Math 32B Discussion Session Week 10 Notes March 14 and March 16, 2017
|
|
- Henry Bell
- 5 years ago
- Views:
Transcription
1 Math 3B iscussion ession Week 1 Notes March 14 and March 16, 17 We ll use this week to review for the final exam. For the most part this will be driven by your questions, and I ve included a practice final exam below as a starting point. After each problem you ll see (olution); clicking this will take you to the relevant solution. This exam is a modified version of a final from tanford s Math 5 course, and several others like it can be found at Because tanford s Math 5 syllabus isn t necessarily exactly the same as ULA s Math 3B syllabus, some of the emphasis and terminology in these exams might be different from ours; use your best discretion. Finally, you should be aware that this is just a sampling of problems that I think you should know how to do, and this should not be considered an indication of the problems you ll see on the exam. In case they re of any help, I ve also reproduced the two tables we made in recent weeks, minus some of the explanations/interpretations. 1
2 Throughout these tables, T represents a positively-oriented unit tangent vector to a curve, n represents a positively-oriented unit normal vector to a surface, and N(u, v) represents a positivelyoriented normal vector whose length depends on our parametrization of the surface. arc length/surface area integral of scalar function f integral of vector field F fluid flowing through or Line Integrals ˆ b a ˆ b a r (t) dt f(r(t)) r (t) dt ˆ ˆ F dr : (F T)ds ˆ F nds ˆ b a ˆ b a F(r(t)) r (t)dt F(r(t)) n(t) r (t) dt urface Integrals G u G v dudv f(g(u, v)) G u G v dudv F d : (F n)d F(G(u, v)) N(u, v)dudv F d F(G(u, v)) N(u, v)dudv FTs Integral across boundary Integral of derivative across solid 31A version, f : R R line integrals, a path from P to Q Green s Theorem tokes Theorem Green s, flux version ivergence Theorem f(b) f(a) f(q) f(p ) F dr F dr (F n)ds W F d ˆ b a ˆ f (x)dx f dr ( F x F ) 1 da curl z (F)dA y curl(f) d ( F1 x + F ) da div(f)da y W div(f) d
3 1. (a) ketch the region R of integration in the following double integral. x xe y5 dydx (b) Express the region R as a horizontally simple region. (c) Evaluate the integral by changing the order of integration. (olution). Given a solid region W R 3, the moment of inertia of W around an axis (line) l R 3 is given by M l d l dv, where d l : R 3 [, ) is the function which gives the distance of a point in R 3 from the axis l. W (a) Let T be a solid cone whose base is the disk x + y 1 in the xy-plane and whose vertex is the point (,, ). et up (but do not evaluate) a triple integral in cylindrical coordinates which computes the moment of inertia of T around the x-axis. (b) Let T be the solid ball of radius 1 centered at (,, ). et up (but do not evaluate) a triple integral in spherical coordinates which computes the moment of inertia of T around the z-axis. (Hint: First consider the change of variables u x, v y, w z.) (olution) 3. onsider the curve parametrized by r(t) (e t, 1 3 e3t + e t ) for t 1. (a) Find the arclength of. (b) Find the x-coordinate of the centroid of. (c) Let be the surface obtained by rotating around the line x. Find the area of. (olution) 4. (a) onsider the change of variables u x 4 + y, v y x. Find the Jacobian (x, y) (u, v). (b) Find the area of the region in the first quadrant enclosed by the ellipses x 4 + y 1 and x 4 + y 4 and the lines y x and y 4x. (olution) 5. Let T be the solid enclosed by the planes z x+y and x+y and the paraboloid z y. et up (but do not evaluate) a triple integral in rectangular coordinates which computes the volume of T by 3
4 (a) regarding T as x-simple (i.e., make x the innermost integral); (b) regarding T as z-simple. (olution) 6. Let be the curve parametrized by r(t) (t, sin t) for t π/. (a) ompute ydx. (b) Let R be the region in the plane bounded by the y-axis, the line y 1, and the curve. Use Green s theorem to find the area of R. (olution) 7. Let be the portion of the cylinder x + y x which lies above the xy-plane and below the surface z x. (a) Write down a parametrization of. Be sure to specify the domain. (b) Find the area of. (olution) 8. Let be the surface traced out by G(u, v) ( [, 1] in the uv-plane. v, u, uv) over the domain [, 1] (a) Find the area of. (Warning: The map G is not one-to-one on. 1 ) (b) Find the x-coordinate of the centroid of. (olution) 9. Let be the portion of the sphere x + y + z 4 which lies above the plane z, and let F(x, y, z) (x 3y + z, 3x + y + z, x + y 3z). ompute the flux of F through in the outward-pointing direction. (olution) 1. Let T be the triangle with vertices A (1, 1, 1), B (3,, ), and (1,, ), parametrized in that order. Let F(x, y, z) (4z, x, y ). (a) Find the area of the triangle. (b) Find the equation of the plane in which the triangle lies. (c) ompute curl(f). (d) ompute F dr. (olution) T 11. Let F(x, y, z) (xy + z, yz + x, zx + y ). 1 I m not entirely convinced that this extra layer of difficulty was intentional on the part of the test-writer; I think the domain may be a typo. If you can t figure out how to deal with the lack of one-to-one-ness, maybe skip down to the solution, find the domain we really want, and then see if you can do the rest of the problem on your own. 4
5 (a) Find a potential for F. (b) Let be the curve parametrized by r(t) (cos t, sin t, t) for t π. F dr. (olution) ompute 1. Let T be the solid region which lies below the surfaces z 1 x and z 1 y and above the xy-plane. (a) Find the volume of T. (b) Find the outward flux through the boundary of T of the vector field F(x, y, z) (zx, z y, 3z z x). (olution) 5
6 olutions 1. (a) The region of interest here is the region where x 1 and x y 1, which we can easily plot: (b) Expressing R as a horizontally simple region means making slices which are parallel to the x-axis, such as the red slice in the above plot. The possible y-values at which we can make such a slice range from to 1, and once y is fixed, x is allowed to vary between and y. o R {(x, y) y 1, x y } gives a horizontally simple description of R. (c) According to Fubini s theorem we have x xe y5 dydx ˆ y xe y5 dxdy [ x ey5 ] y dy 1 y 4 e y5 dy. We may then let u y 5, so that du 5y 4 dy. Notice that our bounds of integration do not change. Then we have x xe y5 dydx 1 1. (a) Here s a plot of the cone we re interested in: e u du e 1 1. In cylindrical coordinates our ranges for r and θ are obvious: we let r vary between and 1, while θ varies between and π. The bounds for z require only a little more work. We know that we should have z, and also that the upper bound for z should depend only on r, and not on θ. In fact, we can write the relationship between z and r as Az + Br 1 for some choices of A and B, since the upper bound for z depends on r in a linear fashion. We know that when r our upper bound for z is, so A 1/. 6
