Math 53: Worksheet 9 Solutions

Transcription

1 Math 5: Worksheet 9 Solutions November 1 1 Find the work done by the force F xy, x + y along (a) the curve y x from ( 1, 1) to (, 9) We first parametrize the given curve by r(t) t, t with 1 t Also note that r (t) 1, t We then have F dr t t 1 t t 1 t(t ), t + t 1, t dt t + t t + t dt ] 1 (81 1) + 6 (7 + 1) (b) the semicircle x + y from (, ) to (, ) moving clockwise We first parametrize the given curve by r(t) sin(t), cos(t) with t π Also note that r (t) cos(t), sin(t) We then have F dr tπ t tπ t tπ sin(t) cos(t), sin(t) + cos(t) cos(t), sin(t) dt 8 sin(t) cos (t) sin (t) sin(t) cos(t) dt 8 sin(t) cos (t) (1 cos(t)) sin(t) dt t 8 ] π cos (t) t + sin(t) + cos(t) π ( 1 1) (π ) For the slinky parametrized by r(t) cos(t), sin(t), t for t π, let its density at point (x, y, z) be given by f(x, y, z) 1 + x + z Find the center of mass of 1

2 Denote the center of mass by ( x, ȳ, z) Note that r (t) sin(t), cos(t), 1 so that r (t) sin (t) + cos (t) + 1 We first find the mass of the slinky We have M f(x, y, z) ds (1 + x(t) + z(t)) r (t) dt (1 + cos(t) + t) dt t + sin(t) + t (π + π ) π (1 + π) We next find the appropriate moments We have M yz xf(x, y, z) ds We conclude that π M xz yf(x, y, z) ds cos(t)(1 + cos(t) + t) dt sin(t) + t + sin(t) + t sin(t) + cos(t) π M xy zf(x, y, z) ds sin(t)(1 + cos(t) + t) dt cos(t) cos(t) t cos(t) + sin(t) t(1 + cos(t) + t) dt t + t sin(t) + cos(t) + t ) (π + 8π π ( 1 + π x ) 1 (1 + π), ȳ 1 π(1 + π/), z 1 + π 1 + π

3 Suppose the position of a particle at time t is r(t) t sin(t), 1 cos(t) with t π Assume that the friction acting on the particle is given by F v where v is the velocity of the particle Find the work done by the friction against the particle We have so that W friction F r (t) 1 cos(t), sin(t) F dr F(t) r (t) dt 1 cos(t), sin(t) 1 cos(t), sin(t) dt 1 cos(t) + cos (t) + sin (t) dt 1 cos(t) dt t sin(t) π Let F xy + y, x + xy and let be the quarter ellipse x + 9y 1 from (1/, ) to (, 1/) (a) By parameterizing the curve, write F dr as a definite integral in terms of t Do not evaluate We can parametrize as r(t) cos(t)/, sin(t)/ with t π/ Note that r (t) sin(t)/, cos(t)/ so that F dr π/ π/ cos(t) sin(t) cos(t) sin (t) 6 + sin (t), cos (t) + 9 sin (t) 9 + cos (t) 1 sin(t) cos(t) sin(t), cos(t) + sin(t) cos (t) 9 (b) Show that F is conservative and find the potential function f for F Let P (x, y) xy + y and Q(x, y) x + xy Observe then that dt dt P y x + y, Q x x + y

4 so that P y Q x In addition, the domain of F is R which is open and simplyconnected We conclude that F is conservative Let f be the corresponding scalar potential for F We then have f x xy + y, f y x + xy (1) The first expression implies that f(x, y) x y + xy + g(y) for some function g This then gives f y x + xy + g (y) omparing this with the second expression in (1) yields g (y) g(y) for some constant We conclude that (c) Use (b) to evaluate the integral in (a) f(x, y) x y + xy + By the Fundamental Theorem of Line Integrals, F dr f dr f(, 1/) (1/, ) 5 (a) Show that a constant force does no work on a particle that moves once around the circle x + y 1 Let F a, b We can parametrize the circle by r(t) cos(t), sin(t) with t π Note that r (t) sin(t), cos(t) so that F dr a, b sin(t), cos(t) dt a sin(t) + b cos(t) dt a cos(t) + b sin(t) Alternatively, since F is constant, it can also be seen that F dr F dr F (r(π) r()) F More generally, since Q x P y for a constant force and the domain (R ) is open and simply-connected, the force is conservative and so we must have F dr (b) Does this also hold for the force F x, y? Using the same parameterization gives F dr cos(t), sin(t) sin(t), cos(t) dt sin(t) cos(t) dt cos(t)

5 Again note that since Q x P y for the given force and the domain (R ) is open and simply-connected, the force is conservative and so we must have F dr (c) How much work will the above fields do if the particle only moves around part of a circle? Suppose the particle only moves from r(t ) to r(t 1 ) For the force from (a), the scalar potential is f(x, y) ax + by + c for some constant c The work done by this force is then f(r(t 1 )) f(r(t )) a(cos(t 1 ) cos(t )) + b(sin(t 1 ) sin(t )) Similarly, for the force from (b), the scalar potential is g(x, y) x + y + for some constant The work done by this force is then g(r(t 1 )) g(r(t )) cos (t 1 ) cos (t ) + (sin (t 1 ) sin (t )) 6 Let F ln(x + y ) (a) Show that F x x + y, y x + y Let f(x, y) ln(x + y ) Then, so that f x F x x + y, f y y x + y x x + y, y x + y (b) Verify that P y Q x on the domain of F, ie, D R {(, )} Since D is not simply-connected, can we conclude that F is not conservative? Let P (x, y) so that P y Q x on D x and Q(x, y) y Observe then that x +y x +y P y xy (x + y ), Q x xy (x + y ) 5

6 Recall that P y Q x and an open simply-connected domain is a sufficient condition for F being conservative but not a necessary condition; in other words, we do not require the domain of F to be necessarily simply-connected for it to be conservative Since we defined F in the first place as the gradient of a scalar field f, it is conservative (c) Show in fact that for any path joining two points A 1 and A and lying entirely in D, the value of F dr depends only on the ratio r r 1 where r i is the distance from A i to the origin Note that F dr ( ) r f dr f A f A1 ln(r) ln(r1) ln ln so the result of the line integral is a function of r r 1 r 1 ( r r 1 ) 6