Vector Calculus Gateway Exam

Transcription

1 Vector alculus Gateway Exam 3 Minutes; No alculators; No Notes Work Justifying Answers equired (see below) core (out of 6) Deduction Grade 5 6 No 1% trong Effort. 4 No core Please invest more time and effort to see 6 benefits. 1 pt prob For each problem without proper technique/justification. Practice Gateway. The structure of the practice exam is identical to the in-class exam. efer to instructions on how to take practice exams. urrent setting you may take the practice exam one time every 45 minutes. Num Topic 1 Line Integrals Green s Theorem 3 Green s Theorem 4 urface Integrals 5 Divergence Theorem 6 toke s Theorem Exam andomization. 1. There are 6 problems on the exam.. One question is randomly selected from each of the topic listed on the left. 3. In addition, each exam problem has randomly generated coefficients. In short, no two exams will have the same problems and/or coefficients. Instructions. This exam will be in-class with the following instructions: 1. Please be patient with server: it takes between 5 3 seconds to generate an exam. Do not click on Take Gateway Test more than once; Do not reload after clicking Take Gateway Test. These actions cause the server to generate multiple copies of the exam, which will cause an error.. lick Preview Test button to verify that your answers are interpreted appropriately. 3. lick Grade Test button before time expires to finish exam and grade it. Webwork will not count your score if you do not finish the exam before time expires. Be careful. Once you click this button, you are finished. Webwork Tips. lick HEE for full listing. You do not need to simplify. For: Enter: + 1/ˆ - 1/ˆ6 Not: Mathematical onstants: ymbol π e Usage pi e ommon Functions: Function x ln x e x Usage sqrt(x) or xˆ(1/) ln(x) eˆ(x) 3! 3! tan 1 (x) atan(x) or arctan(x) 4. You will have (at least) 3 tries; the highest score counts.

2 ontents Vector Operations Div and url Parameterization 3 Line Integral Examples 4 Green s Theorem 6 Green s Theorem F dr x F 1 y dxd y Divergence Theorem div(f)dv F n d A W toke s Theorem F dr curl(f) n d A Green s Theorem Examples 6 urface Integral 7 Divergence Theorem 8 toke s Theorem 9 Idea (Projection). The dot product is the mathematical operation that will allow us to perform a vector projection. ecall how the dot product is defined u v u v cos(θ). Now, fix the tails of u and v at the origin. For a nice diagram, assume v is located on the x-axis and u is also located in the first quadrant. We may use trigonometry if we view u as the hypotenuse of a right triangle and θ as the angle between the u and the v. Thus the component of u that lies along v is the scalar projection of u onto v, scal v u scal v u u cos(θ) u v v The vector projection of u onto v, pr o j v u, is the length (given above) paired with the direction of v: pr o j v u scal v u v v length {}}{ u v v v v }{{} direction u v v v Lastly, recall that work only utilizes the component of the force F in the direction of displacement d, so a projection needs to occur. Hence, the dot product below W F d For example, if you are pushing a lawn mower, the component of the force pushing the lawnmower down into the ground is wasted in the sense that it does not contribute to the work done. Definition (Div). The divergence of F F 1,F,F 3 is given as di v(f) F F 1 x + y + F 3 z