7 When z, our upper bound for r is 1, so B 1. o we have z r. Altogether, the cone is described by θ π, r 1, z r. For the x-axis we re interested in integrating d x y + z, the square of the distance of a point from the x-axis, so we have ˆ π ˆ r I x (y + z )dv (r sin θ + z )rdzdrdθ. T In case you prefer to first describe the cone in rectangular coordinates and then make substitutions, we can do that too. In rectangular coordinates we can desribe T with the inequalities x + y 1 and z x + y. The second inequality may not be so obvious, but consider this. The standard (upwardopening) cone is described by z x + y. We want to turn this upside down and make it sharper, so we have z λ x + y for some λ >. We also need to shift the cone up so that the vertex is at (,, ), so we in fact have z λ x + y. ince we want z whenever x + y 1, we have λ. These two inequalities of course translate to r 1 and z r, and then θ is allowed to fill its usual range from to π. (b) In rectangular coordinates T is described by (x ) + y + z 1. We want to use the transformation indicated by the hint, but we remember that our transformation should map into the region T over which we want to integrate, so we write x u +, y v, z w, and then define G: R 3 R 3 by Now we have I z T d zdv G(u, v, w) (u +, v, w). G (T ) We can quickly check that Jac(G) 1. Also, (d z G(u, v, w)) Jac(G) dudvdw. G (T ) {(u, v, w) u + v + w 1} so we need to integrate d z G (u + ) + v over the unit ball. We now introduce spherical coordinates: u ρ sin φ cos θ, v ρ sin φ sin θ, w ρ cos φ, which then gives dudvdw ρ sin φdρdφdθ. o the moment of inertia is given by I z ˆ π ˆ π ((ρ sin φ cos θ + ) + ρ sin φ sin θ)ρ sin φdρdφdθ. 7
8 3. (a) From the given parametrization we compute r (t) e t, e 3t e t, so r (t) 4e t + e 6t e t + e t e 6t + e t + e t (e 3t + e t ) e 3t + e t. Then the arclength of is given by ˆ 1ds r (t) dt e 3t + e t dt [ 1 3 e3t e t ] 1 ( ) e 3 3 e ( ) (e3 3e + ). (b) We never really talked about centroids of curves, but this has the formula you would expect: x M ˆ y M, where M y xds. Here M is the mass of the curve, which is the same as the arclength if we assume constant density 1. Now let s compute M y : ˆ [ ] 1 1 M y xds (e t ) r (t) dt e 4t + 1dt 4 e4t + t ( ) ( ) e (e4 + 3). Then x M 1 y M (e4 + 3) 1 3 (e3 3e + ). (c) For this part it will probably help to see the curve. On the left we have a plot of (in blue), along with the line x (in orange). On the right we see the surface which results from rotating around the line x. If we fix a value t 1 to get a point on, the rotation around the line x will turn this point into a circle on, such as the green one seen in the figure. We can compute the circumference of this circle if we know its radius, and this value will be the length of the purple radial line in the first plot. Once we know the circumference of the circle generated by a point on, we can integrate across to get the surface area of. In pseudo-math, ˆ ˆ surface area circumference ds π radius ds π R(t) r (t) dt. 8
9 The last integral is the thing we want to compute; we just need to write down R(t), the radius at time t. This is the length of the purple line above, and should be equal to the x-value of r(t), plus the one unit to get from the y-axis to the line x. That is, o R(t) e t + 1. Area π (e t + 1)(e 3t + e t )dt π e t (e 3t + e t )dt + π [ e π + e3 3e ] + π 3 3 (3e4 + e e ). The first integral above was computed in part (b), the second in part (a). 4. (a) We have so (x, y) (u, v) (u, v) (x, y) ( ) (u, v) (x, y) ( ) x/ y det y/x 1 1/x + y x, y x (b) The region R in which we re interested is described by 1 + 4y /x 1 + 4v. R {(x, y) 1 x 4 + y 4, y/x 4}. This is the preimage under our transformation of the region so Area R ˆ 4 {(u, v) 1 u 4, v 4}, 1dA Here s a plot of the region R: (x, y) (u, v) da ˆ 4 ˆ v dudv v [3 arctan(v)]4 3 arctan(8) 3 arctan(4). e 3t + e t dt 9
10 5. (a) ince the innermost integral is in terms of x, we want to bound x between expressions that depend on y and z. We can rewrite the bound z x + y as z y x, and rewrite the bound x + y as x y. o we have z y x y. For this set of bounds to make sense, we must have z y y, so z y. ombining this with the requirement that y z, we have y z y. Finally, we need y y, so y + y. That is, (y + )(y 1), so y 1. The volume of T is then given by vol(t ) ˆ y ˆ y z y y 1dydzdx. (b) We now want to rewrite our bounds so that z lives between expressions depending on x and y. The bounds z x + y and y z immediately give y z x + y. This set of bounds then requires y x+y, so y y x. We also know that x+y, so x y and we see that y y x y. Finally, y y y tells us that y + y, so again we have y 1. o we can also compute the volume of T as vol(t ) 6. (a) Notice that r (t) t, cos t, so ˆ ˆ ydx y, dr ˆ x+y ˆ y y ˆ π/ y y 1dydxdz. sin t, r (t)dt ˆ π/ t sin tdt. We can use integration by parts, letting u t and letting dv sin tdt. Then ˆ ˆ π ydx [ t cos t] π/ ( cos u)du [ sin u] π/. (b) Here s a plot of the region R: 1
11 We re asked to compute the area of R using Green s theorem, which we recall can be done using a number of different integrals: Area xdy ydx 1 xdy ydx. R R R ince we ve already computed ydx in part (a), we ll use the middle option. If we call the top part of R (where y 1) 1 and call the left part (the y-axis), we have (ˆ ˆ ˆ ) Area ydx ydx + ydx + ydx R 1 ( ) + ˆ ( π 4 π /4 1dx + ) π 4. We have ydx because x is constant along. 7. (a) Our cylinder might be more readily parametrized if we rewrite its defining equation to look more familiar. The equation x + y x can be rewritten x x + y x x y 1 (x 1) + y 1. o this is the standard unit-radius cylinder, translated 1 unit in the x-direction. Inspired by cylindrical coordinates, we give the parametrization G(u, v) (cos u + 1, sin u, v). Notice that u and v play the roles usually played by θ and z, respectively, and that the x-coordinate has been bumped up by 1 unit. We also must be clear about the domain of G. ince is bounded below by the xy-plane and above by z x, we have For u and v this means z x. v (cos u + 1). Notice that the requirement that (cos u + 1) imposes no condition on u, so u π. (b) Using the above parametrization, we have so o G u G v 1, and thus surface area G u sin u, cos u, and G v,, 1, i j k G u G v sin u cos u cos u, sin u,. 1 ˆ π ˆ (cos u+1) G u G v dudv 11 ˆ π ˆ (cos u+1) 1dudv 3π.
