This file is /conf/snippets/setheader.pg you can use it as a model for creating files which introduce each problem set.

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1 Jason Hill Math24 3 WeBWorK assignment number Exameview is due : 5/9/212 at 5:am MDT. The (* replace with url for the course home page *) for the course contains the syllabus, grading policy other information. This file is /conf/snippets/setheader.pg you can use it as a model for creating files which introduce each problem set. The primary purpose of WeBWorK is to let you know that you are getting the correct answer or to alert you if you are making some kind of mistake. Usually you can attempt a problem as many times as you want before the due date. However, if you are having trouble figuring out your error, you should consult the book, or ask a fellow student, one of the TA s or your professor for help. Don t spend a lot of time guessing it s not very efficient or effective. Give 4 or 5 significant digits for (floating point) numerical answers. For most problems when entering numerical answers, you can if you wish enter elementary expressions such as 2 3 instead of 8, sin(3 pi/2)instead of -1, e (ln(2)) instead of 2, (2 +tan(3)) (4 sin(5)) 6 7/8 instead of , etc. Here s the list of the functions which WeBWorK understs. You can use the Feedback button on each problem page to send to the professors. 1. ( pts) Library/U/Michigan/hap12ec1/Q31v2.pg Find the equations of planes that just touch the sphere (x 1) 2 + (y 1) 2 + (z 5) 2 = 25 are parallel to (a) The xy-plane: (b) The yz-plane: (c) The xz-plane: The sphere has center at (1,1,5) radius 5. (a) The planes parallel to the xy-plane just touching the sphere are 5 above 5 below the center. Thus, the planes z = 1 z = are both parallel to the xy-plane touch the sphere at the points (1,1,1) (1,1,). (b) The planes parallel to the yz-plane just touching the sphere are 5 to the left of 5 to the right of the center. Thus, the planes x = 6 x = 4 are both parallel to the yz-plane touch the sphere at the points (6,1,5) ( 4,1,5). (c) The planes parallel to the xz-plane just touching the sphere are 5 to the left of 5 to the right of the center. Thus, the planes y = 6 y = 4 are both parallel to the xz-plane touch the sphere at the points (1,6,5) (1, 4,5). z = 1 z = x = 6 x = -4 y = 6 y = ( pts) Library/Michigan/hap12ec5/Q19.pg Find a function f (x,y,z) whose level surface f = 1 is the graph of the function g(x,y) = x + 6y. f (x,y,z) = The graph of g(x,y) = x + 6y is the set of all points (x,y,z) satisfying z = x + 6y, or x + 6y z =. This is a level surface, but we want the surface equal to the constant value 1, not, 1 so we can add 1 to both sides to get x + 6y z + 1 = 1. Thus, f (x,y,z) = x + 6y z + 1 has level surface f = 1 identical to the graph of g(x,y) = x + 6y. Of course, we could also say that the graph of g(x,y) is equivalent to z x + 6y =, so that f = z x 6y + 1. x+6*y-z+1 3. ( pts) Library/U/Michigan/hap12ec6/Q19-hilljb.pg For the function f (x,y) below, determine whether there is a value for c making the function continuous everywhere. If so, find it. { c + y x 4, f (x,y) = 6 y x > 4 c = (If there is no value of