3 Interpretation (Div). In the context of a fluid, the divergence represents the net outflow of a fluid through the boundary. We may also use the terminology flux through the boundary. In D, the flux (net outflow) of F is the component of F normal to the boundary curve about : F nds F dxd y (D Divergence Theorem) In 3D, the flux (net outflow) of F is the component of F normal to the boundary surface of solid T : F nd A F dxd ydz (Divergence Theorem) Definition (url). The curl of F F 1,F,F 3 is given as T i j k curl(f) F F 1 F F 3 Interpretation (url). First, the curl is a vector that indicates the degree of circulation about an axis of rotation. In particular, we have 1. The direction of cur l(f) represents the axis of rotation;. The magnitude of cur l(f) is twice the angular speed of rotation. The circulation of a fluid corresponds to the degree that fluid flows tangent to the boundary. In the plane, the circulation is found by Green s theorem (which is just toke s theorem in the plane): F Tds curl(f) kdxd y x F 1 y dxd y (Green s theorem) ecall, T is the unit tangent vector to. Also, note that k is the normal vector to the plane, so this really is toke s theorem. x y z On a surface, the circulation is found by toke s theorem: F Tds curl(f) nd A Here n is a unit vector normal to the surface. (toke s theorem) Idea (Parameterization). Below you will find multiple examples on how to parameterize curves. Keep in mind that oftentimes the key to computing a line integral is obtaining a parameterization that is easy to work with. Definition (Line Integral). Let F be a continuous vector field on a region region containing a smooth curve. It T is a unit tangent vector to, the line integral over is F Tds emark. In most cases it is inconvenient to parameterize in terms of arclength s, so we reparameterize in terms of t and use it to compute the integral. Below, are the multiple different ways in which we write a line integral: F Tds b a F r (t)dt F 1 dx + F d y + F 3 dz F dr. 3

4 Interpretation (Line Integral). If F is a force, the line integral over, F(r) dr, gives the work done by the force along. Example 1. alculate the line integral F(r) dr for the given data. F y, x, : line segment from (,) to (1,4) olution. We begin by parameterizing the line segment: r(t) (1 t), + t 1,4 t 1,4, t 1 We compose F with r: (F r)(t) (4t), (t) We calculate the derivative along using the above parameterization: r (t) 1,4 The line integral is F(r) dr (4t), (t) 1,4 dt 1t dt 1 3 t (4t) (1) (t) (4)dt olution (variant). We may also use the following form: F dr F 1 dx + F d y We may parametrize the curve using y 4x with d y 4dx. Now, we calculate F(r ) dr y dx x d y [16x 4x ]dx (4x) dx x (4dx) [1x ]dx 1x3 3 Example. alculate the line integral F(r) dr for the given data. 1 4 F e x,e y,e z, : r(t) t, t, t, t olution. ompose F with r: (F r)(t) e t,e t,e t We calculate the derivative along using the above parameterization: The line integral is F(r) dr r (t) 1,t,1 e t,e t,e t 1,t,1 dt e t dt + 4 e t + e t (t) + e t dt e u du [ e t ] + [ e u] 4 (1 e ) + (1 e 4 ) 4

5 Definition (Path Independence). A line integral F dr is path independent in a domain D if for every pair of endpoints A,B D the line integral has the same value for all paths in D. Theorem (Path Independence Equivalence). uppose the line integral F dr is computed for in a simply connected domain D. Then, the following are equivalent: (a) The line integral F dr is path independent; (b) F f, for some scalar-valued function f ; (c) Integration around closed curves in D is ; (d) curl(f) in D. emark. tatement (b) of the above theorem, F f, asserts that we have an exact differential form. We have previously studied the solution of exact differential equations in DEs class with two variables x and y. In the example below we obtain the exact differential form of three variables x, y and z. emark. It is easy to check if the line integral is path independent by verifying curl(f). Example 3. how that the form is exact, and then use the exactness to evaluate the integral I (1,1,1) (,,3) yz sinh(xz)dx + cosh(xz)d y + x y sinh(xz)dz. olution. First, we verify that the form is exact. Below, we show that the curl is the zero vector, î ĵ ˆk F x y z a(x, y, z)i + b(x, y, z)j + c(x, y, z)k, y z sinh(xz) cosh(xz) x y sinh(xz) because a(x, y, z) x sinh(xz) x sinh(xz) b(x, y, z) y sinh(xz) + x yz cosh(xz) y sinh(xz) x yz cosh(xz) c(x, y, z) z sinh(xz) z sinh(xz) ince form is exact, it must be the case that F f, i.e. F 1 f x F f y F 3 f z We integrate to find the scalar-valued function f : F 1 dx yz sinh(xz)dx y cosh(xz) +G 1 (y, z) F d y cosh(xz)d y y cosh(xz) +G (x, z) F 3 dz x y sinh(xz)dz y cosh(xz) +G 3 (x, y) Where G i are constants of integration. It must be the case that G 1 (y, z) G (x, z) G 3 (x, y), so f (x, y, z) y cosh(xz). ince the differential form is exact, we may utilize the following result, B A d f f (B) f (A) to evaluate the line integral. With the following identifications: A (,,3) and B (1,1,1), we have I f (B) f (A) (1)cosh(1) ()cosh() cosh(1).457 5

