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Quiz 10 Name: SOLUTIONS Maths 114 - Calculus II November 30, 2010 Note: In order to receive full credit, you must show work that justifies your answer. 2 2 x 3 1. Show that the vector field F(x, y) R = h3x y + e , 2x yi is conservative, and hence evaluate the line integral C F · dr, where C is the curve parameterized by r(t) = sin2 (2πt)et , cos(2πt) − t , 0 ≤ t ≤ 1. To show that the vector field is conservative, we need to show that the partial y derivative of the x-component of F(x, y) is equal to the partial x derivative of the y-component of F(x, y) (and also note that the function is defined on a ∂ ∂ (3x2 y 2 + ex ) = 6x2 y and ∂x (2x3 y) = 6x2 y. The simply connected region). ∂y domain of F(x, y) is all of R2 , so it is defined on a simply connected region. Therefore F(x, y) is a conservative vector field. Now that we know the vector field is conservative, we are able to apply the fundamental theorem of line integrals. In order to do so, we first need to find a potential function for F(x, y). If fx = 3x2 y 2 + ex and fy = 2x3 y, then R 2 2 f (x, y) = (3x y + ex ) dx = x3 y 2 + ex + C(y), so ∂ 2x3 y = fy = ∂y (x3 y 2 + ex + C(y)) = 2x3 y + C 0 (y). Therefore C 0 (y) = 0, so C(y) is a constant. Any choice of constant will give us a potential function, so we may as well choose the constant to be zero. Thus a potential function is f (x, y) = x3 y 2 + ex . R The fundamental theorem of line integrals says that C F · dr = f (b) − f (a), where a is the starting point of the curve C, and b is the end point. From the parameterization, we see that a = r(0) = h0, 1i and b = r(1) = h0, 0i. Therefore Z F · dr = f (0, 0) − f (0, 1) = 1 − 1 = 0. C 2. Calculate the work done by the force field F(x, y) = hx3 − y, x + ey i on a particle which moves once around the circle x2 + y 2 = 1 in the anticlockwise direction. R 2 2 The work done is C F · dr, where C is the circle RR x + y = 1, positively oriented. By Green’s theorem, this is equal to D Qx − Py dA, where D is the disk bounded by C and P (x, y) = x3 − y, Q(x, y) = x + ey are the components of F(x, y). Therefore the work done is ZZ ZZ ∂ ∂ y 3 (x + e ) − x − y dA = 1 − (−1) dA = 2 Area(D) = 2π. ∂y D D ∂x Quiz 10 Name: SOLUTIONS Maths 114 - Calculus II December 2, 2010 Note: In order to receive full credit, you must show work that justifies your answer. 1. Consider F(x, y) = D 1 , 1 x+y x+y E + y and C : r(t) = ht, et i, 0 ≤ t ≤ 1. (a) Show that F is a conservative vector field. To show that the vector field is conservative, we need to show that the partial y derivative of the x-component of F(x, y)is equal to the partial x ∂ 1 −1 derivative of the y-component of F(x, y). ∂y = (x+y) 2 and x+y 1 ∂ −1 1 + y = (x+y) 2 . Therefore F(x, y) is a conservative vector field . ∂x x+y R (b) Evaluate C F · dr. Because we know that F is conservative, we can apply the fundamental theorem of line integrals. To do so, we first need to find a potential 1 1 and Φy = x+y + y, then function for F. If Φx = x+y R 1 Φ(x, y) = x+y dx = ln |x + y| + C(y), so 1 ∂ 1 + y = Φy = ∂y (ln |x + y| + C(y)) = x+y + C 0 (y). Therefore C 0 (y) = y, x+y 2 so C(y) = y2 + c, where c is a constant. Any choice of constant will give us a potential function, so we may as well put c = 0. Thus a potential 2 function is Φ(x, y) = ln |x + y| + y2 . The fundamental theorem of line integrals says that R F · dr = Φ(b) − Φ(a), where a is the starting point of the curve C, and C b is the end point. From the parameterization, we see that a = r(0) = h0, 1i and b = r(1) = h1, ei. Therefore Z e2 1 F · dr = Φ(1, e) − Φ(0, 1) = ln(1 + e) + − . 2 2 C You could also do this problem without using the fundamental theorem of line integrals, just by using the definition of a line integral. 1 Strictly speaking, we should also show that the function is defined on a simply connected region. The domain of F(x, y) is R2 minus the points where x + y = 0, which is not connected since it consists of two pieces, namely the half-planes x + y < 0 and x + y > 0. However, the curve C is contained in the half-plane x + y > 0, and this is simply connected, so we are OK. R 2. Evaluate C F · dr using Green’s Theorem, where C is the circle x2 + y 2 = 9 oriented clockwise, and where F = hex + sin(x), x + yi . Because C is oriented clockwise, Green’s theorem says that R RR F · dr = − Q − Py dA, where D is the disk bounded by C and x C D x P (x, y) = e + sin(x), Q(x, y) = x + y are the components of F. Therefore ZZ Z ∂ x ∂ (x + y) − (e + sin(x)) dA F · dr = − ∂x ∂y C D ZZ = − 1 − 0 dA D = −Area(D) = −π32 = −9π.