sign in sign up
<<
Home , Y

MATH 0240 -Review for Final Exam

1. Let (, y, ) = x2yz + y3z x2 + z 10, and a = (2, 1, 3). 2 − 2 −2 3 Note: f = (2xyz 2x, x z + 3y z, x y + y + 1) f(a) = (8, 21, 6) ∇ − ∇ (a) Find all stationary points (if any) of f. Set f = 0. Then x2 + y3 + 1 = 0 y = 0 so that z(x2 + 3y2) = 0 z = 0. Now∇z = 0 x = 0, so that y = 1.⇒The6 only stationary point is (0,⇒ 1, 0). ⇒ − − () Find the directional derivative (Duf)(a) of f at a in the direction of 1 = (1, 1, 1) √3 1 35 (Duf)(2, 1, 3) = (8, 21, 6) (1, 1, 1) = · √3 √3 () Find the linearization La of f about a, and use it to give an approximate value of f(1.9, 1.1, 2.8)

L(a) = f(a) + f(a) (x a) = 4 + 8(x 2) + 21(y 1) + 6(z 3) ∇ · − − − − f(1.9, 1.1, 2.8) 4 + 8( 0.2) + 21(0.1) + 6( 0.2) = 4.1. ≈ − − () Find an equation for the tangent plane to the surface f(x, y, z) = 4 at a. a lies on the tangent surface and f(a) = (8, 21, 6) is a normal vector to the ∇ surface. Therefore, pick an arbitrary point (x, y, z) on the surface. Then the vectors (x 2, y 1, z 3) and (8, 21, 6) are perpendicular. So that − − − 8(x 2) + 21(y 1) + 6(z 3) = 0 8x + 21y + 6z = 55. − − − ⇒