6 Definition (Positive Orientation). A curve is said to be positively oriented if we traverse in a counterclockwise manner. emark. If our region has a hole within it with boundary 1, then we traverse 1 on the left, i.e. a clockwise manner. Theorem (Green s Theorem). Let be a piecewise smooth curve positively oriented that encloses simply connected region in the plane. Then, F dr x F 1 y dxd y emark. The following items concern the context of Green s theorem application. The region of integration lives in the x y-plane. Green s theorem is just toke s theorem confined to the plane. As mentioned above, Green s theorem calculates the circulation about by summing at each point along the boundary the component of F tangent to the boundary. In contrast, flux is found by summing at each point along the boundary the component of F normal to the boundary. ee D Divergence theorem above. Application (omputing Area via Green s Theorem). The area A of a region may be computed by the double integral dxd y. If we choose F, x, then F 1 dx + F d y xd y and x F 1 y 1 1, so imilarly, if we choose F y,, we find A dxd y x F 1 y dxd y F dr A dxd y x F 1 y dxd y F dr If we combine both representations we have A 1 (x d y ydx) x d y y dx Example 4. ompute the line integral F dr where F y 3, x 3 + e y, : x + y 5. Assume is positively oriented. olution. We will utilize Green s theorem, F dr x F 1 y d A. 6

7 The circle suggests a change to polar coordinates below. The integrand on the H is x F 1 y 3x ( 3y ) 3(x + y ) 3r. The region is bounded by the downward facing parabola {(r,θ) r 5, θ π} Hence, F dr π 5 π [ 3 4 r 4 3r r dr dθ ] 5 π dθ 3π π 5 3r 3 dr Example 5. Evaluate F(r) dr counterclockwise around the boundary of of the region by Green s Theorem, where F(x, y) x + y, x y : bounded region given by 1 y x. olution. We will utilize Green s Theorem. The region is bounded by the downward facing parabola The integrand in in the double integral is Hence, F dr {(x, y) 1 y x, 1 x 1} x x F 1 x y. y x y d ydx 1 x 4 x 3 + 4x + x 3 dx x 4 + 4x 3 dx + [ x x3 3x ] 1 1 [x y y ] x 1 dx x 3 + x dx + [ ] where we have broken the integral into a sum of even and odd parts, respectively, and used a a g (x)dx for odd function g. Idea (urface Integral). We will consider two cases. ase 1: calar-valued function f. Assume continuous f and smooth surface. Let by parameterized by r(u, v) x(u, v), y(u, v), z(u, v). Assume partial derivatives r u and r v are continuous; then the surface integral is given as f (x, y, z)d f (x(u, v), y(u, v), z(u, v)) r u r v dudv 7