c that works, enter none, be sure that you can explain why there is no such value.) The function f is continuous at all points (x,y) with x 4. o let s analyze the continuity of f at the point (4,a). We have lim f (x,y) = lim(c + y) = c + a (x,y) (4,a) y a x<4 lim f (x,y) = lim(6 y) = 6 a. (x,y) (4,a) y a x>4 o we need to see if we can find one value for c such that c + a = 5 a for all a. This would require that c = 5 2a, but then c would depend on a, which is exactly what we don t want. Therefore, we cannot make the function continuous everywhere. Graphically we can see this by thinking about what the function looks like: for x 4, it is a plane with positive y-slope, passing through the line y =, z = c. For x > 4, it is a plane with negative y-slope, pssing through the line y =, z = y = 6. none

2 4. ( pts) Library/Michigan/hap12ec6/Q15.pg how that the function f (x,y) = x4 y x 8 + y 4. does not have a limit at (,) by examining the following limits. (a) Find the limit of f as (x,y) (,) along the line y = x. f (x,y) = lim (x,y) (,) y=x (b) Find the limit of f as (x,y) (,) along the line y = x 4. lim f (x,y) = (x,y) (,) y=x 4 (Be sure that you are able to explain why the results in (a) (b) indicate that f does not have a limit at (,)! (a) Let us suppose that (x,y) approaches (,) along the line y = x. Then Therefore f (x,y) = f (x,x) = x5 x 8 + x 4 = x x x lim f (x,y) = lim (x,y) (,) x x =. y=x (b) On the other h, if (x,y) approaches 99 99,) the curve y = x 4 we have so f (x,y) = f (x,x 4 ) = x8 x 8 + x 16 = 1 lim f (x,y) = lim f (x,x 4 ) = 1. (x,y) (,) x y=x 4 Thus no matter how close they are to the origin, there will be points (x,y) such that f (x,y) is close to points (x,y) such that f (x,y) is close to 1. o the limit does not exist. 1 lim f (x,y) (x,y) (,) 5. ( pts) Library/Michigan/hap13ec3/Q17.pg (a) Find a vector n perpendicular to the plane z = 8x + 6y. n = (b) Find a vector v parallel to the plane. v = (a) Writing the plane in the form 8x + 6y z = shows that a normal vector is n = 8 i + 6 j k. Any multiple of this vector is also a correct answer. 2 (b) Any vector perpendicular to n is parallel to the plane, so one possible answer is v = 6 i 8 j. Many other answers are possible. 8i+6j-k 6i-8j 6. ( pts) Library/Michigan/hap13ec3/Q31.pg ompute the angle between the vectors ĩ j+ k ĩ+ j+ k. angle = radians (Give your answer in radians, not degrees.) cosθ = ( ĩ j+ k) ( ĩ+ j+ k) ĩ j+ k ĩ+ j+ k = 1 3. o, θ = arccos( 1 3 ) 1.23 radians. arccos(1/3) 7. ( pts) Library/U/Michigan/hap13ec4/Q28.pg uppose v w = 7 v w = 3, the angle between v w is θ (measured in radians). Find (a) tanθ = (b) θ = (a) ince v w = v w cosθ v w = v w sinθ, tanθ = sinθ v w = cosθ v w = 3 7 = (b) Then θ = tan 1 ( ) = /7 arctan(3/7) 8. ( pts) Library/Michigan/hap13ec4/Q13.pg Find an equation for the plane through the points (2,4,4),( 2, 1,),( 2,, 1). The plane is The displacement vector from (2,4,4) to ( 2, 1,) is: a = 4 i 5 j 4 k. The displacement vector from (2,4,4) to ( 2,, 1) is: b = 4 i 4 j 5 k. Therefore a vector normal to the plane is: n = a b = 9 i 4 j 4 k. Using the first point, the equation of the plane can be written as: 9*x-4*y-4*z = -14 9x 4y 4z = 14.