12 8. (a) We noticed that G is not one-to-one on since, for example, ( 1 G, 1 ) ( G 1 ),. If we restrict our attention to the subdomain [, 1] [, 1], G will be one-to-one on the interior of, but will also trace out the same surface as before. We can verify this by explicitly constructing an inverse map: G (x, y, z) ( z x, x ) This inverse map only makes sense on the portion of where x >, but this corresponds precisely with the interior of, where v >. Whether or not we were really supposed to care about these domain scruples is unclear to me, but in any case the domain we care about is [, 1] [, 1]. Once we ve restricted to this domain, we have Area() 1d G u G v dudv. We compute so G u, u, v and G v v,, u, i j k G u G v u v v u u, v, uv.. Then G u G v u 4 + v 4 + 4u v ( u + v ) (u + v ). o the surface area is Area() (u + v )dudv [ u 3 (v + /3)dv [ 3 v3 + 3 v (b) For the x-coordinate of the centroid, we need to compute M yz xd, 3 + uv ] 1 ] 1 dv 4 3. and then we ll have x M yz /M. Here M is the mass of the surface, which we can take to be the area if we assume uniform density 1. Under the parametrization, x so we have ˆ 1 ( ) M yz v G u G v dudv v (u + v ) dudv 1 v,
13 Finally, v (u + v )dudv v (/3 + v )dv [ 9 v3 + 5 v5 [ u v uv ] 1 x 8/45 4 / ] 1 dv Here s a plot of the surface, with the centroid as a red dot. The strange point of view here was chosen so that we could see that the centroid is a point somewhere in the convex hull of our surface, but not actually on. 9. Recall that this flux is computed as a vector surface integral: F d. At this point we could parametrize using our knowledge of spherical coordinates, then compute a normal vector for this parametrization and end up with a big, ugly integral. Instead, let s use the divergence theorem. Our surface is a portion of a sphere, and it intersects the plane z in a circle. We denote by P the solid disk in the plane z which is bounded by this intersection. Then and P together bound a three-dimensional region W, as seen here: 13
14 We recall from the divergence theorem that div(f)dv F d W W F d + F d. P ince div(f) 1 + 3, the integral on the left vanishes and we have F d F d. o instead of integrating F over, we only have to integrate F over P. Now the intersection of and P is the circle where x + y + z 4 and z, so x + y 3. That is, P is a disk of radius 3 centered at (,, ) and contained in the plane z. o we can parametrize P by G(u, v) (u cos v, u sin v, ), where u 3 and v π. The tangent vectors for this parametrization are given by G u cos v, sin v, and G v u sin v, u cos v,, so i j k G u G v cos v sin v,, u. u sin v u cos v Now we have to be a little bit careful. We want our normal vector N(u, v) to be pointing out of the region W, which means that the z-coordinate should be negative along P. ince u 3, we take N(u, v) G u G v. Finally we notice that so F(G(u, v)) u cos v 3u sin v, 3u cos v + u sin v 1, u cos v + u sin v + 3, F(G(u, v) N(u, v) (u cos v + u sin v + 3)( u) (u cos v + u sin v + 3u). We re now ready to integrate: P ˆ π F d ˆ π ˆ 3 P ˆ π [ u cos v + u sin v + 3udu 3 u3 cos v u3 sin v + 3 ] 3 u dv 3 cos v + 3 sin v + 9 dv [ 3 sin v 3 cos v + 9 v ] π ( 3 + 9π ) ( + ) 3 + 9π. o the total outward flux of F through is F d P F d 9π. We could also have done this problem using tokes theorem instead of the divergence theorem. The trick here is to find a vector field G so that curl(g) F. This would give us F d curl(g) d G dr. 14
15 One such G is given by G(x, y, z) 3yz 3xz xy, 3yz + xy, xy + yz xz. (You can verify that curl(g) F.) Then we can parametrize by r(t) ( 3 cos t, 3 sin t, ), t pi and compute F d ˆ π G(r(t)) r (t)dt 9π. I would not recommend this second approach, though, since finding the vector field G can be a messy process. 1. (a) The triangle is spanned by the vectors u B A,, and v A,, 1. We can compute the area of T as 1 u v. We have i j k u v 4,, 4, 1 so Area(T ) (b) The vector u v above gives us a normal vector to the plane P containing T, so we can write 4x y 4z λ for every point (x, y, z) in this plane, for some constant λ. Equivalently, we can write 4x + y + 4z λ. ince we know that, say, (1, 1, 1) is a point in this plane, we have 4(1) + (1) + 4(1) 1 λ, so P is described by 4x + y + 4z 1, or x + y + z 5. (c) We have i j k curl(f) F x y z 4z x y y, 8z, 4x y, 4z, x. 15
16 (d) Let denote the solid triangle contained in P whose boundary is T. theorem tells us that F dr curl(f) d. T Then tokes At first it seems that we ll need to parametrize in order to compute the integral on the right, but remember that vector surface integrals are defined by G d (G n)d, where n is a unit normal vector to. We can easily obtain n: n 4,, 4 4,, 4 1 4,, 4. 6 o we have T F dr curl(f) d 1 y, 8z, 4x 4,, 4 d z + x))d 6 (y 8 5d Area() (a) We re looking for a function f so that f x xy + z, f y yz + x, and f z zx + y. By integrating the first of these conditions we see that f has the form f(x, y, z) x y + xz + g(y, z) (1) for some function g of y and z. We can differentiate this equation with respect to y to see that f y x + g y, so g y yz. Integrating this with respect to y, we have g(y, z) y z + h(z) for some function h of z. Finally, we differentiate (1) with respect to z to find that xz + g z xz + y + h (z). ince we know that f z zx + y, we see that h (z), so h is a constant function. o g(y, z) y z and we see that f(x, y, z) x y + xz + y z is a function with the property that f F. 16
