Math 340 Final Exam December 16, 2006

Transcription

1 Math 34 Final Exam December 6, 6. () Suppose A 3 4. a) Find the row-reduced echelon form of A. 3 4 so the row reduced echelon form is b) What is rank(a)? The rank is two since there are two pivots. c) Find a basis for the null space of A The vectors (x, y, z, w) T in the null space satisfy x+z+4w and y+z w so (x, y, z, w) z(,,, ) + w( 4,,, ). So the two linearly independent vectors (,,, ) T, ( 4,,, ) T form a basis of the null space of A. d) Find a basis for the column space of A. The pivot columns of A form a basis, so (,, ) T, (,, ) T forms a basis of the column space of A. e) Find all solutions X of AX. By either inspection or Gaussian elimination we see that one solution is (,,, ) T so all solutions are of the form (,,, ) T + z(,,, ) T + w( 4,,, ) T.. () Suppose A and B are 3 3 matrices and I is the 3 3 identity matrix. Find ( ) ( ) 3 A I A I and. B B ( ) A I B ( A A + B B ) and ( ) 3 ( ) A I A 3 A + AB + B B B () Recall a square matrix A is skew symmetric if A T A. Show that the skew symmetric matrices are a subspace of R n n. Find a basis for the 3 3 skew symmetric matrices. T so is skew symmetric. If A and B are skew symmetric then (A+B) T A T + B T A + ( B) (A + B) so A + B is skew symmetric. If c is a scalar and A is skew symmetric, then (ca) T ca T c( A) (ca) so ca is skew symmetric. So the

2 skew symmetric matrices form a subspace of R n n. Many of you also correctly showed this by using the formula ent ji (A) ent ij (A). Although nobody stated it this way, what these formulae give you are n(n + )/ homogeneous linear equations in the entries of the matrix, so their solutions form a subspace. 4. (35) onsider the curve parameterized by r(t) 4ti + 3 sintj + 3 costk, t π. a) Find the curvature κ of as a function of t. v(t) 4i + 3 costj 3 sintk and a(t) 3 sintj 3 cos tk. The speed v is a constant, 5. So a T d(speed)/dt. Then a N a a T κ a N / v 3/5. b) Find an equation of the line tangent to the curve at the point r(π/). a 3. One point on the curve is r(π/) (π, 3, ) and a vector in the direction of the curve is r (π/) (4,, 3). So an equation is (x, y, z) (π + 4t, 3, 3t). c) Find 3x ds. 3x ds π 3(4t) v(t) dt π 6t dt 3π. d) Find ydx zdy ydx zdy π (y, z, ) ds π sint 9 cos t dt (3 sint, 3 cost, ) (4, 3 cost, 3 sint) dt π cos t 4.5t +.5 sin(t) ] π sint 4.5( + cos(t)) dt 4.5π + 4 9π/ 5. (5) Let D be the triangular region in the xy plane bounded by the lines x, y, and x + y. Let S be the portion of the surface z x + y lying above D. Let be the boundary of S, oriented counterclockwise when viewed from above. Let F(x.y, z) 3xyi zj. a) Set up completely, but do not evaluate, an integral giving the area of S. ds fx + f y + da 4x + 4y + da. So the area is x 4x + 4y + dydx b) Use Stokes theorem to find F T ds.

3 curlf (,, 3x) so by Stokes theorem, F T ds x S curlf n ds 5x dydx x (,, 3x) ( x, y, ) dydx 5x( x) dx 5/ + 5/3 5/6 6. (3) Let D be the solid region above the surface z x +y, and below the surface z x y. Let S be the boundary of D oriented pointing outward from D. Let F(x, y, z) xi yj + zk. a) Find S F ds. divf so S F ds dv by Gauss theorem. D b) Set up completely, but do not evaluate, integrals giving the volume of D in cartesian, cylindrical, and spherical coordinates. These surfaces intersect where x + y and the shadow of D in the xy plane is the disc x + y. So in cartesian and cylindrical coordinates: volume x x y x x +y dzdydx π r r r dzdrdθ For spherical coordinates there are different limits for ρ above and below the xy plane. Above we have z x y so ρ cosφ ρ, so ρ sin φ + ρ cosφ and by the quadratic formula, ρ cos φ+ cos φ+4sin φ cos φ+ +3sin φ. Similarly sin φ sin φ for φ π/ we have ρ ρ cosφ so ρ cos φ+ +3sin φ. So here are a few ways to compute the volume in spherical coordinates π π/ cos φ+ +3 sin φ π π π ρ sin φ dρdφdθ + cos φ + +3 sin φ π/ π π cos φ+ +3 sin φ π/ cos φ+ +3 sin φ ρ sin φ dρdφdθ ρ sin φ dρdφdθ ρ sin φ dρdφdθ

4 The last expression comes from the observation that D is symmetric about the xy plane so the volume of D is twice the volume of the part of D above the xy plane. 7. () Let D be the region in the first quadrant bounded by xy, xy, x y, and x y 3 xy. Find D x + y da. Let u xy and v x y. Then in the uv( plane the ) region is bounded by u, y x u, v, and v 3 u. (u, v)/ (x, y) det x y so the integral is 3 u (x + y) x y dvdu 3 u dvdu 3 u du 3u u / ] / 3/ 8. () Let F(x, y, z) (3x + y sin(xy))i + (z + x sin(xy))j + (y + 4z)k. Let be the curve parameterized by r(t) t 8 i + t 9 (t + )j e sin(πt) k for t. Find F T ds. Note F is conservative. Trying to solve F gradg we get g/ x 3x + y sin(xy) > g(x, y, z) x 3 cos(xy) + (y, z) z + x sin(xy) g/ y + x sin(xy) + (y, z)/ y > (y, z) yz + D(z) y + 4z g/ z y + D (z) > D(z) z + E so we may let g(x, y, z) x 3 cos(xy) + yz + z. begins at (,, ) and ends at (,, ) so F T ds g(,, ) g(,, ) cos + ( + + ) cos 9. () True or false. (no justification required). a) If A is a square matrix and NS(A) {} then A exists. True, see page 55 or 89-9 of ullen b) (AB) T A T B T. False, (AB) T B T A T. c) det(ab) det(a) det(b). True, see page 5 of ullen

5 d) Any 5 vectors in R 4 are linearly dependent. True, see Thm.7 page 84 of ullen e) Any 5 vectors in R 4 span R 4. False, for example (,,, ), (,,,),(3,3,3,3),(4,4,4,4),(5,5,5,5) span a one dimensional subspace. f) Any 3 vectors in R 4 span a subspace of R 4. True, the span of any collection of vectors is a subspace. You even know in this case that they span a proper subspace of R 4, i.e., a subspace which is definitely smaller than R 4, since the dimension of the span is at most 3. g) If A is a 5 7 matrix then the dimension of NS(A) plus the rank of A is 5. False, it is 7, see page 89 of ullen