Math 023 Fall 2012 Final Review Sheet -solutions - Math 23...

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Unformatted text preview: Math 23, Fall, 2012 December 8, 2012 Solutions to the review sheet for the nal exam 1. For vectors (a) A = 4i + 3j − 2k B = −i + 3k, nd: A − 2B. Solution (b) and = 6i + 3j − 8k. A · B. Solution = −4 + 0 − 6 = −10. You need to be careful to remember that the dot product of any two vectors is a (c) The vector projection Solution projB A of scalar . A onto Here the answer has to be a B. vector , parallel to B. A·B B B·B −10 = (−i + 3k) 10 = i − 3k projB A = . (d) A × B. Solution = 9i − 10j + 3k. Again, it has to be a vector. A. √ √ = 16 + 9 + 4 = 29. (e) The length of Solution 2. Find an equation of the plane Solution P containing the points (1, 0, 3), (−1, 0, 1), and (2, 1, 0). The plane is perpendicular to the cross-product − − − − − − → −− − − − − −−−−−− − − − − −→ (1, 0, 3), (−1, 0, 1) × (1, 0, 3), (2, 1, 0) = (−2i − 2k) × (i + j − 3k) = 2i − 8j − 2k, so the equation is 2(x − 1) − 8(y − 0) − 2(z − 3) = 0, 1 or 2x − 8y − 2z + 4 = 0. 3. What is the distance from the point Solution P0 = (1, 1, 2) There is a formula for the distance from a point to a plane in the text. In N this case, using the normal vector Dist = = = r(t) is a vector-valued (r(t) · r(t)) at t = 0. 4. If Solution 2x + 3y − z + 1 = 0? to the plane P in the plane −→ − N · P P0 N |2 · 1 + 3 · 1 − 1 · 2 + 1| √ 4+9+1 4 √ . 14 function so that (r(t) · r(t)) = 2r(t) · r (t), to the plane, and any point r(0) = 2i + j − k so at and r (0) = i − j + 2k, nd t = 0, (r(t) · r(t)) |t=0 = 2r(0) · r (0) = 2(2i + j − k) · (i − j + 2k) = 2(2 − 1 − 2) = −2. 5. Let r(t) be a vector valued function, and set v(t) := r (t), a(t) := v (t) = r (t) to r(t) · v(t) = 0, show that the be the velocity and acceleration vectors, respectively. If curve traced out by Solution r(t) lies on a sphere centered at the origin. Since the distance from the origin to the point what you have to show is that r(t) · r(t) r(t) r(t) is r(t) = is constant. But that is the same as r(t) · r(t), r(t) 2 = being constant. Dierentiate, and (r(t) · r(t)) = r (t) · r(t) + r(t) · r (t) = 2r(t) · v(t) = 0. Since the derivative of the distance is 0, the distance is constant. 6. The equation Solution the − y2 9 + z2 16 =1 describes what sort of surface? Sketch the surface. This equation describes a hyperboloid of one sheet, with axis of symmetry y -axis. 7. Compute Solution and x2 4 ∂f /∂x and ∂f /∂y for the function f (x, y) = √ x + y sin(x2 y). √ 1 ∂f /∂x = √ sin(x2 y) + 2xy x + y cos(x2 y), 2 x+y √ 1 ∂f /∂x = √ sin(x2 y) + x2 x + y cos(x2 y). 2 x+y 2 8. A heat-eeing bug has been placed on a hot griddle, whose temperature at a point (x, y) T (x, y) = 100 + 2x2 − 3xy + y 2 . The griddle is shaped as a 3 × 3 the measurements x and y taken with the lower-left corner as the origin. dropped at the point (1, 2), in which direction will she move, in order to is given by square, with If the bug is decrease the heat on her feet most rapidly? Solution T, (T ) = She should move in the direction opposite to that of the gradient of increase of T . So, since (4x − 3y)i + (−3x + 2y)j, and at (1, 2), then, (T ) = −2i + j. She should then √ move in the direction u = (2 i−j)/ 5. Note that we always express this direction since the gradient gives the direction of greatest unit as a vector. 