17 (b) ince F has a potential function, this vector field is conservative, so we may apply the fundamental theorem of conservative vector fields. In particular, ˆ F dr f(r(π)) f(r()) f(,, π) f(1,, ) π π. 1. The points (x, y, z) in T must simultaneously satisfy the inequalities z 1 x and z 1 y, so we can write our region as z min(1 x, 1 y ). When x y we have 1 x 1 y, and when x y we have 1 y 1 x. o T is the region between the xy-plane and the plot of the function { 1 x f(x, y), x y 1 y, x y over the region [, 1] [, 1]. We choose this region because this is precisely the region where f(x, y). (a) With the work done above, we see that the volume of T is given by vol(t ) f(x, y)da (1 x )da + 1 (1 y )da, where 1 {(x, y) x y } and {(x, y) x y }. We an describe 1 and in terms of inequalities: { 1 (x, y) x 1 } { and x y x (x, y) y 1 }. y x y o we have vol(t ) ˆ x x (1 x )dydx + x (1 x )dx + ˆ y y (1 y )dxdy y (1 y )dy 4 u (1 u )du. The last equality simply relies on the fact that variable names don t matter. We integrate absolute value piecewise: u (1 u )du ˆ u(1 u )du + [ 1 u u4 ] + ( 1 1 ) ( Altogether, vol(t ) 4 u (1 u )du. Here s a plot of the region T : 17 u(1 u )du [ 1 u 1 4 u4 ) 1. ] 1
18 (b) We can use the divergence theorem to write F d T T div(f)dv. ince div(f) zx xz, the integral on the right will be significantly easier to compute. Indeed, F d dv vol(t ) 4. T o the total flux of F through T is 4. T 18
MATH 52 FINAL EXAM SOLUTIONS
MAH 5 FINAL EXAM OLUION. (a) ketch the region R of integration in the following double integral. x xe y5 dy dx R = {(x, y) x, x y }. (b) Express the region R as an x-simple region. R = {(x, y) y, x y }
More informationMath Review for Exam 3
1. ompute oln: (8x + 36xy)ds = Math 235 - Review for Exam 3 (8x + 36xy)ds, where c(t) = (t, t 2, t 3 ) on the interval t 1. 1 (8t + 36t 3 ) 1 + 4t 2 + 9t 4 dt = 2 3 (1 + 4t2 + 9t 4 ) 3 2 1 = 2 3 ((14)
More informationSurface Area of Parametrized Surfaces
Math 3B Discussion ession Week 7 Notes May 1 and 1, 16 In 3A we learned how to parametrize a curve and compute its arc length. More recently we discussed line integrals, or integration along these curves,
More informationMath 11 Fall 2016 Final Practice Problem Solutions
Math 11 Fall 216 Final Practice Problem olutions Here are some problems on the material we covered since the second midterm. This collection of problems is not intended to mimic the final in length, content,
More informationMath 11 Fall 2007 Practice Problem Solutions
Math 11 Fall 27 Practice Problem olutions Here are some problems on the material we covered since the second midterm. This collection of problems is not intended to mimic the final in length, content,
More informationSOME PROBLEMS YOU SHOULD BE ABLE TO DO
OME PROBLEM YOU HOULD BE ABLE TO DO I ve attempted to make a list of the main calculations you should be ready for on the exam, and included a handful of the more important formulas. There are no examples
More informationSolutions for the Practice Final - Math 23B, 2016
olutions for the Practice Final - Math B, 6 a. True. The area of a surface is given by the expression d, and since we have a parametrization φ x, y x, y, f x, y with φ, this expands as d T x T y da xy
More informationMATHS 267 Answers to Stokes Practice Dr. Jones
MATH 267 Answers to tokes Practice Dr. Jones 1. Calculate the flux F d where is the hemisphere x2 + y 2 + z 2 1, z > and F (xz + e y2, yz, z 2 + 1). Note: the surface is open (doesn t include any of the
More information1 + f 2 x + f 2 y dy dx, where f(x, y) = 2 + 3x + 4y, is
1. The value of the double integral (a) 15 26 (b) 15 8 (c) 75 (d) 105 26 5 4 0 1 1 + f 2 x + f 2 y dy dx, where f(x, y) = 2 + 3x + 4y, is 2. What is the value of the double integral interchange the order
More informationis a surface above the xy-plane over R.
Chapter 13 Multiple Integration Section 13.1Double Integrals over ectangular egions ecall the Definite Integral from Chapter 5 b a n * lim i f x dx f x x n i 1 b If f x 0 then f xdx is the area under the
More informationMath 234 Exam 3 Review Sheet
Math 234 Exam 3 Review Sheet Jim Brunner LIST OF TOPIS TO KNOW Vector Fields lairaut s Theorem & onservative Vector Fields url Divergence Area & Volume Integrals Using oordinate Transforms hanging the
More informationPractice Problems for Exam 3 (Solutions) 1. Let F(x, y) = xyi+(y 3x)j, and let C be the curve r(t) = ti+(3t t 2 )j for 0 t 2. Compute F dr.
