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LINEAR ALGEBRA AND VECTOR ANALYSIS
MATH 22B
Unit 39: Final Exam Practice A
Problems
Problem 39A.1) (10 points): On the graph G in Figure 1 we are given a 1-form F on a graph G = (V, E). a) (3 points) Write the values of the curl dF. As a 2-form it is a function on the set T of triangles. b) (3 points) Compute the “discrete divergence” d∗F , which is a 0-form, a function on the vertices. c) (4 points) Find the value of the Laplacian d∗dF + dd∗F and enter the values near the edges in Figure 2.
Figure 1. A graph with a 1-Form F. Enter here the result for a) and b).
Figure 2. Enter here the result for c).
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Linear Algebra and Vector Analysis
Solution: Here are the pictures after computing the derivatives.
Side re- mark: Greens theorem tells that the sum of dF is the same as the line integral of F along the boundary. This does here not work and the reason is that upper middle trian- gle is oriented clockwise, not counter clockwise. There is no cancellation going on in the interior.
Problem 39A.2) (10 points) Each question is one point: a) Who formulated the law of gravity in the form the partial differential equation div(F ) = 4πσ? b) The expression 5xdxdzdx + 77dydzdy + 3dxdy + 6dydx simplifies to .... c) What value is
S [x, y, z]^ ·^ dS^ if^ S^ is the unit sphere oriented outwards? d) What is the distance between the point (0, 0 , 3) and the xy-plane? e) Is it true that if |r′(t)| = 1 everywhere, then r′′(t) is perpendicular to the velocity r′(t)? f) What is the distortion factor |dr| for the change of coordinates r(u, v) = [− 2 v, 3 u]? g) If r(u, v) parametrizes a surface in R^3 , is it true that ru × (ru × rv) tangent to the surface? h) Yes or no: if (0, 0 , 0) is a maximum of f (x, y, z) then fxx(0, 0 , 0) < 0. i) Write down the quadratic approximation of 1 + x + y + sin(x^2 − y^2 )? j) If S : f (x, y, z) = x^2 + y^2 + z^2 = 1 is oriented outwards, then the flux of ∇f through S is either negative, zero or positive. Which of the three cases is it?
Linear Algebra and Vector Analysis
Solution: a) A. b) k = 1. c) Maxwell equations d) |v · w| ≤ |v| · |w| e) no. f) x cos(xz)dzdxdy. g) r(y, z) = [z^2 − y^3 , y, z]. h) r(t) = [2 cos(t), 0 , 3 sin(t)]. i) Multiply both sides with a ρ. This gives the equation x = z which is a plane. j) |[a, b, c] × [u, v, w]| =
(bu − av)^2 + (cu − aw)^2 + (cv − bw)^2 /2.
Problem 39A.4) (10 points): a) (6 points) Find the equation of the plane which contains the line r(t) = [1+t, 2+t, 3 −t] and which is perpendicular to the plane Σ : x+2y −z = 4. b) (4 points) What is the angle between the normal vectors of Σ and the plane you just found?
Solution: a) The plane contains the point r(0) = [1, 2 , 3], the vector v = [1, 1 , −1] and needs to contain the vector w = [1, 2 , −1] which is normal to the plane Σ. To find the equation of the plane, we have to find the normal vector n = [a, b, c] to this plane. This n can be obtained as a cross product n = v × w = [1, 0 , 1] = [a, b, c]. The equation of the plane is ax + by + dz = d which is x + z = d, where the constant d = 4 is obtained by plugging in the point (1, 2 , 3). So, the equation of the plane is x + z = 4. b) We have to find the angle between w and n, which is can be obtained from cos(α) = (w · n)/(|w||n|) which is 0. The angle between the two vectors is 90 degrees. The two planes are perpendicular to each other. We could have seen this also directly as any plane which contains the normal vector to an other plane is perpendicular to that plane.
Problem 39A.5) (10 points): a) (8 points) Find the critical points of the function f (x, y) = cos(x) + y^5 − 5 y and classify them using the second derivative test. You can assume that 0 ≤ x < 2 π. b) (2 points) Does the function f have a global maximum or a global minimum?
Solution: a) The gradient is ∇f (x, y) = [− sin(x), 5 y^4 − 5] which is zero for x = 0, x = π and y = 1. There are four critical points (0, 1), (0, −1), (π, 1), (π, −1). The Hessian matrix H is
H =
[
− cos(x) 0 0 20 .y^3
]
. It has determinant D = − cos(x)y^3. We have fxx = − cos(x).
We see
point D fxx nature f value (0, −1) 20 -1 max 5 (0, 1) -20 -1 saddle - (π, −1) -20 1 saddle 3 (π, 1) 20 1 min -
b) No, for x = 0 we have the function y^5 − 5 y + 1 which is unbounded both from above and below.