9. Find the tangent plane to the graph of Solution • 9 − x2 − y 2 f (x, y) = at the point (1, 2, 2). There are two formulas that you could use to nd this: One way is to simply remember the way to nd a normal vector to a level surface. If F (x, y, z) = c to that surface. describes a surface, then F is perpendicular Here, you can make the graph a level surface by dening F (x, y, z) := z−f (x, y) = z− 9 − x2 − y 2 . Then the graph is F (x, y, z) = 0, and F = −∂f /∂x i − ∂f /∂y j + k = √ x2 2 i + √ y 2 2 j + k. At 9−x −y 9−x −y N = F |(1,2,2) = 1 i + j + k. Finally, the tangent 2 1 plane is (x − 1) + (y − 2) + (z − 2) = 0, or x + 2y + 2z − 9 = 0. 2 • The other way would be to recall that the tangent plane to a graph z = f (x, y) (1, 2, 2), the normal is is given by − ∂f ∂x (x − x0 ) + − (x0 ,y0 ) ∂f ∂y (y − y0 ) + (z − z0 ) = 0. (x0 ,y0 ) Not coincidentally, that leads to the same equation as in the rst method. 10. A cardboard box has one layer of cardboard for the sides, but a double layer of cardboard on the top and bottom (for the aps). Find the dimensions of the cardboard box which can be made from 12 square feet of cardboard and which holds the most volume. Solution Set x to be the length of the base, y to be the width of the base, and be the height. Then the volume of the box is cardboard needed is S = 4xy + 2xz + 2yz , V = xyz , V 4xy + 2xz + 2yz = 12, x and y and you subject to that constraint. This is clearly a Lagrange multipliers problem (it can be done by substituting for of to counting the top and bottom double for the aps. The constraint in this problem is are to then nd the maximum of z and the square footage of z in terms by using the constraint equation, but that is harder, I think). The equations are: yz = λ(4y + 2z) xz = λ(4x + 2z) xy = λ(2x + 2y). 3 Multiply the rst equation by xyz z(2x + 2y). versions of x, the second by y and the third by z . Set the three λ everywhere, giving: x(4y +2z) = y(4x+2z) = equal, and cancel Alternately, you could solve all three equations for λ λ, then set all three versions λ). A bit of simplifying gives the same equations. The x(4y + 2z) = y(4x + 2z), implies that xz = yz , so z = 0 or x = y . If z = 0, then the box has 0 height, which isn't much of a box and certainly won't maximize the volume. Thus, x = y , and substituting into 2 the last equation gives 4x + 2xz = 4xz , or z = 2x. Then, the only solution is of the form (x, y, z) = (x, x, 2x). The constraint equation S = 12 becomes 4x2 + 2x2x + 2x2x = 12, which of course implies that x = 1. The solution is then that the box is 1 × 1 × 2, with the height the longer dimension. The maximum of equal (eliminating rst equation, volume is then 2 cubic feet. 11. Maximize Solution 9 x + y + 2z on the sphere x2 + y 2 + z 2 = 9. Another Lagrange multipliers problem. Here the constraint is x2 +y 2 +z 2 = and the equations are: 1 = λ2x 1 = λ2y 2 = λ2z 2λ and eliminating λ results in 1/x = 1/y = 2/z , x = y , z = 2x. Plugging into the constraint gives x2 + x2 + 4x2 = 9, or x = ± 3/2. The − choice clearly gives the minimum, so the maximum occurs √ at ( 3/2, 3/2, 6). Solving all three equations for or 12. Find the maximum and minimum of Solution over the disk x2 + y 2 ≤ 1. First, you have to look for critical points