8(x 2) + 21(y 1) + 6(z 3) = 0 8x + 21y + 6z = 55.

Transcription

1 MATH 24 -Review for Final Exam. Let f(x, y, z) x 2 yz + y 3 z x 2 + z, and a (2,, 3). Note: f (2xyz 2x, x 2 z + 3y 2 z, x 2 y + y 3 + ) f(a) (8, 2, 6) (a) Find all stationary points (if any) of f. et f. Then x 2 + y 3 + y so that z(x 2 + 3y 2 ) z. Now z x, so that y. The only stationary point is (,, ). (b) Find the directional derivative (D u f)(a) of f at a in the direction of u 3 (,, ) (D u f)(2,, 3) (8, 2, 6) (,, ) (c) Find the linearization L a of f about a, and use it to give an approximate value of f(.9,., 2.8) L(a) f(a) + f(a) (x a) 4 + 8(x 2) + 2(y ) + 6(z 3) f(.9,., 2.8) 4 + 8(.2) + 2(.) + 6(.2) 4.. (d) 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, 2, 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, z 3) and (8, 2, 6) are perpendicular. o that 8(x 2) + 2(y ) + 6(z 3) 8x + 2y + 6z uppose x x(t), where x (x, y, z), with x (t ) (, 2, 3), and that f(x ) (5, 3, 2), where x x(t ). Find (f x) (t ). By the hain Rule: (f x) (t ) f(x(t )) x (t ) (5, 3, 2) (, 2, 3) Let f(x, y) x 2 2xy + 4y 3 be defined on the triangular region T with vertices (, ), (2, ), and (2, 4). Find the points on T at which f takes on its maximum or its minimum value, and compute these values. et f (2x 2y, 2x+2y 2 ) (, ). Then x y and 2y+2y 2 y, 6. Points are (, ) and (/6, /6) to consider. Now take the boundary y : f(x, ) x 2 gives (, ) and (2, ) x : f(, y) 2y + 4y 3, f (, y) 2 + 2y 2 gives additional points (, / 3) and (2, 4).

2 y 2x: f(x, 2x) 32x 3 3x 2, f (x, 2x) 96x 2 6x gives additional point (/6, /8). (x, y) f(x, y) (, ) (/6, /8) /256 (/6, /6) /8 (2, ) 4 (2, / 3) 4 8/(3 3) (2, 4) 244 f has a minimum on T at (/6, /8) with f(/6, /8) /255 f has a maximum on T at (2, 4) with f(2, 4) Use the econd Derivative Test to classify the stationary points of the function f(x, y) x 3 + 3xy 2 3x +. f(x) (3x 2 + 3y 2 3, 6xy) (, ) {(x, y) : x 2 + y 2 and 6xy }. o either x y ± or y x ±. Points are (±, ) and (, ±). f xx 6x, f yy 6x, f xy 6. f xx f yy (f xy ) 2 36(x 2 y 2 ). (, ±): f xx f yy (f xy ) 2 36 <. Therefore saddle points. (, ): f xx f yy (f xy ) 2 36 > with f xx >. Therefore a local minimum point. (, ): f xx f yy (f xy ) 2 36 > with f xx <. Therefore a local maximum point. 5. Use Lagrange Multipliers to locate the point(s) on the curve x 2 y 6 that is (are) closest to the origin. Optimization function: D x 2 +y 2. onstraint: x 2 y 6. et D λ so that: 2x λ2xy 2y λx 2 Knowing that on x 2 y 6, x we have λy so that λ /y. ubstitution into the secod equation gives 2y 2 x 2. Now substitute 2y 2 for x 2 back into x 2 y 6 giving 2y 3 6 y 2 and x ±2 2. The points on x 2 y 6 closest to the origin are (2 2, 2) and ( 2 2, 2). Their actual distance is Find the dimensions and cost of the least expensive box with volume 54 ft 3, where the bottom of the box costs $5/ft 2, the top costs $3/ft 2, and the sides cost $2/ft 2. 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 2x xyz 54 xyz Therefore: x 3, y 3, z 6, and the cost is $28.

3 7. Let D {(x, y, z) R 3 : x, y, z >, x + y + z 9} and consider the integral I(D) D xyz dx dy dz. Use the change of variables u x, v y, w z, where (x, y, z) D to compute I(D). Note that (u, v, w) ( x, y, z) (x, y, z) (u 2, v 2, w 2 ). o that Jacobian det 2u 2v 2w 8uvw Now note that x + y + z 9 with x, y, z >, u 2 + v 2 + w 2 9 which is the first octant of a sphere of radius 3. o: I(D) D dx dy dz xyz B ( ( ) 4 and I(D) 8 8) 3 π33 36π. uvw B (8uvw) du dv dw 8 du dv dw 8. Let B {(x, y, z) R 3 : x, y, z, x 2 + y 2 + z 2 9}. Use spherical coordinates to compute I(B) (x 2 + y 2 + z 2 ) 5 dx dy dz. B B is the first octant of a sphere of radius 3 so ρ 3, θ π/2 and φ π/2. I(B) 9. Evaluate the iterated integral π/2 π/2 3 I ρ 2 sin φ dρ dφ dθ 33 3 x e x/y2 dy dx. ( π 2 ) This says that as x goes from to, y goes from x to. If we change the order of integration, we have: as y goes from to, x will go from to y 2. o that: I x e x/y2 dy dx y 2 e x/y2 dx dy y 2 e x/y2 y 2 dy y 2 (e ) dy 3 (e ).

