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MATH 241 Lecture 27: L27 2018-04-13 math 241

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26 Jul 2018

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University of Maryland

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Course

Professor

Christian Rosendal

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Q: is this conservative? (3) conservative vector fields defn: a vector field f is conservative if there is some (cid:1858) within (cid:1858)=f . Then (cid:1858) is potential fctn for f . ex. I e is there some (cid:1858) within (cid:1858)(cid:1876)=(cid:1877) and (cid:1858)(cid:1877)=(cid:1876) (cid:882) so not conservative! ex. is (cid:4666)(cid:1876),(cid:1877),(cid:1878)(cid:4667)=(cid:884)(cid:1876)(cid:1878)(cid:2835) (cid:1878)(cid:2836) +(cid:4666)(cid:1876)2 (cid:1877)+(cid:884)(cid:1878)(cid:4667) cons. well = ( (cid:883) (cid:4666) (cid:883)(cid:4667))(cid:2835) (cid:4666)(cid:884)(cid:1876) (cid:884)(cid:1876)(cid:4667)(cid:2836) +(cid:4666)(cid:882) (cid:882)(cid:4667) since =(cid:882) and we can plug anything into , it is cons! Need f with (cid:1858)(cid:3051)=(cid:884)(cid:1876)(cid:1878) (cid:4666)(cid:4667) (cid:1858)(cid:3052)= (cid:1878) (cid:4666)(cid:4667) (cid:1858)(cid:3053)=(cid:1876)2 (cid:1877)+(cid:884)(cid:1878) (cid:4666)(cid:4667) (i) tells us (cid:1858)(cid:4666)(cid:1876),(cid:1877),(cid:1878)(cid:4667)=(cid:1876)2(cid:1878)+(cid:1859)(cid:4666)(cid:1877),(cid:1878)(cid:4667) (iv) says (cid:1858)(cid:3052)=(cid:882)+(cid:1859)(cid:3052)(cid:4666)(cid:1877),(cid:1878)(cid:4667) but this =(cid:4666)(cid:4667) We"ll do 4 (cid:374)e(cid:449) i(cid:374)tegrals various theorems which connect them. 15. 2 line integrals (of fctns. and vector fields) (1) line integrals of functions. Situation: suppose c is a curve in space (2d or 3d) (think of c as a thin wire) The curve has density given by f(x,y,z) or f(x,y) (3d or 2d)

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Calculus

ISBN: 9781319050733

Single Variable Calculus: Early Transcendentals

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Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S