in the interior. You look for those by nding where and f (x, y) = x2 − 3y 2 −6y = 0, f = 0, 2xi − 6yj = 0. These two equations, 2x = 0 by x = y = 0. That point, (0, 0), is a critical which is can only be solved point, but is neither a local maximum nor a local minimum. It is a saddle point, which you can see from the discriminant D = (2)(−6) = −12. This leaves the boundary points, which gives us another Lagrange multipliers problem. There the constraint is x2 + y 2 = 1, and so the equations are 2x = λ2x and − 6y = λ2y. x = 0 or λ = 1, and from the second equation, either y = 0 or λ = −3. If x = 0, then the only points satisfying the constraint are when y = ±1, (0, 1) and (0, −1). When λ = 1, then necessarily y = 0 because λ = −3. Since y = 0, then x = ±1, so (1, 0) and (−1, 0) are the points in that case. If y = 0 we already have that, and if λ = −3 we have to have x = 0 which From the rst equation, either again we already covered. There are only those 4 possible points, and f (±1, 0) = 1 f (0, ±1) = −3. 4 (1, 0) (0, −1). So, the maximum value has to be 1, achieved at −3, minimum value is √ 1−x2 1 0 0 13. Evaluate achieved at (0, 1) and and (−1, 0), and the x dy dx. Solution √ 1 1−x2 1 1 − x2 x dx x dy dx = 0 0 0 1 1 = − (1 − x2 )3/2 3 1 = . 3 0 1 1 3 0 x x sin(y )dy dx. 14. Evaluate Solution Reverse the order of integration, and 1 1 0 1 x sin(y 3 )dy dx = 0 x = = y x sin(y 3 )dx dy 0 1 1 y 2 sin(y 3 )dy 2 0 1 1 (− cos(y 3 ))|1 = (1 − cos(1)). 0 6 6 15. Find the center of mass of the portion of the circle quadrant. Assume that the region is homogeneous. Solution x2 + y 2 ≤ 1 which is in the rst Hint: Use symmetry. The hint was supposed to help you notice that x = y. `Homogeneous' means that we can take the density to be constant, and it might as well be 1 since the constant will cancel top and bottom. So, y=x= √ 1 1−x2 0 0 x dy dx 4 = . π/4 3π It's nice to have already done that integral two problems ago. 16. Find the volume under the graph of the function (x − 1)2 + y 2 ≤ 1 above the xy -plane. of the circle which is Solution in the xy -plane. f (x, y) = x + y and above the inside Be sure to consider only that portion Here it's best to use polar coordinates. Be careful, though, since the function dips below the below the line xy -plane when x + y < 0, which y = −x in the circle. Then, π/2 V ol = 2 cos θ r2 (cos θ + sin θ)dr dθ = −π/4 0 5 means that the region doesn't go 8 3 π/2 −π/4 cos4 θ + cos3 θ sin θdθ. This is a nontrivial, but standard integral: 8 3 π/2 cos4 θ + cos3 θ sin θdθ = −π/4 = = = = = = 17. Find the mass of the region and x + y + 2z = 2, Solution R cos4 θdθ + −π/4 π/2 8 3 −π/4 π/2 2 3 8 3 π/2 cos3 θ sin θdθ −π/4 1 (1 + cos(2θ)) 2 2 dθ − π/2 1 + 2 cos(2θ) + −π/4 2 3 2 3 3π 4 1 cos4 (θ) 4 1 + 2 cos(2θ) + cos2 (2θ) dθ − −π/4 2 3 8 3 ρ(x, y, z) 2−y π 1 +0+ 2 2 5 + . 6 is given by x = 0, y = 0, z = 0, ρ(x, y, z) = y . 1−x/2−y/2 2 0 2−y 0 0 2 1−x/2−y/2 yz|0 = dxdy 2−y y (1 − x/2 − y/2) dxdy = 0 0 2 2−y = 0 y − xy/2 − y 2 /2 dxdy 0 2−y 1 1 xy − x2 y − xy 2 dy 4 2 0 0 2 1 1 (2 − y) y − (2 − y)2 y − (2 − y) y 2 dy 4 2 0 2 1 3 1 2y − y 2 − y + y 2 − y − y 2 + y 3 dy 4 2 0 2 1 3 y − y 2 + y dy 4 0 2 1 2 1 3 1 y − y + y4 2 3 16 0 1 . 3 2 = = = = = = 18. Find the integral of z2 = 4(x2 + y 2 ) and f (x, y, z) = x2 + y 2 + z the plane z = 2. 