18-----yellowfinalsol[1]

18-----yellowfinalsol[1] - yellow MATH 32A FINAL EXAM...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Sheet1 Page 1 yellow MATH 32A FINAL EXAM December 13, 2007 LASTNAME FIRST NAME IDNO. YourTA: To receive credit, you must write your answer in the space provided. DO NOT WRITE BELOW THIS LINE 1 (20 pts) 4 (20 pts) 2 (20 pts) 5 (20 pts) 3 (20 pts) 6 (20 pts)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Sheet1 Page 2 TOTAL FOR WRITTEN PROBLEMS 1 PROBLEM 1 (20 Points) The plane xz + y+ =1 23 intersects the x, y, and z axes in points P, Q, and R. (A)Determine P, Q, and R. (B)Find the area of the triangle .PQR. Solution: (A)The points are P =(2,0,0),Q=(0,1,0),R =(0,0,3) (B)The area of .PQR is one-half the length of the cross product: .. PQ PR =(j -2i) (3k -2i)=3i +2k +6j Err:509 Thus, the area is: .. p 11 7 || PQ PR||=32 62
Background image of page 2
Sheet1 Page 3 22 Err:520 22 2 2 PROBLEM 2 (20 Points) Thetemperature atposition(x,y,z)in a room is oC f(x,y,z)= xy +2z +17 A woman walksdown a spiral staircasein the middle of the room. She holds a thermometer whose position at time t (seconds)is tt t c(t)= hcos ,sin ,8- 333 (t in seconds). How fast is temperature reading on the thermometer changing at t =6 s. Solution: We have .f = hy,x,2i 1 t 1 t 1 c (t)= h- sin , cos ,- 3 33 33 At t = 6, c(6)= (x,y,z)= h1,0,8-2i .f = h0,1,2i
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Sheet1 Page 4 c (6)= h0, 11 ,- 33 By the ChainRule forpaths, the rate of change of tempera ture is d (t) dt The temperature reading at t =6 is changing at the rate 11 1 ,- i = - oC/s 33 3 3 4 PROBLEM 3 (20 Points) Findthe maximum value of f(x,y,z)= xyz, subject to the constraint 222 g(x,y,z)=9x + y + z =243 Solution: Use the method of Lagrange multipliers: .f = hyz,xz,xyi = .g = h18x, 2y,2zi Case 1: x,y,z are all non-zero. Then yz xz xy f(c(t))= .f h c .f h c ' (6)= h0,1,2ihh0,
Background image of page 4
Sheet1 Page 5 Err:520 18x 2y 2z This yields 2 22 2 y =9x, z =9x Plug into the constraint: 222 22 9x + y + z =9x 2 +9x 2 +9x =243 . x =243/27 =9 f(3,9,9)=3(9)(9)=243 Case2: Atleastone of x,y,z is zero. Inthis casef(x,y,z)= xyz =0. We conclude that the maximum value of f = xyz subject
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 17

18-----yellowfinalsol[1] - yellow MATH 32A FINAL EXAM...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online