Name: Print TA Name: Math 20C. Final Exam June 15, 2006
Section Number: Section Time:
No calculators or any other devices are allowed on this exam. Write your solutions clearly and legibly; no credit will be given for illegible solutions. Read each question carefully. If any question is not clear, ask for clarification. Answer each question completely, and show all your work. 1. (10 points) Find the plane through the point P0 = (2, 1, -1) which is perpendicular to the planes 2x + y - z = 3 and x + 2y + z = 2.
The plane is determined by its normal vector n and a point. We choose the point to be P0 = (2, 1, -1). The normal vector can be computed as n = n 1 n2 , n1 = 2, 1, -1 , n1 = 1, 2, 1 .
where n1 and n2 are the normal vectors to the planes 2x + y - z = 3 and x + 2y + z = 2, respectively. Then, n= i j k 2 1 -1 1 2 1 = (1 + 2), -(2 + 1), (4 - 1) n = 3, -3, 3 .
We can pick up any vector proportional to 3, -3, 3 as the normal vector to the plane, for example a simpler one is n = 1, -1, 1 . Then, the equation of the plane is (x - 2) - (y - 1) + (z + 1) = 0 x-y+z =0 .
2. (8 points) Decide whether the
x4 - y 2 exists. Give reasons your answer. (x,y)(0,0) x4 + y 2 lim
Consider the path given by the line x = 0, then lim x4 - y 2 -y 2 = lim = lim -1 = -1. x4 + y 2 y0 +y 2 y0
(0,y)(0,0)
Consider the path given by the line y = 0, then x4 - y 2 x4 = lim 4 = lim 1 = 1. x0 x x0 (x,0)(0,0) x4 + y 2 lim Therefore, the limit does not exist.
3. (8 points) Does the function f (x, y, z) = e3x+4y cos(5z) satisfy the Laplace equation fxx + fyy + fzz = 0? Give reasons your answer.
fx = 3e3x+4y cos(5z), fxx = 9e3x+4y cos(5z), therefore,
fy = 4e3x+4y cos(5z), fyy = 16e3x+4y cos(5z),
fzz = -25e3x+4y cos(5z),
fz = -5e3x+4y sin(5z)
fxx + fyy + fzz = (9 + 16 - 25)e3x+4y cos(5z) = 0
fxx + fyy + fzz = 0 .
4. (10 points) Find the linear approximation L(x, y) of the function f (x, y) = 6 - x2 - y 2 at the point (1, 1). Use this approximation to estimate the value of f (0.8, 1.1).
f (x, y) = fx (x, y) = fy (x, y) =
6 - x2 - y 2 , -x , 6 - x2 - y 2 -y , 6 - x2 - y 2
6 - 2 = 2, -1 1 fx (1, 1) = =- , 2 6-2 -1 1 fy (1, 1) = =- . 2 6-2 f (1, 1) =
1 1 L(x, y) = - (x - 1) - (y - 1) + 2 . 2 2 1 1 1 1 41 L(0.8, 1.1) = - (-0.2) - (0.1) + 2 = (0.1) + 2 = 2 + = . 2 2 2 20 20 L(0.8, 1.1) = 41 . 20
5. (10 points) Find the local maxima, local minima and saddle points of the function f (x, y) = x3 + y 3 + 3x2 - 3y 2 - 8.
f = 3x2 + 6x, 3y 2 - 6y = 0, 0
3x(x + 2) = 0 3y(y - 2) = 0
so x = 0 or x = -2, while y = 0 or y = 2. Then, there are four stationary points given by 0), (0, fxx = 6x + 6, Therefore, D = fxx fyy - fxy D(0, 0) = -36, D(0, 2) = 36, D(-2, 0) = 36, D(-2, 2) = -36, fxx (0, 2) = 6, fxx (-2, 0) = -6,
2
(0, 2),
(-2, 0),
(-2, 2). fxy = 0.
fyy = 6y - 6,
= fxx fyy = 36(x + 1)(y - 1). (0, 0) saddle point , (0, 2) local minimum , (0, 2) local maximum , (-2, 2) saddle point .
Evaluating it at each stationary point we get:
6. (10 points) Use Lagrange multipliers to find the maximum and minimum values of the 1 1 1 1 function f (x, y) = - + subject to the constraint 2 + 2 = 1. x y x y
Denote g(x, y) = are
1 x2
+ y12 - 1, so the constraint is g = 0. The Lagrange multipliers equations f = g, 1 1 ,- x2 y 2 = - and g = 0. 2 1 = - 3, 2 x x 2 1 - = - . y2 y3 x = -y. x = 2.
2 2 ,- x3 y 3
Then x and y must be nonzero, so, x = -2, 1 1 + =1 x2 x2 Then, y = y = 2
Then, using this information in the constraint we have 2 =1 x2
2, that is, the points to consider are ( 2, - 2), (- 2, 2). 1 1 2 f ( 2, - 2) = - + = - = - 2, 2 (- 2) 2 1 2 1 f (- 2, 2) = - + = = 2. (- 2) 2 2
Therefore, we conclude that ( 2, - 2) is a minimum of f , (- 2, 2) is a maximum of f .
3
7. Consider the integral
D
f (x, y) dA =
0
2(1- x ) 3 -2 q 2 1- x2
3
f (x, y) dy dx.
(a) (8 points) Sketch the region of integration. (b) (8 points) Switch the order of integration in the above integral. (c) (8 points) Compute the integral
D
f (x, y) dA for the case f (x, y) = xy.
(a)
y 3 2
3
x
-2
(b)
0 3 0
f (x, y) dA =
D -2
q 2 1- y2
2
2
f (x, y) dx dy +
0 0
3(1- y ) 2
f (x, y) dx dy.
(...