6 4 π/2 1 1 1 θ + sin(4θ) + 2 4 6 −π/4 π π π π 1 − +0 + 0 − − + sin(− ) + 2 4 2 2 4 y dzdxdy 0 1 √ 2 θ + sin(2θ) + M ass = 0 −π/4 1 1 (1 + cos(4θ)) dθ + 2 6 ρ = y, 2 π/2 2 0− 3 in space bounded by the planes if the density Since the density is π/2 8 3 over the region R bounded by the cone + 1 6 Solution In cylindrical coordinates, x2 + y 2 + z dV R 2 1 2π = . . . (r2 + z)rdzdrdθ 2r 0 0 6π . 5 = f (x, y, z) = x2 + y 2 + z 2 over the region R bounded z 2 = 4(x2 + y 2 ) and above by the sphere x2 + y 2 + z 2 = 4. 19. Find the integral of cone Solution r2 + z = 2r, In cylindrical coordinates, since the cone is z2 = 4, or z= 2 √ 4− 2 and the top becomes r2 , √ √ 2/ 5 2π 2 x + y + z dV 4−r2 = R 0 . . . = 0 (r2 + z 2 )rdzdrdθ 2r √ 64 (5 − 2 5)π. 25 In spherical coordinates it's a bit easier. The cone becomes the top just 2 2 arctan( 1 ) 2 2π x + y + z dV = 0 R = (2π) = = 0 25 5 2 (ρ2 )ρ2 sin φdρdφdθ 0 arctan( 1 ) 2 20. For the vector eld 64π 1 1 − cos arctan( ) 5 2 √ 64 (5 − 2 5)π, 25 F = x2 i + (3xy + z)j + (z 2 + x)k, nd: div(F) Solution div(F) = 2x + 3x + 2z = 5x + 2z. curl(F) Solution curl(F) = −i − j + 3yk. 7 sin φdφ 0 though the answer is the same, of course. (b) φ = arctan( 1 ) 2 ρ=2 2 (a) below by the and 21. Find the line integral C 2x dx + (x + y) dy , where C is the circle of radius 2, centered at the origin, traversed counterclockwise. Solution This is much easier if you use Green's theorem: 2x dx + (x + y) dy = 1dA C R = Area = 4π. x2 dx + dy C is the (upside-down) V-shaped curve which is composed of the line segment from (1, 0) to (2, 3), followed by the line segment from (2, 3) to (3, 0). 22. Find the line integral Solution C where 2 2 C x dx + dy = C1 x dx + dy + C2 rst segment, C2 the other. C1 has description x = C2 Directly: has description x = 2 + t, y = 3 − 3t, t ∈ [0, 1]. 1 x2 dx + dy = So, 1 ((1 + t)2 + 3)dt + ((2 + t)2 − 3)dt 0 0 C x2 dx + dy , where C1 is the 1 + t, y = 3t, t ∈ [0, 1], and = 26/3. Using theorems: It should be noted that the vector eld is conservative, y) = x2 i + j, so x2 dx + dy = x3 /3 + y C = = in the presence of a force (3,0) (1,0) 1 3 1 3 3 − 1 3 3 26 . 3 C described F = x2 i + y 2 j. 23. Find the work done in moving along the path t ∈ [0, 4π], (x3 /3+ by G(t) = ti + sin(t)j, Solution W x2 dx + y 2 dy = C 4π = (t2 + sin2 (t) cos(t))dt 0 = 24. Evaluate x-axis and 64π 3 . 3 xy dx + ex dy , where C is the curve that goes from (0, 0) 2 returns from (2, 0) to (0, 0) on the parabola y = 2x − x . C 8 to (2, 0) on the Solution Again, you should be thinking of Green's theorem. To do the exponential integrals, use integration by parts. xy dx + ex dy = ex − x dA C R 2x−x2 2 = ex − x dydx 0 0 2 = (ex − x) 2x − x2 dx 0 2 = 2xex − x2 ex − 2x2 + x3 dx 0 2 (2xex − 2ex ) − x2 ex − 2xex + 2ex − x3 + 3 2 1 4e2 − 2e2 − 4e2 − 4e2 + 2e2 − 8 + 16 3 4 16 − +4 +4 3 = = = = 25. Find C 2 1 4 x 4 0 − ((−2) − (2)) 8 . 