1. Let F(x, y) xyi+(y 3x)j, and let be the curve r(t) ti+(3t t 2 )j for t 2. ompute F dr. Solution. F dr b a 2 2 F(r(t)) r (t) dt t(3t t 2 ), 3t t 2 3t 1, 3 2t dt t 3 dt 1 2 4 t4 4. 2. Evaluate the line
More informationLine and Surface Integrals. Stokes and Divergence Theorems
Math Methods 1 Lia Vas Line and urface Integrals. tokes and Divergence Theorems Review of urves. Intuitively, we think of a curve as a path traced by a moving particle in space. Thus, a curve is a function
More informationPractice problems **********************************************************
Practice problems I will not test spherical and cylindrical coordinates explicitly but these two coordinates can be used in the problems when you evaluate triple integrals. 1. Set up the integral without
More informationFinal exam (practice 1) UCLA: Math 32B, Spring 2018
Instructor: Noah White Date: Final exam (practice 1) UCLA: Math 32B, Spring 218 This exam has 7 questions, for a total of 8 points. Please print your working and answers neatly. Write your solutions in
More informationLINE AND SURFACE INTEGRALS: A SUMMARY OF CALCULUS 3 UNIT 4
LINE AN URFAE INTEGRAL: A UMMARY OF ALULU 3 UNIT 4 The final unit of material in multivariable calculus introduces many unfamiliar and non-intuitive concepts in a short amount of time. This document attempts
More informationCalculus III. Math 233 Spring Final exam May 3rd. Suggested solutions
alculus III Math 33 pring 7 Final exam May 3rd. uggested solutions This exam contains twenty problems numbered 1 through. All problems are multiple choice problems, and each counts 5% of your total score.
More informationDO NOT BEGIN THIS TEST UNTIL INSTRUCTED TO START
Math 265 Student name: KEY Final Exam Fall 23 Instructor & Section: This test is closed book and closed notes. A (graphing) calculator is allowed for this test but cannot also be a communication device
More informationHOMEWORK 8 SOLUTIONS
HOMEWOK 8 OLUTION. Let and φ = xdy dz + ydz dx + zdx dy. let be the disk at height given by: : x + y, z =, let X be the region in 3 bounded by the cone and the disk. We orient X via dx dy dz, then by definition
More informationJim Lambers MAT 280 Summer Semester Practice Final Exam Solution. dy + xz dz = x(t)y(t) dt. t 3 (4t 3 ) + e t2 (2t) + t 7 (3t 2 ) dt
Jim Lambers MAT 28 ummer emester 212-1 Practice Final Exam olution 1. Evaluate the line integral xy dx + e y dy + xz dz, where is given by r(t) t 4, t 2, t, t 1. olution From r (t) 4t, 2t, t 2, we obtain
More informationMath 233. Practice Problems Chapter 15. i j k
Math 233. Practice Problems hapter 15 1. ompute the curl and divergence of the vector field F given by F (4 cos(x 2 ) 2y)i + (4 sin(y 2 ) + 6x)j + (6x 2 y 6x + 4e 3z )k olution: The curl of F is computed
More information(a) 0 (b) 1/4 (c) 1/3 (d) 1/2 (e) 2/3 (f) 3/4 (g) 1 (h) 4/3
Math 114 Practice Problems for Test 3 omments: 0. urface integrals, tokes Theorem and Gauss Theorem used to be in the Math40 syllabus until last year, so we will look at some of the questions from those
More informationSections minutes. 5 to 10 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed.
MTH 34 Review for Exam 4 ections 16.1-16.8. 5 minutes. 5 to 1 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed. Review for Exam 4 (16.1) Line
More informationSolutions to the Final Exam, Math 53, Summer 2012
olutions to the Final Exam, Math 5, ummer. (a) ( points) Let be the boundary of the region enclosedby the parabola y = x and the line y = with counterclockwise orientation. alculate (y + e x )dx + xdy.
More informationLINE AND SURFACE INTEGRALS: A SUMMARY OF CALCULUS 3 UNIT 4
LINE AN URFAE INTEGRAL: A UMMARY OF ALULU 3 UNIT 4 The final unit of material in multivariable calculus introduces many unfamiliar and non-intuitive concepts in a short amount of time. This document attempts
More informationMath 53 Spring 2018 Practice Midterm 2
Math 53 Spring 218 Practice Midterm 2 Nikhil Srivastava 8 minutes, closed book, closed notes 1. alculate 1 y 2 (x 2 + y 2 ) 218 dxdy Solution. Since the type 2 region D = { y 1, x 1 y 2 } is a quarter
More informationPractice problems. 1. Evaluate the double or iterated integrals: First: change the order of integration; Second: polar.
Practice problems 1. Evaluate the double or iterated integrals: R x 3 + 1dA where R = {(x, y) : 0 y 1, y x 1}. 1/ 1 y 0 3y sin(x + y )dxdy First: change the order of integration; Second: polar.. Consider
More information(b) Find the range of h(x, y) (5) Use the definition of continuity to explain whether or not the function f(x, y) is continuous at (0, 0)
eview Exam Math 43 Name Id ead each question carefully. Avoid simple mistakes. Put a box around the final answer to a question (use the back of the page if necessary). For full credit you must show your
More informationPractice problems. m zδdv. In our case, we can cancel δ and have z =
Practice problems 1. Consider a right circular cone of uniform density. The height is H. Let s say the distance of the centroid to the base is d. What is the value d/h? We can create a coordinate system
More information18.1. Math 1920 November 29, ) Solution: In this function P = x 2 y and Q = 0, therefore Q. Converting to polar coordinates, this gives I =
Homework 1 elected olutions Math 19 November 9, 18 18.1 5) olution: In this function P = x y and Q =, therefore Q x P = x. We obtain the following integral: ( Q I = x ydx = x P ) da = x da. onverting to
More informationArchive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma
Archive of Calculus IV Questions Noel Brady Department of Mathematics University of Oklahoma This is an archive of past Calculus IV exam questions. You should first attempt the questions without looking
More informationName: Date: 12/06/2018. M20550 Calculus III Tutorial Worksheet 11
1. ompute the surface integral M255 alculus III Tutorial Worksheet 11 x + y + z) d, where is a surface given by ru, v) u + v, u v, 1 + 2u + v and u 2, v 1. olution: First, we know x + y + z) d [ ] u +
More informationVector Calculus Gateway Exam
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
More informationMath 20C Homework 2 Partial Solutions
Math 2C Homework 2 Partial Solutions Problem 1 (12.4.14). Calculate (j k) (j + k). Solution. The basic properties of the cross product are found in Theorem 2 of Section 12.4. From these properties, we
More informationMAC2313 Final A. (5 pts) 1. How many of the following are necessarily true? i. The vector field F = 2x + 3y, 3x 5y is conservative.
