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Final Exam: Solutions
Problem 1. Given vectors u = (1, 0 , 2), v = (2, − 1 , 3), and w = (− 2 , 1 , 0), find (a) the vector u − v; (b) the unit vector in the direction of w; (c) the projection of u − v onto the direction of w.
Solution. (a) u − v = (1 − 2 , 0 − (−1), 2 − 3) = (− 1 , 1 , −1). (b) ‖w‖ =
(−2)^2 + 1^2 + 0^2 =
- The unit vector in the direction
of w is (^) ‖ww‖ =
−√ 2 5 ,^ √^1 5 ,^0
(c) the projection of u − v onto the direction of w is
(u−v)·
w ‖w‖
Problem 2. Identify and sketch the following surfaces: (a) 2x^2 + y^2 − z^2 = 0; (b) x^2 = z; (c) x^2 + 2y^2 + z^2 − 4 y = 0.
Solution. (a) 2x^2 + y^2 − z^2 = 0 is equivalent to z^2 = 2x^2 + y^2 , i.e., z^2 = x 2 (1/√2)^2 +^ y
(^2) , which is an equation of an elliptic cone.
(b) x^2 = z is an equation of a parabolic cylinder. (c) x^2 + 2y^2 + z^2 − 4 y = 0 is equivalent to x^2 + 2(y^2 − 2 y + 1) + z^2 = 2, i.e., x
2 ( √ 2)^2 + (y^ −^ 1)
(^2) + z^2 ( √ 2)^2 = 1, which is an equation of an ellipsoid centered at (0, 1 , 0).
Problem 3. Use an appropriate form of the chain rule to find (a) dzdt , where z = 5x^2 y, x = t^3 , y = t^2 ; (b) ∂ω∂u and ∂ω∂v , where ω = 4x − y + 2z, x = u sin v, y = v sin u, z = sin u sin v.
Solution. (a) dzdt = ∂z∂xdxdt + ∂z∂ydydt , thus
dz dt
= 10xy · 3 t^2 +5x^2 · 2 t = 10·t^3 ·t^2 · 3 t^2 +5·(t^3 )^2 · 2 t = 30t^7 +10t^7 = 40t^7.
(b) ∂ω∂u = ∂ω∂x∂x∂u + ∂ω∂y∂u∂y , thus
∂ω ∂u
= 4 sin v − v cos u + 2 cos u sin v. 1
∂ω ∂v =^
∂ω ∂x
∂x ∂v +^
∂ω ∂y
∂y ∂v , thus
∂ω ∂v
= 4u cos v − sin u + 2 sin u cos v.
Problem 4. Use the Lagrange multiplier method to find the maxi- mum sum of x^2 + y^2 + z^2 if x + 2y + 2z = 12.
Solution. The problem can be formulated as follows: find the maximum value of f (x, y, z) = x^2 + y^2 + z^2 under the constraint g(x, y, z) = x + 2y + 2z − 12 = 0. The Lagrange (vector) equation ∇f = λ∇g is equivalent to the three scalar equations
fx = λgx, fy = λgy, fz = λgz ,
i.e.,
2 x = λ · 1 , 2 y = λ · 2 , 2 z = λ · 2.
Eliminating λ, we get 2x = y = z. Using the constraint equation we get x + 2 · 2 x + 2 · 2 x − 12 = 0, i.e., 9x = 12. Therefore, x = 129 = 43 , y = z = 2 · 43 = 83. The maximum value of f (x, y, z) = x^2 + y^2 + z^2 is, therefore,
( 4 3
Problem 5. Use spherical coordinates to find the mass of the solid bounded below by the cone z =
x^2 + y^2 and above by the sphere x^2 + y^2 + z^2 = 9 if its density is given by δ(x, y, z) = x^2 + y^2 + z^2.
Solution. The solid has a shape of “ice-cream cone”. It can be de- scribed in spherical coordinates by the inequalities 0 ≤ θ ≤ 2 π (because it is axial symmetric about the z-axis), 0 ≤ φ ≤ π 4 (because the conic surface has an equation φ = π 4 in spherical coordinates), and 0 ≤ ρ ≤ 3 (because the radius of a spherical cap is 3). Since x^2 + y^2 + z^2 = ρ^2 , the density function in spherical coordinates is δ(ρ, φ, θ) = ρ^2. Therefore,
is
W =
∫ Q(1,1)
P (0,1)
F · dr =
0
F(x(t), y(t)) · r′(t) dt
0
(f (x(t), y(t))x′(t) + g(x(t), y(t))y′(t)) dt =
0
t^2 + 1
t t^2 + 1
dt
0
t^2 + 1
dt = − tan−^1 (t)
∣^1
0
= − tan−^1 (1) = −
π 4