2. Suppose x = x(), where x = (x, y, z), with x0(t0) = (1, 2, 3), and that f(x0) = (5, 3, 2), where x0 = x(t0). Find (f x)0(t0). ∇ ◦ By the Chain Rule: (f x)0(t0) = f(x(t0)) x0(t0) = (5, 3, 2) (1, 2, 3) = 17. ◦ ∇ · · 3. Let f(x, y) = x2 2xy + 4y3 be defined on the triangular region T with vertices (0, 0), (2, 0), and −(2, 4). Find the points on T at which f takes on its maximum or its minimum value, and compute these values. Set f = (2x 2y, 2x+12y2) = (0, 0). Then x = y and 2y+12y2 = 0 y = 0, 1 . ∇ − − − ⇒ 6 Points are (0, 0) and (1/6, 1/6) to consider. Now take the boundary y = 0: f(x, 0) = x2 gives (0, 0) and (2, 0) 3 2 x = 1: f(1, y) = 1 2y + 4y , f 0(1, y) = 2 + 12y = 0 gives additional points − − (1, 1/√3) and (2, 4). 3 2 2 y = 2x: f(x, 2x) = 32x 3x , f 0(x, 2x) = 96x 6x = 0 gives additional point (1/16, 1/8). − − (x, y) f(x, y) (0, 0) 0 (1/16, 1/8) 1/256 − (1/6, 1/6) 1/108 − (2, 0) 4 (2, 1/√3) 4 8/(3√3) − (2, 4) 244 f has a minimum on T at (1/16, 1/8) with f(1/16, 1/8) = 1/255 f has a maximum on T at (2, 4) with f(2, 4) = 244. − 4. Use the Second Derivative Test to classify the stationary points of the function f(x, y) = x3 + 3xy2 3x + 1. 2 2− 2 2 f(x) = (3x + 3y 3, 6xy) = (0, 0) (x, y) : x + y = 1 and 6xy = 0 . So either∇ x = 0 y = −1 or y = 0 x = ⇒1.{Points are ( 1, 0) and (0, 1). } ⇒  ⇒  2 2 2 fxx = 6x, fyy = 6x, fxy = 6. fxxfyy (fxy) = 36(x y ). 2 − − (0, 1): fxxfyy (fxy) = 36 < 0. Therefore saddle points.  − 2 − (1, 0): fxxfyy (fxy) = 36 > 0 with fxx > 0. Therefore a local minimum point. − 2 ( 1, 0): f f (f ) = 36 > 0 with f < 0. Therefore a local maximum point. − xx yy − xy xx 5. Use Lagrange Multipliers to locate the point() on the curve x2y = 16 that is (are) closest to the origin. Optimization function: D = x2 +y2. Constraint: C = x2y 16. Set D = λ C so that: − ∇ ∇ 2x = λ2xy 2y = λx2 Knowing that on x2y = 16, x = 0 we have 1 = λy so that λ = 1/y. Substitution into the secod equation gives 2y2 6 = x2. Now substitute 2y2 for x2 back into x2y = 16 giving 2y3 = 16 y = 2 and x = 2√2. The points on x2y = 16 closest to the origin are (2√2, 2)⇒and ( 2√2, 2). Their actual distance is 2√3. − 6. Find the dimensions and cost of the least expensive box with volume 54 ft3, where the bottom of the box costs $5/ft2, the top costs $3/ft2, and the sides cost $2/ft2. 8y + 4z = λyz 8xy + 4xz = λxyz 4xz = 4yz y = x 8x + 4z = λxz 8xy + 4yz = λxyz 8xy = 4xz z = 2y 4x + 4y = λxy 4xz + 4yz = λxyz ⇒ ⇒ xyz = 54 ⇒ 2x3 = 54 xyz = 54 xyz = 54 Therefore: x = 3, y = 3, z = 6, and the cost is $128. 7. Let D = (x, y, z) R3 : x, y, z > 0, x + y + z 9 and consider the integral { ∈ ≤ } 1 I(D) = dx dy dz. Z Z ZD √xyz Use the change of variables u = √x, = √y, = √z, where (x, y, z) D to 2 2 ∈2 compute I(D). Note that (u, v, w) = (√x, √y, √z) (x, y, z) = (u , v , w ). So ⇒ that 2u 0 0 Jacobian = det  0 2v 0  = 8uvw 0 0 2w     Now note that x + y + z 9 with x, y, z > 0, u2 + v2 + w2 9 which is the first ≤ ⇒ ≤ octant of a sphere of radius 3. So: 1 1 I(D) = dx dy dz = (8uvw) du dv dw = 8 du dv dw Z Z ZD √xyz Z Z ZB uvw Z Z ZB 1 4 and I(D) = 8 π33 = 36π. 8 3  8. Let B = (x, y, z) R3 : x, y, z 0, x2 + y2 + z2 9 . Use spherical coordinates to compute{ ∈ ≥ ≤ } I(B) = (x2 + y2 + z2)5 dx dy dz. Z Z ZB B is the first octant of a sphere of radius 3 so 0 ρ 3, 0 θ π/2 and ≤ ≤ ≤ ≤ 0 φ π/2. ≤ ≤ π/2 π/2 3 313 π I(B) = ρ12 sin φ dρ dφ dθ = 0 0 0 Z Z Z 13  2  9. Evaluate the iterated integral

1 1 2 I = ex/y dy dx. Z0 Z√x This says that as x goes from 0 to 1, y goes from √x to 1. If we change the order of integration, we have: as y goes from 0 to 1, x will go from 0 to y2. So that:

1 1 1 2 1 2 1 2 y 2 2 y 1 I = ex/y dy dx = ex/y dx dy = y2ex/y dy = y2( 1) dy = (e 1). 0 √x 0 0 0 0 0 − 3 − Z Z Z Z Z Z