3 9. ( pts) Library/Michigan/hap14ec3/Q3.pg Find the equation of the tangent plane to at the point (,4,78). z = We have z = e x + y + y z = e x + y + y The partial derivatives are x = e x ln(e) = 1, (x,y)=(,4) (x,y)=(,4) y = 1 + 3y 2 = 49. (x,y)=(,4) (x,y)=(,4) o the equation of the tangent plane is z = (x ) + 49(y 4) = 78 + x + 49(y 4). 78+x+49*(y-4) 1. ( pts) Library/Michigan/hap14ec4/Q69.pg (a) What is the rate of change of f (x,y) = 3xy + y 2 at the point (4,4) in the direction v = 2i + 2 j? f v = (b) What is the direction of maximum rate of change of f at (4,4)? direction = (Give your answer as a vector.) (c) What is the maximum rate of change? maximum rate of change = We see that f = 3yi + (3x + 2y) j, so at the point (4,4), we have f = 12i + 2 j. (a) The directional derivative is f (4,4) v 2i + 2 j = (12i + 2 j) = 64. v 8 8 (b) The direction of maximum rate of change is f (4,4) = 12i + 2 j. (c) The maximum rate of change is i+2j f (4,4) = If 11. ( pts) Library/Michigan/hap14ec6/Q8.pg z = (x + 4y)e y, x = u, y = ln(v), find / u / v. The variables are restricted to domains on which the functions are defined. / u = / v = ince z is a function of two variables x y which are functions of two variables the two chain rule identities which apply are: u = x x u + y y u so so u = (ey )(1) + (4e y + (x + 4y)e y ln(e)) = e ln(v) 1. v = x x v + y y v v = (ey )() + (4e y + (x + 4y)e y ln(e)) ( 1 v ) ( 1 = 4e ln(v) + (u + 4ln(v))e ln(e)) ln(v) v. ((eˆ[ln(v)]))*((1)) ((4*eˆ[ln(v)]+[u+4*ln(v)]*eˆ[ln(v)]*ln(e)))*((1/v)) 12. ( pts) Library/Michigan/hap15ec1/Q7.pg Find the critical points for the function f (x,y) = x 2 2xy + 3y 2 12y classify each as a local maximum, local minimum, saddle point, or none of these. critical points: (give your points as a comma separated list of (x,y) coordinates.) classifications: (give your answers in a comma separated list, specifying maximum, minimum, saddle point, or none for each, in the same order as you entered your critical points) To find the critical points, we solve f x = f y = for x y. These equations are f x = 2x 2y =, f y = 2x + 6y 12 =. We see from the first equation that x = y. ubstituting this into the second equation shows that y = 3. The only critical point is (3,3). We then have D = ( f xx )( f yy ) ( f xy ) 2 = (2)(6) ( 2) 2 = 8.

4 ince D > f xx = 2 >, the function f has a local minimum at the point (3,3). (3,3) minimum 13. ( pts) Library/Michigan/hap15ec2/Q13.pg Does the function f (x,y) = x y3 + 8y 2 5x have a global maximum global minimum? If it does, identify the value of the maximum minimum. If it does not, be sure that you are able to explain why. Global maximum? (Enter the value of the global maximum, or none if there is no global maximum.) Global minimum? (Enter the value of the global minimum, or none if there is no global minimum.) uppose x is fixed. Then for large values of y the sign of f is determined by the highest power of y, namely y 3. Thus, f (x,y) as y, there can be no global maximum. For y we similarly have f (x,y), so that there is no global minimum either. none none 