MAC2313 Final A (5 pts) 1. How many of the following are necessarily true? i. The vector field F = 2x + 3y, 3x 5y is conservative. ii. The vector field F = 5(x 2 + y 2 ) 3/2 x, y is radial. iii. All constant
More informationFinal Exam Review Sheet : Comments and Selected Solutions
MATH 55 Applied Honors alculus III Winter Final xam Review heet : omments and elected olutions Note: The final exam will cover % among topics in chain rule, linear approximation, maximum and minimum values,
More informationMath 31CH - Spring Final Exam
Math 3H - Spring 24 - Final Exam Problem. The parabolic cylinder y = x 2 (aligned along the z-axis) is cut by the planes y =, z = and z = y. Find the volume of the solid thus obtained. Solution:We calculate
More informationMcGill University April 16, Advanced Calculus for Engineers
McGill University April 16, 2014 Faculty of cience Final examination Advanced Calculus for Engineers Math 264 April 16, 2014 Time: 6PM-9PM Examiner: Prof. R. Choksi Associate Examiner: Prof. A. Hundemer
More informationPractice problems. 1. Evaluate the double or iterated integrals: First: change the order of integration; Second: polar.
Practice problems 1. Evaluate the double or iterated integrals: x 3 + 1dA where = {(x, y) : 0 y 1, y x 1}. 1/ 1 y 0 3y sin(x + y )dxdy First: change the order of integration; Second: polar.. Consider the
More informationProblem 1. Use a line integral to find the plane area enclosed by the curve C: r = a cos 3 t i + b sin 3 t j (0 t 2π). Solution: We assume a > b > 0.
MATH 64: FINAL EXAM olutions Problem 1. Use a line integral to find the plane area enclosed by the curve C: r = a cos 3 t i + b sin 3 t j ( t π). olution: We assume a > b >. A = 1 π (xy yx )dt = 3ab π
More informationOne side of each sheet is blank and may be used as scratch paper.
Math 244 Spring 2017 (Practice) Final 5/11/2017 Time Limit: 2 hours Name: No calculators or notes are allowed. One side of each sheet is blank and may be used as scratch paper. heck your answers whenever
More informationIn general, the formula is S f ds = D f(φ(u, v)) Φ u Φ v da. To compute surface area, we choose f = 1. We compute
alculus III Test 3 ample Problem Answers/olutions 1. Express the area of the surface Φ(u, v) u cosv, u sinv, 2v, with domain u 1, v 2π, as a double integral in u and v. o not evaluate the integral. In
More information********************************************************** 1. Evaluate the double or iterated integrals:
Practice problems 1. (a). Let f = 3x 2 + 4y 2 + z 2 and g = 2x + 3y + z = 1. Use Lagrange multiplier to find the extrema of f on g = 1. Is this a max or a min? No max, but there is min. Hence, among the
More informationM273Q Multivariable Calculus Spring 2017 Review Problems for Exam 3
M7Q Multivariable alculus Spring 7 Review Problems for Exam Exam covers material from Sections 5.-5.4 and 6.-6. and 7.. As you prepare, note well that the Fall 6 Exam posted online did not cover exactly
More informationName: SOLUTIONS Date: 11/9/2017. M20550 Calculus III Tutorial Worksheet 8
Name: SOLUTIONS Date: /9/7 M55 alculus III Tutorial Worksheet 8. ompute R da where R is the region bounded by x + xy + y 8 using the change of variables given by x u + v and y v. Solution: We know R is
More informationSOLUTIONS TO THE FINAL EXAM. December 14, 2010, 9:00am-12:00 (3 hours)
SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN December 14, 2010, 9:00am-12:00 (3 hours) 1) For each of (a)-(e) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please
More informationMath Vector Calculus II
Math 255 - Vector Calculus II Review Notes Vectors We assume the reader is familiar with all the basic concepts regarding vectors and vector arithmetic, such as addition/subtraction of vectors in R n,
More informationPRACTICE PROBLEMS. Please let me know if you find any mistakes in the text so that i can fix them. 1. Mixed partial derivatives.
PRACTICE PROBLEMS Please let me know if you find any mistakes in the text so that i can fix them. 1.1. Let Show that f is C 1 and yet How is that possible? 1. Mixed partial derivatives f(x, y) = {xy x
More information( ) ( ) ( ) ( ) Calculus III - Problem Drill 24: Stokes and Divergence Theorem
alculus III - Problem Drill 4: tokes and Divergence Theorem Question No. 1 of 1 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed () Pick the 1. Use
More informationGreen s, Divergence, Stokes: Statements and First Applications
Math 425 Notes 12: Green s, Divergence, tokes: tatements and First Applications The Theorems Theorem 1 (Divergence (planar version)). Let F be a vector field in the plane. Let be a nice region of the plane
More informationMath 23b Practice Final Summer 2011
Math 2b Practice Final Summer 211 1. (1 points) Sketch or describe the region of integration for 1 x y and interchange the order to dy dx dz. f(x, y, z) dz dy dx Solution. 1 1 x z z f(x, y, z) dy dx dz
More informationMATH 332: Vector Analysis Summer 2005 Homework
MATH 332, (Vector Analysis), Summer 2005: Homework 1 Instructor: Ivan Avramidi MATH 332: Vector Analysis Summer 2005 Homework Set 1. (Scalar Product, Equation of a Plane, Vector Product) Sections: 1.9,
More informationMATH 52 FINAL EXAM DECEMBER 7, 2009
MATH 52 FINAL EXAM DECEMBER 7, 2009 THIS IS A CLOSED BOOK, CLOSED NOTES EXAM. NO CALCULATORS OR OTHER ELECTRONIC DEVICES ARE PERMITTED. IF YOU NEED EXTRA SPACE, PLEASE USE THE BACK OF THE PREVIOUS PROB-
More informationG G. G. x = u cos v, y = f(u), z = u sin v. H. x = u + v, y = v, z = u v. 1 + g 2 x + g 2 y du dv
1. Matching. Fill in the appropriate letter. 1. ds for a surface z = g(x, y) A. r u r v du dv 2. ds for a surface r(u, v) B. r u r v du dv 3. ds for any surface C. G x G z, G y G z, 1 4. Unit normal N
More informationMath Exam IV - Fall 2011
Math 233 - Exam IV - Fall 2011 December 15, 2011 - Renato Feres NAME: STUDENT ID NUMBER: General instructions: This exam has 16 questions, each worth the same amount. Check that no pages are missing and
More information7a3 2. (c) πa 3 (d) πa 3 (e) πa3
1.(6pts) Find the integral x, y, z d S where H is the part of the upper hemisphere of H x 2 + y 2 + z 2 = a 2 above the plane z = a and the normal points up. ( 2 π ) Useful Facts: cos = 1 and ds = ±a sin
More informationVector Calculus. Dr. D. Sukumar
Vector Calculus Dr. D. Sukumar Space co-ordinates Change of variable Cartesian co-ordinates < x < Cartesian co-ordinates < x < < y < Cartesian co-ordinates < x < < y < < z < Cylindrical Cylindrical Cylindrical
More informationName: Instructor: Lecture time: TA: Section time:
Math 222 Final May 11, 29 Name: Instructor: Lecture time: TA: Section time: INSTRUCTIONS READ THIS NOW This test has 1 problems on 16 pages worth a total of 2 points. Look over your test package right
More informationLet s estimate the volume under this surface over the rectangle R = [0, 4] [0, 2] in the xy-plane.