10. Suppose that the surface S is a portion of the plane 2x + 3y + 4z = 5 such that its projection onto the (z, x)-plane is a region D whose area A(D) is 3. Compute the surface area A(S) of S. Explain your answer. The normal to the plane is (2, 3, 4). A unit normal in this direction is 1 1 = √29 (2, 3, 4), so a surface area vector for S is √29 (2, 3, 4)dσ where dσ is the length of the surface area vector. Now take the projection of this onto the unit normal to D which is the vector (0, 1, 0) and you get 1 3 dA = (2, 3, 4) (0, 1, 0)dσ = dσ. Therefore, dσ = √29 and √29 · √29 3 √29 √29 A(S) = dσ = dA = A(D) = √29 Z Z ZD 3 3 11. A piece of a thin metal in the shape of the surface S = (x, y, z) R3 : x2 + y2 + z2 = 1, z > 0 has a constant density of 1 gr/cm2. Find its mass {and center∈ of mass (x, y, z). } With a density of 1 gr/cm2, and a surface area of a hemisphere, the mass is m = 2πgr. By symmetry you have x = y = 0. Need to determine z. Consider the unit normal to the sphere which is ~n = (x, y, z) (length of 1 and normal to surface). Then for D the unit circle x2 +y2 = 1 on the z = 0 plane, a unit normal to D is ~ so that dA = dx dx = ~n ~kdσ = zdσ. Therefore, dσ = 1 dA and: · z 1 1 1 1 z = zdσ = dA = π = m Z ZS 2π Z ZD 2π 2 and (x, y, z) = (0, 0, 1/2). 12. Use the Gradient Test to decide whether or not the given vector field F, in its suitable domain of definition D, is conservative, and if so also find its potential f, i.e. a scalar field f so that F = f on D. Here x stands for (x, y) or for (x, y, z) in D. ∇ 1 (a) F(x) = (y2 x2, 2xy), x = 0 (x2 + y2)2 − − 6 y2 x2 2xy P = − and = − (x2 + y2)2 (x2 + y2)2 ∂P (x2 + y2)2(2y) (y2 x2)2(x2 + y2)(2y) 2y(y2 3x2) = − − = − − ∂y (x2 + y2)4 (x2 + y2)3 ∂Q (x2 + y2)2( 2y) + 2xy2(x2 + y2)(2x) 2y(y2 3x2) = − = − − ∂x (x2 + y2)4 (x2 + y2)3 Therefore F is conservative. (b) F(x) = (yz, xz, xy)exyz. P = yzexyz, Q = xzexyz, = xyexyz. You can just see that if you differentiate xyz f = e with respect to x you get fx = P , likewise fy = Q and fz = R. Without checking the Gradient Test, try to determine f by taking f(x, y, z) = yzexyzdx = exyz + (y, z) xyz Now differenRtiate with respect to y and get fy = xze + h0(y, z) Since Q = xzexyz, h(y, z) has no term of y so h(y, z) = h(z). Now differentiate xyz xyz f(x, y, z) = e + h(z) with respect to z and fz = xye + h0(z). Since R = xyexyz, h(z) has no term of z. Therefore f(x, y, z) = exyz is a potential for F. (c) F(x) = (x2ey+z, y2ex+z, z2ex+y). ∂P ∂Q Note that = x2ey+z and = y2ex+z and these are not equal so you need ∂y ∂x not go any further to say that F is not conservative (there does not exist a potential f such that f = F. ∇ 1 (d) F(x) = (x, 1), x2 + 2y > 0. √x2 + 2y x 1 P = and Q = √x2 + 2y √x2 + 2y ∂P x ∂Q x = and = . F is conservative. ∂y −(x2 + 2y)3/2 ∂x −(x2 + 2y)3/2 x 2 f(x, y) = 2 dx = x + 2y + h(y). Now differentiate with respect Z √x + 2y 1 q 1 to y and f = + h0(y). Since Q = , h(y) has no y term and y √x2 + 2y √x2 + 2y f(x, y) = √x2 + 2y is a potential for f = F. ∇ (e) F(x) = (x + y + z)(1, 1, 1). P = Q = R = x + y + z. Py = Qx = 1 = Pz = Rx = Qz = Ry = 1. F is conservative. 1 2 f(x, y, z) = 2 x + xy + xz + h(y, z) so that fy = x + h0(y, z) which implies 1 2 h0(y, z) = y + z which in turn says that h(y, z) = 2 y + yz + k(z). 1 2 1 2 So far we have f(x, y, z) = 2 x +xy+xz+ 2 y +yz+k(z). Take fz = x+y+k0(z) 1 2 and we see that k0(z) = z so k(z) = 2 z . 1 Finally: f(x, y, z) = x2 + y2 + z2 + xy + xz + yz is a potential for F 2   13. Evaluate the following line integrals I(C): (a) I(C) = xy2 , where C is the line segment from ( 2, 3) to (1, 4). C − − A parametrizationZ for the line segment is (x, y) = ( 2 + 3t, 3 7t), 0 t 1. − − ≤ ≤ ds = r0(t) dt = √58dt so that: I(C) = || || 1 1 5√58 ( 2 + 3t)(3 7t)2√58 dt = √58 ( 18 + 11t 224t2 + 147t3) dt = Z0 − − Z0 − − − 12