14. ( pts) Library/U/Michigan/hap15ec3/Q43.pg For each value of λ the function h(x,y) = x 2 + y 2 λ(2x + 8y 15) has a minimum value m(λ). (a) Find m(λ) m(λ) = (Use the letter L for λ in your expression.) (b) For which value of λ is m(λ) the largest, what is that maximum value? λ = maximum m(λ) = (c) Find the minimum value of f (x,y) = x 2 + y 2 subject to the constraint 2x + 8y = 15 using the method of Lagrange multipliers evaluate λ. minimum f = λ = (How are these results related to your result in part (b)?) (a) The critical points of h(x,y) occur where h x (x,y) = 2x 2λ =, h y (x,y) = 2y 8λ =. The only critical point is (x,y) = (λ,4λ) it gives a minimum value for h(x,y). That minimum value is m(λ) = h(λ,4λ) = λ λ 2 λ(2λ + 32λ 15) 4 or m(λ) = 17λ λ. (b) The maximum value of m(λ) = 17λ λ occurs at a critical point, where m (λ) = 4λ + 15 =. At this point, λ = m( ) = (c) We want to minimize f (x,y) = x 2 +y 2 subject to the constraint g(x,y) = 15, where g(x,y) = 2x + 8y. The method of Lagrange multipliers has us solve 2x = 2λ, 2y = 8λ, 2x + 8y = 15. We can see that this gives the same solutions for x y as we obtained in part (a), then, plugging in to the third equation, we have 2λ + 32λ = 15. Thus we get the same λ as before: λ = , so that the minimum value of f is f (1λ, 4λ), where λ = , or f min = Note that the two question have the same answer. This makes sense, because in the first problem the largest minimum value will clearly occur when 2x + 8y = 15, which is the constraint in the third equation. 15*L - (2*2 + 8*8)*Lˆ2/4 2*15/(2*2 + 8*8) 15*15/(2*2 + 8*8) 15*15/(2*2 + 8*8) 2*15/(2*2 + 8*8) 15. ( pts) Library/Michigan/hap16ec2/Q13.pg onsider the shaded region in the graph below. Write f da on this region as an iterated integral: f da = b d a c f (x,y)d d, where a =, b =, c =, d =. This region lies between x = x = 6 between the lines y = 2x y = 12, so the iterated integral is x f (x,y)dydx. Alternatively, we could have set up the integral as follows: 12 y 2 f (x,y)dxdy.

5 y x 6 2*x ( pts) Library/Michigan/hap16ec4/Q7.pg For each of the following, set up the integral of an arbitrary function f (x,y) over the region in whichever of rectangular or polar coordinates is most appropriate. (Use t for θ in your expressions.) (a) The region With a =, b =, c =, d =, integral = b d a c d d (b) The region With a =, b =, c =, d =, integral = b d a c d d (a) ince this is a rectangular region, we use artesian coordinates. This gives f (x,y)dydx. (b) ince this is a partially-circular region, we use polar coordinates. This gives π/ f(x,y) y x f (r cos(θ),r sin(θ))r dr dθ. 5 4 r*f(r*cos(t),r*sin(t)) r t 17. ( pts) Library/Michigan/hap16ec5/Q9.pg Evaluate the triple integral of f (x,y,z) = cos(x 2 + y 2 ) over the the solid cylinder with height 6 with base of radius 2 centered on the z axis at z =. Integral = We have 2π 2 f dv = (cos(r 2 )r dr dθdz. Integrating, W 2π 2 (cos(r 2 )r dr dθdz = 2π 1 2 sin(r2 ) 1 2 (sin(4)) 2πdz = 6π(sin(4)). 