3 x3 dx + (x + ey )dy , where C is the circle of radius 1 centered at the point (1, 0), oriented positively. Solution Once again, use Green's theorem, x3 dx + (x + ey )dy = C where 26. Evaluate R 1 dxdy = π, R is of course the inside of the circle. −y dx C x2 +y 2 + x dy over the ellipse x2 +y 2 C given by x2 /4 + y 2 /9 = 1. Hint: This is a lot easier if you use Green's theorem, in the two-curve case. Solution Since x ∂/∂x( x2 +y2 ) = y 2 −x2 , which is the (x2 +y 2 )2 double-integral part of Green's theorem is 0. same as ∂/∂y( x2−y 2 ), +y integral is 0, though. Instead, since you have to remove a little circle 0 to nd a region R the This doesn't imply that the line over which the vector eld is dened C around and dierentiable, you apply the 2-curve version and have that the integral over the outside plus that on the little circle (with the backwards orientation since it is the inside curve) is the integral over R C where C of 0, which is 0. Then, that implies that x dy −y dx + x2 + y 2 x2 + y 2 = C −y dx x dy + , x2 + y 2 x2 + y 2 is a small circle centered at the origin, small enough to be completely inside the ellipse. But, this integral can be done directly, using y = sin(t). You get 2π for that, so C −y dx x dy + 2 2 + y2 x x + y2 9 = 2π. x = cos(t) and 27. Which of the following vector elds are conservative? potential function. For those which are, nd a The domain of the vector eld is assumed to be all points for which the functions make sense. (a) F = 3x2 yi + x3 + 2x + y j. Solution Not conservative, since ∂ 3x2 y ∂y = 3x2 , ∂ x3 + 2x + y ∂x (b) but = 3x2 + 2 F = (y + 1)i + (x + 1)j. Solution f (x, y) = xy + x + y + c. Conservative, and You nd the potential function by integrating the rst component and checking against the second component: gration may f (x, y) = (y + 1)dx = xy + x + k(y) (the "constant" of intedepend on y ). Then dierentiate with respect to y , and it must be the second component: ∂f ∂y ∂ (xy + x + k(y)) ∂y = x + k (y) which must = be = x + 1, so (c) k (y) = 1, or k(y) = y + c. F = ex cos yi − ex sin yj. Solution Conservative, since −ex sin y , and F ∂(−ex sin y)/∂x = −ex sin y , F= x i x2 +y 2 Solution The centroid + ∂(ex cos y)/∂y = is dened over the whole plane, which is simply-connected. The potential function is clearly (d) and f (x, y) = ex cos(y). y j. x2 +y 2 Conservative, and of a surface S f= ln(x2 +y 2 ) 2 + c. is found just as for other regions, the point (x, y, z) given by x := x dS S dS S y := 28. Find the centroid of the hemisphere y dS S dS S z := z dS . S dS S x2 + y 2 + z 2 = 1, z ≥ 0. You really should only have to do one surface integral. Solution By symmetry, x=0 and y = 0. Also, the denominator S dS = 2π , half the surface area of a sphere of radius 1. So, all we have to nd is the numerator for z, S z dS . But, since the surface is a graph, 10 z = 1 − x2 − y 2 , over the disk A given by x2 + y 2 ≤ 1, 1 − x2 − y 2 z dS = S A 1 dA n·k 1 − x2 − y 2 = −2x 2 1 − x2 − y 2 1+ A A = dA, = π. + dA 2 + −2y 2 1 − x2 − y 2 x2 y2 + dA 1 − x2 − y 2 1 − x2 − y 2 1 − x2 − y 2 1 + = ∂z ∂y 2 ∂z ∂x A So, 2 1 − x2 − y 2 1 + = yes, those square roots cancel! A z = 1. 2 29. Find the surface integral of the function x+y+z =1 Solution that is in the rst octant f (x, y, z) = x2 + y 2 over that part of the plane (so x ≥ 0, y ≥ 0, and z ≥ 0). S f dS . You have to do this surface integral directly, but it is not too bad. z = 1 − x − y. surface is of course a graph, = = 2 ∂z ∂x dS = √ The element ∂z ∂x + x = 0, y = 0, and + 1dxdy (−1)2 + (−1)2 + 1dxdy 3dxdy x + y = 1, 1 x2 + y 2 dS = 0 S = √ 1−x = = = = √ 1 3 √ √ √ √ 1 3 3 3 3 3 11 x and y being in the region so x2 + y 2 √ 3dydx 0 0 = of surface area is 2 That part in the rst octant would correspond to bounded by dS The 1 x2 y + y 3 3 1−x dx 0 1 (1 − x)3 dx 3 0 1 1 1 x2 − x3 + − x + x2 − x3 dx 3 3 0 2 3 1 4 1 1 2 1 x − x + x− x 3 3 3 2 0 2 1 1 1 − + − 3 3 3 2 1 . 