(b) I(C) = (2 + xz)xyexz dx, +x2exz dy + x3yexz dz , where C is given by C Z { 2 } x(t) = (t2 + 1, tet , cos πt), 0 t 1. ≤ ≤ Note the difficulty of F: see if there is a potential for F: xz 2 xz 2 xz 3 xz Py = (2+xz)xe = Qx, Pz = x ye )+(2+xz)x ye = Rx, and Qz = x e = 2 xz Ry. F is conservative and a potential for F is f(x, y, z) = x ye (Easy way to determine this is start with Q and determine x2exzdy = x2yexz + h(x, z) 1 Then f dx = f(2, e, 1) f(1, 0, 1) = 4e− R ZC ∇ · − − (c) I(C) = F N ds, where F(x) = (x2 y2, 2xy) and C is described by C · − x(t) = 2(cosZ (2t), sin (2t)), π/4 t π/2. ≤ ≤ F Nds = (F1, F2) (dy, dx) so: · · − I(C) = (x2 y2)dy 2xydx ZC − − π/2 = (4 cos2 2t 4 sin2 2t)(4 cos 2t) 8 cos 2t sin 2t( 4 sin 2t) dt 4 Zπ/ − − − π/2 = (16 cos3 2t + 16 sin2 2t cos 2t) dt 4 Zπ/ π/2 π/2 = 16 (cos 2t)(cos2 2t + sin2 2t) dt = 16 cos 2t dt 4 4 Zπ/ Zπ/ 1 π/2 = 16 sin 2t = 8( 1) = 8 2 π/4 − − 14. Evaluate I(C) = F dx ZC · where F(x) = (x2 + y, z, z4 + 2x), and C is described by x(t) = (cos t, sin t, cos t + sin t), 0 t 2π. ≤ ≤ If you do this calculation straightforwardly it is quite a bit of work but notice you have a nice closed region so much will evaluate to zero.

2π I(C) = ( cos2 t + sin t)( sin t) + (cos t + sin t)(cos t) 0 − − Z  +(cos t + sin t)4( sin t + cos t) + 2 cos t( sin t) + 2 cos2 t dt − −  2π 2 2π 2 2 2π Note: 0 cos t sin t dt = 0 ( sin t + cos t) dt = 0 sin t cos t dt = 2π 4 − 0 (cosRt + sin t) ( sin t +R cos t) dt = 0. R − 2π 2π TheR only evaluation left is cos2 t dt = (1 + cos 2t) dt = 2π. 0 0 F Z 2 4 Z 1 3 1 5 This is partly because = (x , 0, z ) + (y, z, 2x) and f = 3 x + 5 x is a potential for (x2, 0, z4) and partly because z = x + y along C.

15. Evaluate I(S) = ( F) N dσ Z ZS ∇ × · where S is the portion of the surface 36z = 108 4x2 9y2 for which z 2, and − − ≥ the normal is outward, and where F = (F1, F2, F3) is given by 2 2 3 3 2 z2 F1(x, y, z) = 3x yz + yz , F2(x, y, z) = x z + xz + z , F3(x, y, z) = xye . Use Stokes Theorem: ( F) N dσ = F dx where C is the boundary: for S ∇ × · C · z = 2, 4x2 + 9y2 = 36Z whicZ h is an ellipse withZ major axis of 3 in the x direction, minor axis of 2 in the y direction and on the plane z = 2.

