This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: yellow MATH 32A FINAL EXAM December 13, 2007 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) 4 (20 pts) 2 (20 pts) 5 (20 pts) 3 (20 pts) 6 (20 pts) TOTAL FOR WRITTEN PROBLEMS 1 2 PROBLEM 1 (20 Points) The plane x 2 + y + z 3 = 1 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) , Q = (0 , 1 , 0) , R = (0 , , 3) (B) The area of △ PQR is onehalf the length of the cross product: → PQ × → PR = ( j − 2 i ) × (3 k − 2 i ) = 3 i + 2 k + 6 j = ( 3 , 6 , 2 ) Thus, the area is: 1 2  → PQ × → PR  = 1 2 radicalbig 3 2 + 6 2 + 2 2 = 7 2 3 PROBLEM 2 (20 Points) The temperature at position ( x, y, z ) in a room is f ( x, y, z ) = xy + 2 z + 17 o C A woman walks down a spiral staircase in the middle of the room. She holds a thermometer whose position at time t (seconds) is c ( t ) = ( cos t 3 , sin t 3 , 8 − t 3 ) ( t in seconds). How fast is temperature reading on the thermometer changing at t = 6 π s. Solution: We have ∇ f = ( y, x, 2 ) c ′ ( t ) = (− 1 3 sin t 3 , 1 3 cos t 3 , − 1 3 ) At t = 6 π , c (6 π ) = ( x, y, z ) = ( 1 , , 8 − 2 π ) ∇ f = ( , 1 , 2 ) c ′ (6 π ) = ( , 1 3 , − 1 3 ) By the Chain Rule for paths, the rate of change of temper ature is d dt f ( c ( t )) = ∇ f · c ′ ( t ) The temperature reading at t = 6 π is changing at the rate ∇ f · c ′ (6 π ) = ( , 1 , 2 ) · ( , 1 3 , − 1 3 ) = − 1 3 o C/s 4 PROBLEM 3 (20 Points) Find the maximum value of f ( x, y, z ) = xyz , subject to the constraint g ( x, y, z ) = 9 x 2 + y 2 + z 2 = 243 Solution: Use the method of Lagrange multipliers: ∇ f = ( yz, xz, xy ) = λ ∇ g = λ ( 18 x, 2 y, 2 z ) Case 1: x, y, z are all nonzero. Then yz 18 x = xz 2 y = xy 2 z This yields y 2 = 9 x 2 , z 2 = 9 x 2 Plug into the constraint: 9 x 2 + y 2 + z 2 = 9 x 2 + 9 x 2 + 9 x 2 = 243 ⇒ x 2 = 243 / 27 = 9 We obtain x = ± 3,...
View
Full Document
 Spring '08
 GANGliu
 Math, Derivative, Multivariable Calculus, Gradient

Click to edit the document details