6*pi*sin(2*2) r=2 r= dθdz = 18. ( pts) Library/U/Michigan/hap16ec7/Q19.pg In this problem we use the change of variables x = 3s + t, y = s 2t to compute the integral (x + y)da, where is the parallelogram formed by (,), (6,2), (8, 2), (2, 4). First find the magnitude of the Jacobian, (x,y) (s,t) =. Then, with a =, b =, c =, d =, (x + y)da = b a Given we have (x, y) (s, t) = x s y s d c ( s+ t+ )dt ds = x = 3s +t y = s 2t, x t y t = = 7, hence (x,y) (s, t) = 7. We therefore get (x + y)da = ((3s +t) + (s 2t)) (x, y) T (s, t) dsdt = (4s t)(7)dsdt = (7(4s t))dsdt, T where T is the region in the st-plane corresponding to. Now, we need to find T. Because x = 3s +t y = s 2t T 2π

6 we can use the boundaries of the region to find the boundaries of T. Using the four vertices of the parallelogram, we the edges y = 3 1 x, y = 2x, y = 1 3 (x 2) 4 y = 2(x 6)+2. o, from the above transformation, the boundaries are t =, s =, t = 2 s = 2. Therefore 2 2 (x + y)da = 7(4s t) dsdt = 84. (Alternately, we could of course also write the integral as 2 2 7(4s t) dsdt.) 3*2 + 1 (2*6 + 2)/(1 + 3*2) (2-3*-4)/(3*2 + 1) to the line of intersection: 7*(3 + 1) i j k 7*(1-2) n 1 n 2 = (2*6 + 2)*(2-3*-4)*( *2-3*-4 + 2*(6 + 3* *-4))/(2*(1 + 3*2)ˆ2) 19. ( pts) Library/Michigan/hap16ec7/Q13.pg Find a number a so that the change of variables s = x + ay,t = y transforms the integral dxdy over the parallelogram in the xy-plane with vertices (,), (1,), ( 14,11), ( 4,11) into an integral (x, y) T (s, t) dsdt over a rectangle T in the st-plane. a = What is (x,y) (s,t) in this case? (x,y) (s,t) = Inverting the change of variables gives x = s at, y = t. The four edges of are y =,y = 11,y = 11 x,y = 11(x 1) The change of variables transforms the edges to t =,t = 11,t = s + 11 at,t = s at These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (onstant). This happens when the t terms drop out, or a = With a = 11 the change of variables gives (x, y) T (s, t) dsdt over the rectangle T : t 11, s 1. The jacobian (x,y) (s,t) is (x, y) (s, t) = = *-14/ ( pts) Library/Michigan/hap17ec1/Q47.pg (a) Find a vector parallel to the line of intersection of the planes 2x + 3y z = 2x 5y 4z = 7. v = (b) how that the point (1, 1, 1) lies on both planes. Then find a vector parametric equation for the line of intersection. r(t) = (a) Normal vectors to the two planes are n 1 = 2 i + 3 j k n 2 = 2 i 5 j 4 k. The vector n 1 n 2 is perpendicular to both planes parallel = 17ĩ + 1 j 4 k. (b) To check that the point (1, 1, 1) lies on the planes, substitute into each equation. 2x + 3y z = (2)(1) + (3)( 1) + ( 1)( 1) =, 2x 5y 4z = ( 2)(1) + ( 5)( 1) + ( 4)( 1) = 7. Thus, the point lies on both planes. Then a vector parametric equation of the line is r(t) = (1 17t)i + (1t 1) j (1 + 4t)k. -17i+1j-4k (1,-1,-1)+(-17*i+1*j-4*k)*t 21. ( pts) Library/Michigan/hap17ec5/Q27.pg Find parametric equations for the sphere centered at the origin with radius 6. Use the parameters s t in your answer. x(s,t) =, y(s, t) =, z(s, t) =, where s t. We use spherical coordinates with φ = s θ = t as the two parameters. ince the radius is 6, we can take x = 6cos(t)sin(s), y = 6sin(t)sin(s), z = 6cos(s), with 6*cos(t)*sin(s) 6*sin(t)*sin(s) 6*cos(s) 2*pi s π t 2π.