2 F dx = (3x2yz + yz2)dx + (x3z + xz3 + z2)dy + (xyez )dz ZC · ZC Since z = 2 and thus dz = 0 this reduces to:

F dx = (6x2y + 4y)dx + (2x3 + 8x + 4)dy ZC · ZC Since the ellipse C is a simple closed curve use Green’s Theorem: ∂ ∂ (6x2y+4y)dx+(2x3+8x+4)dy = (2x3 + 8x + 4) (6x2y + 4y) dx dy ZC Z ZE ∂x − ∂y !

= (6x2 + 8 6x2 4) dx dy = 4 dx dy = 4(A(E)) = 4(6π) = 24π. Z ZE − − Z ZE 16. Let S be the portion of the sphere x2 + y2 + z2 = 9 for which z 0, and the normal ≥ is upward, and let F = (F1, F2, F3) be given by 2 z2 3 2 2 2 F1(x, y, z) = 3x + y e + z , F2(x, y, z) = 6y + 2x z , F3(x, y, z) = x + z. Use the Divergence Theorem to compute the flux ΦS(F) = (F N)dσ. S · Careful on this one. The Divergence Theorem applies to theZ Zentire surface of the region bounded and this surface does not include the ”cap” on the bottom of the hemisphere. So to use the Divergence Theorem, calculate the flux over the entire surface then subtract the flux over the ”cap”. In other words, call H the surface of the covered hemisphere and T the base of the hemisphere and B the interior region of the hemisphere:

(F N) dσ = (F N) dσ (F N) dσ Z ZS · Z ZH · − Z ZT ·

(F N) dσ = FdV Z ZH · Z Z ZB ∇ · 1 4 = 10 dV = 10 π(3)3 = 180π Z Z ZB 2 3  On the ”cap”, T , the outward normal is k, z = 0 and x2 + y2 = 9 so that: − (F N) dσ = F(x, y, 0) ( k)dσ Z ZT · Z ZT · − 2π 3 = x2 dx dy = r2 cos2 θr dr dθ Z ZT − Z0 Z0 − 2π 81 = cos2 θ dθ 0 Z − 4 81 2π 1 1 81π = + cos 2θ dθ = . − 4 Z0 2 2  − 4 81 Therefore Φ (F) = 180π + π. S 4 17. Suppose S is any portion of the plane ax + by + cz = d, c > 0, whose area is A and that is positively oriented. Let F = (α, β, γ) be a constant vector field. Compute the flux ΦS(F) = (F N) dσ Z ZS · Recall, a normal to the plane ax + by + cz = d is (a, b, c). A unit normal in 1 this direction (and we want this direction since c > 0 is N (a, b, c). √a2 + b2 + c2 Therefore: F N 1 dσ = 2 2 2 (a, b, c) (α, β, γ)dσ Z ZS · Z ZS √a + b + c · aα + bβ + cγ = 2 2 2 dσ Z ZS √a + b + c aα + bβ + cγ aα + bβ + cγ = A(S) = A √a2 + b2 + c2 √a2 + b2 + c2

2 2 18. Let F = (P, Q), where P (x, y) = 8ex + 3xy 3x2y and Q(x, y) = 8ey + 4xy + 3x2y, 2 2 − and let C be the circle x + y = 2 described in the positive direction. Use Green’s Theorem to evaluate the circulation

ΨC (F) = F dx ZC ·

∂ 2 ∂ 2 F dx = (8ey + 4xy + 3x2y) (8ex + 3xy 3x2y) dx dy ZC · Z ZS ∂x − ∂y − ! = (4y + 6xy 3x + 3x2) dx dy Z ZS − √2 2π = 4r sin θ + 6r2 cos θ sin θ 3r cos θ + 3r2 cos2 θ r dr dθ 0 0 − Z Z   2π √2 √2 2π 1 1 = 3r3 cos2 θ dθ = 3r3 + cos 2θ dθ dr Z0 Z0 Z0 Z0 2 2  √2 3 = 3πr3 dr = π(√2)4 = 3π. Z0 4