8 The parameterization transforms the integral over to integration in the uv-plane. An infinitesimal patch on the surface has area approximated by the parallelogram formed by r u,r v. The area of the parallelogram may be computed with the cross product: d r u r v dudv Note: the single bars above denote the magnitude of the vector. ase : Vector-valued function F. (Flux Integral) Assume F is continuous and is a smooth surface. Let by parameterized by r(u, v) x(u, v), y(u, v), z(u, v). Assume partial derivatives r u and r v are continuous; then the surface integral is given as F n d F n r u r v dudv (Flux Integral) emark. The area of a surface may be computed with 1 as the integrand: A() 1d r u r v dudv emark. The unit normal vector is n(u, v) r u r v r u r v and the normal vector N(u, v) r u r v. We may rewrite the flux integral as F n d F n r u r v dudv F ru r v r u r v r u r v dudv F (r u r v )dudv F N(u, v)dudv Theorem (Divergence Theorem). Let W be a closed, simply connected region with smooth boundary. Further, let F has continuous partial derivatives, then div(f)dv F n d A. (1) W emark. The unit normal vector n is chosen so that it points away from the solid not into the solid. Example 6. Evaluate the surface integral W F n d A, where F 13y + 7z 3, x 3 + cos(z 3 ), xz, W : x + y z, x. olution. We will utilize the Divergence theorem. The divergence of F is div(f ) + + x r cos(θ). The solid given by the paraboloid W is best rewritten in cylindrical coordinates. It may be written in each of the following ways W {(r,θ, z) r z, r, π/ θ π/} 8

9 or Hence, Or, W W div(f )dv W {(r,θ, z) r z, π/ θ π/, z } W π/ π/ div(f )dv π/ r cos(θ) r drdθdz π/ cos(θ)dθ [ 3 r r 5 W π/ π/ 5/ 15 ] r cos(θ) r drdθdz cos(θ)dθ r r cos(θ) dzdrdθ r dzdr r ( r )dr r [ 5 5 ] z π/ z π/ r z 3/ drdz 3 dz r cos(θ) drdθdz Example 7. Evaluate the surface integral F n d A by the Divergence theorem, where F x y, yz, xz, : surface of cone x + y 4z, z. olution. We will utilize the Divergence theorem. The solid given by the cone T is best rewritten in cylindrical coordinates T The divergence of F is T {(r,θ, z) r z, θ π, z } div(f ) y + z + x r (cos(θ) + sin(θ)) + z. Hence, T div(f )dv (r cos(θ) + r sin(θ) + z)r drdθdz T π z π π π (cos(θ) + sin(θ))r + zr drdθdz 8 3 (cos(θ) + sin(θ))z3 + 4 z3 dθdz z (z) dθdz z 3 dz 4π π 9

10 Theorem (toke s Theorem). Let be a piecewise smooth orientable surface with smooth boundary. Then, F dr curl(f) n d A () Example 8. alculate the line integral F(r) dr by toke s Theorem for the given F and. Assume the z-component of the surface normal to be positive, where F y, x, z + x, : triangle with vertices (,,),(1,,),(1,1,) olution. We utilize tokes s theorem. The curl is computed as curl(f), 1,x y. The surface normal is perpendicular to the x y-plane: n 1,, 1,,,,1. We may characterize as {(x, y, z) y x, x 1, z }. We may now compute the integral F dr curl(f) n d A x y d A x y y x dx x3 3 x x dx 1 1 3, 1,x y,,1 d A x y d y dx x x dx Example 9. Evaluate the line integral F(r) dr by using toke s Theorem for the given F and. F x, y, y, : boundary curve of paraboloid z 9 x y, z 9 Assume is oriented counterclockwise when viewed from above. olution. Parameterize the paraboloid r(u, v) u, v,9 u v, {(u, v) u + v 9} (u, v) The surface normal is i j k N r u r v 1 u 1 v u,v,1, 1

11 This vector is oriented upward and consistent with orientation of. The curl is computed as i j k curl(f) x y z y,, v,, x y y We may now compute the integral by ultimately changing to polar coordinates F dr curl(f) N du dv v,, u,v,1 dudv 4uv dudv 4 4 π π 3 cos(θ) sin(θ)dθ 3 r cos(θ)sin(θ)r dr dθ [ sin r 3 (θ) dr 4 ] π [ r 4 4 ]