Problem 39A.6) (10 points): a) (5 points) Use the Lagrange method to find the maximum of f (x, y) = y^2 − x under the constraint g(x, y) = x + x^3 − y^2 = 2. b) (5 points) The Lagrange equations fail to find the maximum of f (x, y) = y^2 − x under the constraint g(x, y) = x^3 − y^2 = 0. Still, the Lagrange theorem still allows you to find the maximum. How?
Solution: a) The Lagrange equations are
− 1 = λ(1 + 3x^2 ) 2 y = λ(− 2 y) x + x^3 − y^2 = 2
The first two equations give 2y = (1 + 3x^2 )2y. This implies y = 0 and from the constraint x = 1 b) The Lagrange theorem allows allows for the gradient of g to be zero. This also implies that the gradient of f and g are parallel. This happens here at (0, 0).
Problem 39A.7) (10 points): a) (6 points) Find the tangent plane at the point P = (4, 2 , 1 , 1) of the surface x^2 − 2 y^2 + z^3 + w^2 = 2. b) (4 points) Parametrize the line r(t) which passes through P which is perpendicular to the hyper surface at that point. Then find (r(1) + r(−1))/2.
Problem 39A.10) (10 points): a) Integrate the function f (x, y) = x+x^2 −y^2 over the region 1 < x^2 +y^2 < 4 , xy > 0. b) Find the surface area of r(t, s) = [cos(t) sin(s), sin(t) sin(s), cos(s)] 0 ≤ t ≤ 2 π, 0 ≤ s ≤ t/2.
Solution: a) The region consists of two quarter rings. One is in the first quadrant, the second is in the third quadrant. Make a picture. ∫ 2 1
∫ (^) π/ 2 0 (r^ cos(θ) +^ r
(^2) cos(2θ))r dθ + ∫^2 1
∫ (^3) π/ 2 π (r^ cos(θ) +^ r
(^2) cos(2θ))r dθ = 0.
b) We know the value of |rt × rs| = sin(s) already as this is just a sphere of radius 1. We now integrate (^) ∫ 2 π
0
∫ (^) t/ 2
0
sin(s)dsdt =
∫ (^2) π
0
1 dt = 2π.
Problem 39A.11) (10 points): Let E be the solid x^2 + y^2 ≥ z^2 , x^2 + y^2 + z^2 ≤ 9 , y ≥ |x|. a) (7 points) Integrate (^) ∫ ∫ ∫
E
x^2 + y^2 + z^2 dxdydz.
b) (3 points) Let F be a vector field F = [x^3 , y^3 , z^3 ] Find the flux of F through the boundary surface of E, oriented outwards.
Figure 4. The solid in Problem 10.
Linear Algebra and Vector Analysis
Solution: a) The region is best described in spherical coordinates. The φ angle goes from π/4 to 3 π/4. The θ angle goes from π/4 to 3π/4. The radius ρ goes from zero to 3. The integral is (^) ∫ 3 π/ 4
π/ 4
∫ (^3) π/ 4
π/ 4
0
ρ^2 ρ^2 sin(φ) dρdφdθ.
The answer is (π/2)(3^5 /5)
2 = 243 π
b) Since the divergence is 3x^2 + 3y^2 + 3z^2 the result is just three times the result found
in a). It is 729 π
Problem 39A.12) (10 points): What is the line integral of the force field F (x, y, z, w) = [1, 5 y^4 + z, 6 z^5 + y, 7 w^6 ]T^ + [y − w, 0 , 0 , 0]T^ along the path r(t) = [t^3 , sin(6t), cos(8t), sin(6t)] from t = 0 to t = 2π. Hint. We have written the field by purpose as the sum of two vector fields.
Solution: Use the fundamental theorem of line integrals for the first part of the vector field and compute the integral directly for the second. But we see that y − w is zero on the path so that the result is just f (r(2π)) − f (r(0)), where f = x + y^5 + z^6 + w^7 + yz. We have r(2π) = [(2π)^3 , 0 , 1 , 0] and r(0) = [0, 0 , 1 , 0]. The result is (2π)^3.
Problem 39A.13) (10 points): Find the area of the region |x|^2 /^5 + |y|^2 /^5 ≤ 1. Use an integral theorem.
Solution: Use the parametrization r(t) = [cos(t)^5 , sin(t)^5 ] and the vector field F (x, y) = [0, x]. The area is the line integral ∫ (^2) π
0
[0, cos(t)^5 ] · [5 cos(t)^4 sin(t), 5 sin(t)^4 cos(t)]
which asks to integrate
∫ (^2) π 0 5 sin(t)
(^4) cos(t) (^6) dt. This can be done using double angle
formulas and gives 15π/128. We integrate cos(t)^55 − sin(t)^4 cos(t) which is the same than integrating (cos(2t) + 1) sin(2t)^4 /32 and only the integral sin(2t)^4 /32 survives. This gives the result 15π/128.
Linear Algebra and Vector Analysis
Figure 5. The boundary of the surface is made of two circles r(t, 0) and r(t, 7 π/2). The picture gives the direction of the velocity vectors of these curves (which in each case might or might not be compatible with the orientation of the surface).
