is - 2( x - 1) - 3( y - 2) - 4 z = 0 or 2 x + 3 y + 4 z = 8 (B) The line through (1 , 0 , 0) and (0 , 2 , 1) has the parametriza- tion r ( t ) = h 1 - t, 2 t,t i We solve for t in the equation 2(1 - t )+3(2 t )+4 t = 8 2 - 2 t +6 t +4 t = 8 8 t = 6 t = 3 / 4 Since r ( 3 4 ) = h 1 4 , 3 2 , 3 4 i , we ﬁnd that the point Q = ( 1 4 , 3 2 , 3 4 ) lines on the intersection of the plane and the line.

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4 PROBLEM 2 (20 Points) Suppose that f ( x,y ) is a function of x and y , and that x = u 2 - v 2 , y = uv . Use the table of values ∂f ∂x and ∂f ∂y to compute ∂f ∂u ± ± ± ± ( u,v )=(1 , 1) and ∂f ∂v ± ± ± ± ( u,v )=(1 , 0) x y ∂f ∂x ( x,y ) ∂f ∂y ( x,y ) 0 0 2 -3 1 0 5 1 0 1 -1 -4 1 1 3 2 Answer : ∂f ∂u ± ± ± ( u,v )=(1 , 1) Answer : ∂f ∂v ± ± ± ( u,v )=(1 , 0)
5 Solution: (A) When ( u,v ) = (1 , 1), we have x = u 2 - v 2 = 0 , y = uv = 1 The table gives the values ∂f ∂x ± ± ± ( x,y )=(0 , 1) = - 1 , ∂f ∂y ± ± ± ( x,y )=(0 , 1) = - 4 By the Chain Rule: ∂f ∂u = ∂f ∂x ∂x ∂u + ∂f ∂y ∂y ∂u = ∂f ∂x (2 u ) + ∂f ∂y ( v ) Evaluate at ( u,v ) = (1 , 1) and ( x,y ) = (0 , 1): ∂f ∂u ± ± ± ( u,v )=(1 , 1) = ( - 1)(2) + ( - 4)(1) = - 6 (B) When ( u,v ) = (1 , 0), we have x = u 2 - v 2 = 1 , y = uv = 0 The table gives the values ∂f ∂x ± ± ± ( x,y )=(1 , 0) = 5 , ∂f ∂y ± ± ± ( x,y )=(1 , 0) = 1 By the Chain Rule: ∂f ∂v = ∂f ∂x ∂x ∂v + ∂f ∂y ∂y ∂v = ∂f ∂x ( - 2 v ) + ∂f ∂y ( u ) Evaluate at ( u,v ) = (1 , 0) and ( x,y ) = (0 , 1): ∂f ∂v ± ± ± ( u,v )=(1 , 0) = (5)(0) + (1)(1) = 1

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PROBLEM 3 (20 Points) Let r ( t ) = h cos 3 t, 0 , sin 3 t i (A) Show that || r 0 ( t ) || 2 = 9 cos 2 t sin 2 t (B) Find a formula for the curvature κ ( t ) at time t . Answer
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YellowFinalSolutions - yellow MATH 32A Final Exam LAST NAME...

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