1. Consider the function f (x, y) = y 3 + 2x2 y − xy 2 − 2x3 + 3x − 3y . At P (1, −1), find ~ ~ a) the directional √ derivative of f in the direction 2i − j I Ans: −4 5 b) the maximum directional derivative √ I Ans: 4 5 c) the direction of minimum directional derivative I Ans: h8, −4i 2. Examine the function for relative extrema and saddle points. (a) f (x, y) = 3x2 + 3xy + y 2 + 3x − 4 I Ans: (−2, 3, f (−2, 3)) is a relative minimum. (b) f (x, y) = 3x2 − 3y 2 − 4x + 2y − 8 I Ans: 32 , 13 , f 23 , 13 is a saddle point. y−x 3. The distance from a point (x, y) to the line y = x in the xy-plane is given by f (x, y) = √ . Use 2 the Lagrange method to find the point (x, y) on the parabola x2 − y + 2 = 0 which is closest to the line. 1 9 , I Ans: 2 4 4. Determine the equation of the tangent plane to the surface R(u, v) = h3u2 + 8uv, 3v 2 , 3u + vi at the point (−5, 3, 2). I Ans: 18(x + 5) + 12(y − 2) + 26(z − 2) = 0 5. Evaluate the following double integrals. (Sketch the region of integration in each case.) Z 1Z 0 x (a) dxdy 2 0 −1 (xy + 2) 1 I Ans : − ln 2 (see notes) 2 Z 8Z 2 1 (b) dydx √ 4 3x y + 1 0 √ I R : 3 x ≤ y ≤ 2, 0 ≤ x ≤ 2 ⇐⇒ R : 0 ≤ x ≤ y 3 , 0 ≤ y ≤ 2 Z 0
8Z 2 √ 3
x
1 dydx = y4 + 1
Z 0
2 Z y3 0
1 dx dy = y4 + 1
Z 0
2
3 Z 2 x
y y3 1 dy = dy = ln 17. 4+1 y 4 + 1 0 y 4 0
Z (c)
1Z
√
2−x2
p 2 − x2 − y 2 dydx 0 x √ I R : x ≤ y ≤ 2 − x2 , 0 ≤ x ≤ 1 ⇐⇒ R :
Z 0
1Z
√
2−x2
Z p 2 − x2 − y 2 dydx =
π/2 Z
π/4
x
≤ θ ≤ π4 , 0 ≤ r ≤
π 4
√
√
2
2p
2 − r2 rdr dθ =
0
Z
π/2
π/4
√
√2 3
1 2π 2 2
− (2 − r ) dθ = . 3 3 0
6. Set up an iterated double integral that represents the area of the paraboloid z = 9 − x2 − y 2 above the xy-plane. Z 3 Z √9−x2 p 1 + 4x2 + 4y 2 dy dx I Ans: √ −3
− 9−x2
p 7. Set up an iterated double integral that represents the volume of the cone z = 2 x2 + y 2 above the region in the xy-plane enclosed by y = x2 − 1 and y = x + 1. Z 2 Z x+1 p I Ans: 2 x2 + y 2 dy dx −1
x2 −1
8. Set up an iterated double integral that represents the area of the sphere x2 + y 2 + z 2 = 4z that lies inside the paraboloid zv = x2 + y 2 . !2 !2 Z √3 Z √9−x2 u u −x −y t + p dy dx I Ans: 1+ p √ √ 4 − x2 − y 2 4 − x2 − y 2 − 3 − 9−x2 9. Determine the surface area of the portion of the inverted cone z = 3 − lying outside cylinder x2 + y 2 = 1. Z 2π Zthe 3√ √ 2 r dr dθ = 8 2π I Ans: 0
endenumerate
1
p x2 + y 2 above the xy-plane