7 pi 22. ( pts) Library/Michigan/hap17ec5/Q31.pg onsider the cone shown below. If the height of the cone is 9 the base radius is 4, write a parameterization of the cone in terms of r = s θ = t. x(s,t) =, y(s, t) =, z(s, t) =, with s t. ince the parameterization is specified to be in terms of the radius r angle θ, we find x, y z in terms of the parameters r = s θ = t. We have with x = scost, y = ssint, z = 9(1 1 4 s), s 4 t 2π. There are, of course, other parameterizations, but they will not be in terms of r θ, as required in this problem. s*cos(t) s*sin(t) 9 - (9/4)*s 4 2*pi 23. ( pts) Library/Michigan/hap18ec1/Q9.pg alculate the line integral of the vector field F = 4ĩ 3 j along the line from the point (1,) to the point (7,). The line integral = ince F is a constant vector field the curve is a line, F d r = F r, where r = 6 i. Therefore, F d r = (4ĩ 3 j) 6 i = 24. (7-1)*4 24. ( pts) Library/Michigan/hap18ec2/Q13.pg Find F d r for F = 5y i (siny) j on the curve counterclockwise around the unit circle starting at the point (1,). F d r = The curve is parameterized by so, Thus, F d r = r = cost i + sint j, for t 2π, 2π 2π = -1*5*pi r (t) = sint i + cost j. (5sint i sin(sint) j) ( sint i + cost j)dt ( 2sin 2 t sin(sint)cost)dt = 5 2 (sint cost t) + cos(sint) 2π = 5π 25. ( pts) Library/Michigan/hap18ec3/Q31.pg For the vector field G = (ye xy + 3cos(3x + y)) i + (xe xy + cos(3x + y)) j, find the line integral of G along the curve from the origin along the x-axis to the point (5,) then counterclockwise around the circumference of the circle x 2 + y 2 = 25 to the point (5/ 2,5/ 2). G d r = ince G = (e xy + sin(3x + y)), the line integral can be calculated using the Fundamental Theorem of Line Integrals: F d r = e xy (5/ 2,5/ 2) +sin(3x + y) = e 25/2 +sin(6 2). c 5/sqrt(2) (.) 26. ( pts) Library/U/Michigan/hap18ec4/Q18.pg alculate ((8x + 7y) i + (4x + 4y) j) d r where is the circular path with center (a, b) radius m, oriented counterclockwise. Use Green s Theorem. ((8x + 7y) i + (4x + 4y) j) d r = Green s theorem gives (((8x+7y) i+(4x+4y) j) d r = = (4-7)*pi*mˆ2 da = Area of = πm 2. x (4x+4y) y (8x+7y)dA 7

8 27. ( pts) Library/Michigan/hap18ec4/Q15.pg Use Green s Theorem to calculate the circulation of F = 2xy i around the rectangle x 3, y 8, oriented counterclockwise. circulation = By Green s Theorem, with representing the interior of the square, ( F d r = x () ) y (2xy) da = 2x da. Thus the circulation is F d r = -2*3ˆ2*8/ xdydx = ( pts) Library/Michigan/hap19ec1/Q7.pg Find the flux of the constant vector field v = 2ĩ + 4 j 2 k through a square plate of area 9 in the zx-plane oriented in the positive y-direction. flux = On the surface, d A = j da, so only the j component of v contributes to the flux: Flux = v d A = ( 2ĩ + 4 j 2 k) j da = 4 9 = 36. 9*4 29. ( pts) Library/Michigan/hap19ec1/Q21.pg alculate the flux of the vector field F = 6 i + x 2 j 4 k, through the square of side 4 in the plane y = 5, centered on the y-axis, with sides parallel to the x z axes, oriented in the positive y-direction. flux = The only contribution to the flux is from the j-component, since d A = jdxdz on the square,, we have Flux = (6 i+x 2 j 4 k) d A = 1*4*2*(2)ˆ3/ x 2 j jdxdz = 3. ( pts) Library/Michigan/hap19ec2/Q3.pg alculate the flux integral. (3 i + 5z k) d A where is z = x 2 + y 2 with x 3, y 3, oriented upward. (3 i + 5z k) d A = ince z = x 2 + y 2, we have z x = 2x z y = 2y, so d A = ( 2x i 2y j + k)dxdy. Thus 3 3 (3 i+5z k) d A = (3 i+5(x 2 +y 2 ) k) ( 2x i 2y j+ k)dxdy, 2 2 x = 3 3 ( 6x+5x 2 +5y 2 )dxdy = 3 (x x3 + 5xy 2 ) 