6 x2 − x3 + 2 dA 30. Find the ux of the vector eld y2 + (z + Solution 1)2 F = xi + yj + z 2 k outward across the sphere (x − 1)2 + = 4. This one cries out for the divergence theorem. F · n dS = · FdV, S where V V ·F = 1+1+2z = 2+2z , so, using zdV = z · V ol(V ) (and similarly, V V ydV = y · V ol(V ) xdV = x · V ol(V ), taking the center-of-mass integrals with density set is the inside of the ball. In this case, the trick that and V to 1). In this case we know that the center of mass (density 1) is the center of the sphere, so F · n dS = 2 + 2zdV S V = 2 V ol(V ) + 2z V olV = (2 + 2(−1))V olV = 0. 31. Find the ux of the vector eld F = x2 i − 3yj + z 2 k to n) across the surface S which (1, 2, 0), (3, 0, 1) and (3, 2, 1). Solution upward (positive k component is the inside of the rectangle with vertices (1, 0, 0), This one cannot be done using Stokes' theorem, or any other of those the- orems. We have to solve the integral directly. First, though, we have to nd the S is then the part of that plane lying inside the rectangle. The plane has equation 1 2x − 4z − 2 = 0, or z = 2 x − 1 . 2 equation of the plane through those 4 points. Using the formula on p. 1088 of the book: F · n dS = S = = = = 32. Find the line integral C is (x − 1)2 + y 2 = 1, z and the curve C ∂z ∂z x2 + − (−3y) + z 2 dxdy ∂x ∂y A 1 1 1 2 − x2 + (0) (−3y) + x− dxdy 2 2 2 A 2 3 1 1 1 − x2 − x + dxdy 4 2 4 1 0 1 1 1 3 2 − x3 − x2 + x 12 4 4 1 22 − . 3 − F · dr, F = (sin(x) + y 2 )i + x2 j + xzk, x + y + z = 0 with the cylinder if the vector eld is the intersection of the plane arbitrary. The curve C is oriented positively (counter-clockwise) when viewed from above. Solution First, apply Stokes' theorem, to get F · dr = C × F · dS, S 12 where S is the plane x + y + z = 0, z = −x − y , and (x − 1)2 + y 2 ≤ 1, and or, as a graph, part of the plane inside the cylinder, so i ×F = j k ∂ ∂x ∂ ∂y x2 we want that ∂ ∂z (sin(x) + y 2 ) xz = 0i − zj + (2x − 2y) k. ∂z/∂x = −1 and ∂z/∂y = −1, and R is the region 2 2 disk (x − 1) + y ≤ 1, and of course z = −x − y , So, since just the F · dr = under S, which is × F · dS C S (0i zj + (2x − 2y) k) · dS = S 0 − = R ∂z ∂x −z − ∂z ∂y + (2x − 2y) dxdy (−(−x − y) (+1) + (2x − 2y)) dxdy = R (3x − y) dxdy = R = 3xArea(R) − yArea(R) = 3(1)π − (0)π = 3π. 33. Find the line integral and the curve the plane Solution x= C √ F·dr, if F is the vector eld F = (sin(x)+y)i+ey j+(x+y 2 )k, 2 2 2 curve of intersection between the sphere x + y + z = 4 and C is the 3. Again apply Stokes' theorem. i ×F = j k ∂ ∂x ∂ ∂y ey ∂ ∂z (sin(x) + y) = 2yi − j − k, and the surface S √ x = 3 inside the circle y 2 + z 2 = 1, √ x = g(y, z) where g(y, z) is the constant 3, a plane is that part of the plane (that surface is the graph parallel to the (x + y 2 ) yz -plane) so F · dr = × F · dS C S (2yi − j − k) · idydz = S = 2yArea(S) = 0. F = xi + yj + zk enclosed by S . 34. Show that the ux of the vector eld surface S is three times the volume 13 outward across any closed Solution By the divergence theorem, if V is the region enclosed by (xi + yj + zk) · n dS = S · (xi + yj + zk) dV V = 3 dV V = 3 V ol(V ). 14 S, ...
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