Math 54 - Vector Calculus Notes 3. - 3. Double Integrals Consider f(x, y) = 8 x y. Let s estimate the volume under this surface over the rectangle R = [, 4] [, ] in the xy-plane. Here is a particular estimate:
More informationNo calculators, cell phones or any other electronic devices can be used on this exam. Clear your desk of everything excepts pens, pencils and erasers.
Name: Section: Recitation Instructor: READ THE FOLLOWING INSTRUCTIONS. Do not open your exam until told to do so. No calculators, cell phones or any other electronic devices can be used on this exam. Clear
More informationFinal exam (practice 1) UCLA: Math 32B, Spring 2018
Instructor: Noah White Date: Final exam (practice 1) UCLA: Math 32B, Spring 2018 This exam has 7 questions, for a total of 80 points. Please print your working and answers neatly. Write your solutions
More informationMath 212-Lecture 20. P dx + Qdy = (Q x P y )da. C
15. Green s theorem Math 212-Lecture 2 A simple closed curve in plane is one curve, r(t) : t [a, b] such that r(a) = r(b), and there are no other intersections. The positive orientation is counterclockwise.
More informationMath 234 Final Exam (with answers) Spring 2017
Math 234 Final Exam (with answers) pring 217 1. onsider the points A = (1, 2, 3), B = (1, 2, 2), and = (2, 1, 4). (a) [6 points] Find the area of the triangle formed by A, B, and. olution: One way to solve
More informationx + ye z2 + ze y2, y + xe z2 + ze x2, z and where T is the
1.(8pts) Find F ds where F = x + ye z + ze y, y + xe z + ze x, z and where T is the T surface in the pictures. (The two pictures are two views of the same surface.) The boundary of T is the unit circle
More informationSummary of various integrals
ummary of various integrals Here s an arbitrary compilation of information about integrals Moisés made on a cold ecember night. 1 General things o not mix scalars and vectors! In particular ome integrals
More informationDisclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.
Disclaimer: This is meant to help you start studying. It is not necessarily a complete list of everything you need to know. The MTH 234 final exam mainly consists of standard response questions where students
More informationMATH H53 : Final exam
MATH H53 : Final exam 11 May, 18 Name: You have 18 minutes to answer the questions. Use of calculators or any electronic items is not permitted. Answer the questions in the space provided. If you run out
More informationMath 6A Practice Problems II
Math 6A Practice Problems II Written by Victoria Kala vtkala@math.ucsb.edu SH 64u Office Hours: R : :pm Last updated 5//6 Answers This page contains answers only. Detailed solutions are on the following
More information1. (a) (5 points) Find the unit tangent and unit normal vectors T and N to the curve. r (t) = 3 cos t, 0, 3 sin t, r ( 3π
1. a) 5 points) Find the unit tangent and unit normal vectors T and N to the curve at the point P 3, 3π, r t) 3 cos t, 4t, 3 sin t 3 ). b) 5 points) Find curvature of the curve at the point P. olution:
More informationReview problems for the final exam Calculus III Fall 2003
Review problems for the final exam alculus III Fall 2003 1. Perform the operations indicated with F (t) = 2t ı 5 j + t 2 k, G(t) = (1 t) ı + 1 t k, H(t) = sin(t) ı + e t j a) F (t) G(t) b) F (t) [ H(t)
More informationMajor Ideas in Calc 3 / Exam Review Topics
Major Ideas in Calc 3 / Exam Review Topics Here are some highlights of the things you should know to succeed in this class. I can not guarantee that this list is exhaustive!!!! Please be sure you are able
More informationJim Lambers MAT 280 Fall Semester Practice Final Exam Solution
Jim Lambers MAT 8 Fall emester 6-7 Practice Final Exam olution. Use Lagrange multipliers to find the point on the circle x + 4 closest to the point (, 5). olution We have f(x, ) (x ) + ( 5), the square
More informationPractice problems ********************************************************** 1. Divergence, curl
Practice problems 1. Set up the integral without evaluation. The volume inside (x 1) 2 + y 2 + z 2 = 1, below z = 3r but above z = r. This problem is very tricky in cylindrical or Cartesian since we must
More informationMath 3435 Homework Set 11 Solutions 10 Points. x= 1,, is in the disk of radius 1 centered at origin
Math 45 Homework et olutions Points. ( pts) The integral is, x + z y d = x + + z da 8 6 6 where is = x + z 8 x + z = 4 o, is the disk of radius centered on the origin. onverting to polar coordinates then
More information18.02 Multivariable Calculus Fall 2007
MIT OpenourseWare http://ocw.mit.edu 18.02 Multivariable alculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.02 Lecture 30. Tue, Nov
More information1 4 (1 cos(4θ))dθ = θ 4 sin(4θ)
M48M Final Exam Solutions, December 9, 5 ) A polar curve Let C be the portion of the cloverleaf curve r = sin(θ) that lies in the first quadrant a) Draw a rough sketch of C This looks like one quarter
More informationMath 2E Selected Problems for the Final Aaron Chen Spring 2016
Math 2E elected Problems for the Final Aaron Chen pring 216 These are the problems out of the textbook that I listed as more theoretical. Here s also some study tips: 1) Make sure you know the definitions
More informationMath 265H: Calculus III Practice Midterm II: Fall 2014
Name: Section #: Math 65H: alculus III Practice Midterm II: Fall 14 Instructions: This exam has 7 problems. The number of points awarded for each question is indicated in the problem. Answer each question
More informationSolutions to old Exam 3 problems
Solutions to old Exam 3 problems Hi students! I am putting this version of my review for the Final exam review here on the web site, place and time to be announced. Enjoy!! Best, Bill Meeks PS. There are
More information1. If the line l has symmetric equations. = y 3 = z+2 find a vector equation for the line l that contains the point (2, 1, 3) and is parallel to l.