3 3 = y 2 dy = 18y y3 = *3ˆ2*3 + 5*3ˆ3*3/3 + 5*3*3ˆ3/3 31. ( pts) Library/Michigan/hap19ec2/Q11.pg ompute the flux of the vector field F = 5x i + 5y j through the surface, which is the part of the surface z = 36 (x 2 + y 2 ) above the disk of radius 6 centered at the origin, oriented upward. flux = Writing the surface as z = f (x,y) = 36 x 2 y 2, we have d A = ( f x i f y j + k)dxdy. = (2x i + 2y j + k)dxdy. Thus, F d A = 5(x i+y j) (2x i+2y j + k) dx dy = 1(x 2 +y 2 )dxdy. onverting to polar coordinates, we have flux = 2π 6 pi*5*1296 1r 2 r dr dθ = (2π)( )r4 = 648π. 32. ( pts) Library/U/Michigan/hap19ec3/Q9b.pg Find the surface area of the region on the plane z = 8x + 6y such that x 15 y 1 by finding a parameterization of the surface then calculating the surface area. A parameterization is x(s,t) =, y(s, t) =, z(s, t) =, with s t. Then, the surface area = A parameterization of is dz = x = 4 s, = y = 3 3. t, z = 8s + 6t, for s 15, t 1. We compute 8 so that Then urface area = r s r t = ( i + 8 k) ( j + 6 k) = 8 i 6 j + k, s t r s r t = 11. da = r s r 1 t da = t= 15 s= dy 11dsdt = 15 11

9 8*s + 6*t 15 1 sqrt(8*8 + 6*6 + 1)*15*1 33. ( pts) Library/Michigan/hap19ec3/Q1.pg ompute the flux of the vector field F = z k through the parameterized surface, which is oriented toward the z-axis given, for s 3, t 4, by x = 5s + 5t, y = 5s 5t, z = s 2 +t 2. flux = ince is given by r(s,t) = (5s + 5t) i + (5s 5t) j + (s 2 +t 2 ) k, we have r s = 5 i + 5 j + 2s k r s r t = i j k 5 5 2s 5 5 2t r t = 5 i 5 j + 2t k, = (1s+1t) i+(1s 1t) j 5 k. ince the i component of this vector is positive for < s < 3, < t < 4, it points away from the z-axis, so has the opposite orientation to the one specified. Thus, we use so we have F d A = = d A = r s r t dsdt, (s 2 +t 2 ) k (s 2 +t 2 )dsdt = = 5 (9 + 3t 2 )dt = 5(9t +t 3 2*5*5*3*4*(3*3 + 4*4)/3 ( ) (1s + 1t) i + (1s 1t) j 5 k 4 ( s st2 ) s=3 dt s= = 5( ) = ( pts) Library/U/Michigan/hap2ec1/Q17.pg A smooth vector field F has div F(3,2,1) = 3. Estimate the flux of F out of a small sphere of radius.5 centered at the point (3,2,1). flux ince div F(3,2,1) is the flux density out of a small region surrounding the point (3,2,1), we have o div F(3,2,1) Flux out of small region around (3,2,1) Volume of region. Flux out of region (div F(3,2,1)) Volume of region 9 = π(.5)3 = *4*pi*(.5)ˆ3/3 35. ( pts) Library/Michigan/hap2ec1/Q1.pg ( ) y i + x j onsider div (x 2 + y 2. ) (a) Is this a vector or a scalar?? (b) ( alculate ) it: div y i+x j = + = (x 2 +y 2 ) ( ) (a) div y i+x j is a scalar (x 2 +y 2 ) (b) We have ( ) y i + x j 2xy div (x 2 + y 2 = ) (x 2 + y 2 ) 2 + ( 2)xy (x 2 + y 2 ) 2 =. scalar 2*1*x*y/(xˆ2 + yˆ2)ˆ(1+1) -2*1*x*y/(xˆ2 + yˆ2)ˆ(1+1) 36. ( pts) Library/Michigan/hap2ec2/Q7.pg Use the Divergence Theorem to calculate the flux of the vector fields F 1 = z i +3x k F 2 = z i +(8 y) j +3x k through the surface given by the sphere of radius a centered at the origin with outwards orientation. Be sure that you are able to explain your answers geometrically. With W giving the interior of the sphere, F 1 d A = W dv = dsdt F 2 d A = W dv = The divergence of the fields are Hence, div F 1 = x (z) + y () + (3x) =. div F 2 = x (z) + y (8 3y) + (3x) =. F 2 d A = F 1 d A = dv =, W dv = dv = 4πa 3. W W This makes sense, because, the vector field F 1 is flowing around the y-axis is therefore always tangent to the sphere, so that the flux is always zero. F 2, however, adds a component in the y direction that points in for y > out (for sufficiently large negative y) for y <, therefore results in a non-zero flux.

