MH2800/MAS183 - Linear Algebra and Multivariable Calculus

Transcription

1 MH28/MAS83 - Linear Algebra and Multivariable Calculus SEMESTER II EXAMINATION 2-22 Solved by Tao Biaoshuai taob@e.ntu.edu.sg QESTION Let A Solve the homogeneous linear system Ax and write down the general solution in parametric form. 2. Write down a basis for the null space of A. SOLTION. Consider the augmented matrix After performing row operations we obtain. Since the number of variables (which is 4) minus the rank (which is 2) is 2, we should have 2 parameters in the general solution. If we let x 2 u, x 3 t, then x 4 x 3 t and x x 2 + x 4 u + t. Thus, one of the possible parametric forms for general solution can be (x, x 2, x 3, x 4 ) (u + t, u, t, t).

2 2. The null space of A is just the solution space of Ax. Since all its solutions can be represented as (x, x 2, x 3, x 4 ) u(,,, ) + t(,,, ), a basis of this null space can be {(,,, ), (,,, )}. QESTION 2 Consider the function f(x, y, z) 2x 2 + 4y z, and let S be the level surface given by f(x, y, z). Furthermore, let C be the curve given by x t, y t, z t 2 + 4t, t [, 5, and let P be the point of intersection between the level surface S and the curve C.. Find the coordinates of the point P. 2. Find an equation for the tangent plane to S at P. 3. Find parametric equations for the tangent line l to C at P. 4. Find the directional derivative of f at P in the direction along l in which f increases. SOLTION. We just need to solve the system of equations: 2x 2 + 4y z x t y t z t 2 + 4t Therefore, the coordinates of P is (2,, 2). x 2 y z 2 t 2 2. The surface can be represented as z 2x 2 +4y. Let the intersection plane be z ax+by +c. Then the coefficients a, b are just the partial derivatives z z x and y at P. Therefore, a z x 4x 8 b z P y 4. P 2

3 So the plane is z 8x + 4y + c. Since the plane surely passes through P, we have c, which implies c 8. Thus, the equation for the tangent plane to S at P is z 8x + 4y The line l must be paralleled to the vector (x, y, z ) at P. At P, we have (x, y, z ) (,, 2t + 4) (,, 8). So the parametric equations for l must be of form (t+a, t+b, 8t+c). Taking into consideration that it passing through (2,,2), one of the possible choices of a, b, c can be a, b, c 4. So one of the possible parametric equations for l can be (x, y, z) (t +, t, 8t + 4). Notice that the answer is not unique. 4. Among the two directions, (,, 8) and (,, 8), of l, the first one (,, 8) is the one in which f increases, since the substitution of l (which is (t+, t, 8t+4)) into f (which is f(t) 2(t+) 2 +4t (8t+4) 2t 2 2) is a increasing function of t for t [, 5. Let n be the unit vector pointing at the direction along l in which f increases. Then n (,, 8) (,, 8) The directional derivative of f at P along n is n f f n ( f x, f P y, z P x ) n P (4x, 4, ) (,, 8) (8, 4, ) (,, 8) 4. QESTION 3 Let [ cos 2θ sin 2θ A sin 2θ cos 2θ.. Determine if A is orthogonally diagonalizable. Justify your answer. 2. Show that for any θ R,u (cos θ, sin θ) is an eigenvector of A. What is the corresponding eigenvalue? 3. Find another eigenvector v of A, such that {u, v} is linearly independent. What is the eigenvalue corresponding to v? 4. If the answer to part is yes, write down an orthogonal matrix P and a diagonal matrix D such that D P T AP. 3

4 SOLTION. Since A is symmetric, it is orthogonally diagonalizable. 2. To find the eigenvalues, A Iλ λ 2 cos 2 2θ sin 2 2θ λ ±. Noticing that [ cos 2θ sin 2θ sin 2θ cos 2θ [ cos θ sin θ [ and [ cos 2θ + sin 2θ sin 2θ cos 2θ + [ sin θ cos θ [ we can see that u is an eigenvector of A and its corresponding eigenvalue is. 3. As shown in 2, another eigenvector is ( sin θ, cos θ) which corresponds to the eigenvalue Recall that D s diagonal line consists of all the eigenvalues and P s columns consists of all the corresponding eigenvectors, we have [ [ cos θ sin θ D P sin θ cos θ, QESTION 4 Compute the integrals:. ( ) x 3 dy dx. x 2 y 2. x2 + y 2 dxdydz, R where the region of integration R, is the volume bounded by the cone x 2 + y 2 z 2 and the plane z. SOLTION 4

5 . Let D be the region bounded by y x 2,x and y. Inside D, y varies from to, and for each given y, x varies from to y in D. By Fubini s theorem, we have ( ( y ) x 3 dy dx x 2 y D [ x 3 dxdy y x 4 y 4 dy y [ arcsin(y 3 ) π 2 24 x 3 y dx ) dy y 2 4 y dy 2. We switch to polar coordinate (x, y, z) (r cos θ, r sin θ, z) and compute the Jacobian: cos θ r sin θ J sin θ r cos θ r. Inside the region R, z varies from to. For fixed z, r varies from to z, and for fixed z and r, θ changes from to 2π. Therefore, z 2π x2 + y 2 dxdydz (r cos θ)2 + (r sin θ) 2 Jdθdrdz R z 2π 2 3 πz3 dz π r 2 dθdrdz z 2πr 2 drdz QESTION 5 Let C be the circle in R 3 given by x 2 + y 2, z, oriented anticlockwise around the origin in the xy-plane, and let. Calculate curl F. 2. Is F a conservative vector field? F(x, y, z) (, x, sin 2 y). 3. Suppose F represents a force vector field. What is the work done by F along C (one lap around the circle)? SOLTION. curl F i j k x y z x sin 2 y (2 sin y cos y,, ) 5

6 2. A conservative field must be irrotational. However, as shown in, curl F, so F is not conservative. 3. The parametric equation for C can be (cos θ, sin θ, ) for θ varies from to 2π. Thus, the work done is given by W F dr C 2π 2π (, cos θ, sin 2 (sin θ)) ( sin θ, cos θ, )dθ [ sin 2θ cos 2 θ + θ 4 2 2π π QESTION Let S be the surface S {(x, y, z) : x 2 + y 2 + 2(z ) 2, z } with outwards (away from the origin) pointing unit normal. Let F be the vector field F(x, y, z) (x 3 sin z, y 3 e z, 3y 2 e z + 3x 2 cos z).. Calculate div F. 2. Calculate the surface integral S F ds. SOLTION. div F x (x3 sin z) + y ( y3 e z ) + z (3y2 e z + 3x 2 cos z) 3x 2 sin z 3y 2 e z + (3y 2 e z 3x 2 sin z) 2. Let be the surface (actually a disk) given by {(x, y, z) : x 2 + y 2 4, z } with downwards (point at negative z direction) pointing unit normal. Then S is a closed surface which bounds some region R. By Divergence Theorem and noticing that div F, we have F ds + F d div Fdxdydz. S R

7 Therefore, F ds F d S (x 3 sin, y 3 e, 3y 2 e + 3x 2 cos ) (,, )dxdy 3x 2 + 3y 2 dxdy 3r 2 rdrdθ 2 2π 3r 3 dθdr 2 πr 3 dr 24π 7