. If the line l has symmetric equations MA 6 PRACTICE PROBLEMS x = y = z+ 7, find a vector equation for the line l that contains the point (,, ) and is parallel to l. r = ( + t) i t j + ( + 7t) k B. r
More informationPage Problem Score Max Score a 8 12b a b 10 14c 6 6
Fall 14 MTH 34 FINAL EXAM December 8, 14 Name: PID: Section: Instructor: DO NOT WRITE BELOW THIS LINE. Go to the next page. Page Problem Score Max Score 1 5 5 1 3 5 4 5 5 5 6 5 7 5 8 5 9 5 1 5 11 1 3 1a
More informationMTH 234 Exam 2 November 21st, Without fully opening the exam, check that you have pages 1 through 12.
Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages 1 through 12. Show all your work on the standard
More informationThe Divergence Theorem Stokes Theorem Applications of Vector Calculus. Calculus. Vector Calculus (III)
Calculus Vector Calculus (III) Outline 1 The Divergence Theorem 2 Stokes Theorem 3 Applications of Vector Calculus The Divergence Theorem (I) Recall that at the end of section 12.5, we had rewritten Green
More informationThe Divergence Theorem
Math 1a The Divergence Theorem 1. Parameterize the boundary of each of the following with positive orientation. (a) The solid x + 4y + 9z 36. (b) The solid x + y z 9. (c) The solid consisting of all points
More informationMATH 228: Calculus III (FALL 2016) Sample Problems for FINAL EXAM SOLUTIONS
MATH 228: Calculus III (FALL 216) Sample Problems for FINAL EXAM SOLUTIONS MATH 228 Page 2 Problem 1. (2pts) Evaluate the line integral C xy dx + (x + y) dy along the parabola y x2 from ( 1, 1) to (2,
More informationWORKSHEET #13 MATH 1260 FALL 2014
WORKSHEET #3 MATH 26 FALL 24 NOT DUE. Short answer: (a) Find the equation of the tangent plane to z = x 2 + y 2 at the point,, 2. z x (, ) = 2x = 2, z y (, ) = 2y = 2. So then the tangent plane equation
More informationNote: Each problem is worth 14 points except numbers 5 and 6 which are 15 points. = 3 2
Math Prelim II Solutions Spring Note: Each problem is worth points except numbers 5 and 6 which are 5 points. x. Compute x da where is the region in the second quadrant between the + y circles x + y and
More information(a) The points (3, 1, 2) and ( 1, 3, 4) are the endpoints of a diameter of a sphere.
MATH 4 FINAL EXAM REVIEW QUESTIONS Problem. a) The points,, ) and,, 4) are the endpoints of a diameter of a sphere. i) Determine the center and radius of the sphere. ii) Find an equation for the sphere.
More informationPractice Final Solutions
Practice Final Solutions Math 1, Fall 17 Problem 1. Find a parameterization for the given curve, including bounds on the parameter t. Part a) The ellipse in R whose major axis has endpoints, ) and 6, )
More informationPage Problem Score Max Score a 8 12b a b 10 14c 6 6
Fall 2014 MTH 234 FINAL EXAM December 8, 2014 Name: PID: Section: Instructor: DO NOT WRITE BELOW THIS LINE. Go to the next page. Page Problem Score Max Score 1 5 2 5 1 3 5 4 5 5 5 6 5 7 5 2 8 5 9 5 10
More informatione x2 dxdy, e x2 da, e x2 x 3 dx = e
STS26-4 Calculus II: The fourth exam Dec 15, 214 Please show all your work! Answers without supporting work will be not given credit. Write answers in spaces provided. You have 1 hour and 2minutes to complete
More informationf. D that is, F dr = c c = [2"' (-a sin t)( -a sin t) + (a cos t)(a cost) dt = f2"' dt = 2
SECTION 16.4 GREEN'S THEOREM 1089 X with center the origin and radius a, where a is chosen to be small enough that C' lies inside C. (See Figure 11.) Let be the region bounded by C and C'. Then its positively
More informationMTH 234 Solutions to Exam 2 April 13, Multiple Choice. Circle the best answer. No work needed. No partial credit available.
MTH 234 Solutions to Exam 2 April 3, 25 Multiple Choice. Circle the best answer. No work needed. No partial credit available.. (5 points) Parametrize of the part of the plane 3x+2y +z = that lies above
More informationThe Divergence Theorem
The Divergence Theorem 5-3-8 The Divergence Theorem relates flux of a vector field through the boundary of a region to a triple integral over the region. In particular, let F be a vector field, and let
More informationPage Points Score Total: 210. No more than 200 points may be earned on the exam.
Name: PID: Section: Recitation Instructor: DO NOT WRITE BELOW THIS LINE. GO ON TO THE NEXT PAGE. Page Points Score 3 18 4 18 5 18 6 18 7 18 8 18 9 18 10 21 11 21 12 21 13 21 Total: 210 No more than 200
More informationWithout fully opening the exam, check that you have pages 1 through 12.
Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages 1 through 12. Show all your work on the standard
More informationPast Exam Problems in Integrals, Solutions
Past Exam Problems in Integrals, olutions Prof. Qiao Zhang ourse 11.22 December 7, 24 Note: These problems do not imply, in any sense, my taste or preference for our own exam. ome of the problems here
More informationWithout fully opening the exam, check that you have pages 1 through 12.
MTH 34 Solutions to Exam November 9, 8 Name: Section: Recitation Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through.
More informationTom Robbins WW Prob Lib1 Math , Fall 2001
Tom Robbins WW Prob Lib Math 220-2, Fall 200 WeBWorK assignment due 9/7/0 at 6:00 AM..( pt) A child walks due east on the deck of a ship at 3 miles per hour. The ship is moving north at a speed of 7 miles
More informationSolution. This is a routine application of the chain rule.
EXAM 2 SOLUTIONS 1. If z = e r cos θ, r = st, θ = s 2 + t 2, find the partial derivatives dz ds chain rule. Write your answers entirely in terms of s and t. dz and dt using the Solution. This is a routine
More information