10 -3 4*-3*pi*aˆ3/3 37. ( pts) Library/Michigan/hap2ec3/Q2.pg ompute the curl of the vector field F = zi + 7y j 4xk. curl = We have i j k curl(zi + 7y j 4xk) = j x y z 7y 4x = j. 38. ( pts) Library/Michigan/hap2ec3/Q19.pg Three small squares, 1, 2, 3, each with side.5 centered at the point (4,8,1), lie parallel to the xy-, yz- xzplanes, respectively. The squares are oriented counterclockwise when viewed from the positive z-, x-, y-axes, respectively. A vector field G has circulation around 1 of.75, around 2 of.25, around 3 of Estimate curl G at the point (4,8,1) curl G The vector curl G has its component in the x-direction given by (curl G) x irculation around small square around x-axis Area inside square = irculation around 2 =.25 Area inside 2 (.5) 2 = 1. imilar reasoning leads to (curl G) y irculation around 3 Area inside 3 = 1.25 (.5) 2 = 5. (curl G) z irculation around 1 Area inside 1 =.75 (.5) 2 =. Thus, 1i+5j-3k curl G 1ĩ + 5 j 3 k. 39. ( pts) Library/U/Michigan/hap2ec4/Q17.pg The figure below open cylindrical can,, sting on the xyplane. ( has a bottom sides, but no top.) 1 The side of is given by x 2 + y 2 = 9, its height is 3. (a) Give a parametric equation, r(t) for the rim,. r(t) = (You must use angle bracket notation enter a 3-vector), with t. (b) If is oriented outward downward, find curl ( 5y i + 5x j + 6z k) d A. curl ( 5y i + 5x j + 6z k) d A = (a) The equation of the rim,, is x 2 + y 2 = 9, z =. This is a circle of radius 3 centered on the z-axis, lying in the plane z = 2. We can parameterize this as r(t) = 3cos(t)i 3sin(t) j + 3k, t 2π. (b) Use tokes Theorem, with oriented clockwise when viewed from above: curl ( 5y i + 5x j + 6z k) d A = ( 5y i + 5x j + 6z k) d r. ince is horizontal, the k component does not contribute to the integral. The remaining vector field, 5y i + 5x j, is tangent to, of constant magnitude 5y i + 5x j = 15 on, points in the opposite direction to the orientation. Thus curl ( 5y i + 5x j + 6z k) d A = ( 5y i + 5x j) d r = 15 Length of curve = 15 2π3 = 9π. 3*cos(t)i-3*sin(t)j+3k 2*pi -2*pi*5*3ˆ2 4. ( pts) Library/Michigan/hap2ec4/Q15.pg Find F d r where is a circle of radius 3 in the plane x+y+z = 4, centered at (4,1, 1) oriented clockwise when viewed from the origin, if F = 4z j + 6y k F d r = ince i j k curl F = x y 4z 6y = 1 i, writing for the disk in the plane enclosed by the circle, tokes Theorem gives F d r = curl F d A = 1 i d A. Now d A = nda, where n is the unit vector perpendicular to the plane, so n = 1 ( i + j + k). 3 Thus F d r = + j + k 1 1 i i da = da = 1 Area of disk = 1 π3 2 = *pi*3*3/sqrt(3)

11 Generated by the WeBWorK system c WeBWorK Team, Department of Mathematics, University of ochester 11