4 . uppose that the surface 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. ompute the surface area A() of. Explain your answer. The normal to the plane is (2, 3, 4). A unit normal in this direction is n 29 (2, 3, 4), so a surface area vector for 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 (,, ) and you get da 29 (2, 3, 4) (,, )dσ 3 29 dσ. Therefore, dσ 29 3 and A() dσ D da 3 A(D) 29. A piece of a thin metal in the shape of the surface {(x, y, z) R 3 : x 2 + y 2 + z 2, z > } has a constant density of gr/cm 2. Find its mass m and center of mass (x, y, z). With a density of gr/cm 2, and a surface area of a hemisphere, the mass is m 2πgr. By symmetry you have x y. Need to determine z. onsider the unit normal to the sphere which is n (x, y, z) (length of and normal to surface). Then for D the unit circle x 2 +y 2 on the z plane, a unit normal to D is k so that da dx dx n kdσ zdσ. Therefore, dσ da and: z z zdσ da m 2π D 2π π 2 and (x, y, z) (,, /2). 2. 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. (a) F(x) (x 2 + y 2 ) 2 (y2 x 2, 2xy), x P y2 x 2 2xy and Q (x 2 + y 2 ) 2 (x 2 + y 2 ) 2 P y (x2 + y 2 ) 2 (2y) (y 2 x 2 )2(x 2 + y 2 )(2y) 2y(y2 3x 2 ) (x 2 + y 2 ) 4 (x 2 + y 2 ) 3 Q x (x2 + y 2 ) 2 ( 2y) + 2xy2(x 2 + y 2 )(2x) 2y(y2 3x 2 ) (x 2 + y 2 ) 4 (x 2 + y 2 ) 3 Therefore F is conservative.

5 (b) F(x) (yz, xz, xy)e xyz. P yze xyz, Q xze xyz, R xye xyz. You can just see that if you differentiate f e xyz with respect to x you get f x P, likewise f y Q and f z R. Without checking the Gradient Test, try to determine f by taking f(x, y, z) yze xyz dx e xyz + h(y, z) Now differentiate with respect to y and get f y xze xyz + h (y, z) ince Q xze xyz, h(y, z) has no term of y so h(y, z) h(z). Now differentiate f(x, y, z) e xyz + h(z) with respect to z and f z xye xyz + h (z). ince R xye xyz, h(z) has no term of z. Therefore f(x, y, z) e xyz is a potential for F. (c) F(x) (x 2 e y+z, y 2 e x+z, z 2 e x+y ). Note that P y x2 e y+z and Q x y2 e x+z and these are not equal so you need not go any further to say that F is not conservative (there does not exist a potential f such that f F. (d) F(x) x2 + 2y (x, ), x2 + 2y >. P x x2 + 2y and Q x2 + 2y x Q and 3/2 P y (x 2 + 2y) f(x, y) to y and f y x x. F is conservative. (x 2 + 2y) 3/2 x x x2 + 2y dx 2 + 2y + h(y). Now differentiate with respect x2 + 2y + h (y). ince Q f(x, y) x 2 + 2y is a potential for f F., h(y) has no y term and x2 + 2y (e) F(x) (x + y + z)(,, ). P Q R x + y + z. P y Q x P z R x Q z R y. F is conservative. f(x, y, z) 2 x2 + xy + xz + h(y, z) so that f y x + h (y, z) which implies h (y, z) y + z which in turn says that h(y, z) 2 y2 + yz + k(z). o far we have f(x, y, z) 2 x2 +xy+xz+ 2 y2 +yz+k(z). Take f z x+y+k (z) and we see that k (z) z so k(z) 2 z2. Finally: f(x, y, z) ( x 2 + y 2 + z 2) + xy + xz + yz is a potential for F 2 3. Evaluate the following line integrals I():

6 (a) I() xy 2 ds, where is the line segment from ( 2, 3) to (, 4). A parametrization for the line segment is (x, y) ( 2 + 3t, 3 7t), t. ds r (t) dt 58dt so that: I() (b) I() ( 2 + 3t)(3 7t) 2 58 dt 58 ( 8 + t 224t t 3 ) dt {(2 + xz)xye xz dx, +x 2 e xz dy + x 3 ye xz dz}, where is given by x(t) (t 2 +, te t2, cos πt), t. Note the difficulty of F: see if there is a potential for F: P y (2+xz)xe xz Q x, P z x 2 ye xz )+(2+xz)x 2 ye xz R x, and Q z x 3 e xz R y. F is conservative and a potential for F is f(x, y, z) x 2 ye xz (Easy way to determine this is start with Q and determine x 2 e xz dy x 2 ye xz + h(x, z) Then f dx f(2, e, ) f(,, ) 4e (c) I() F N ds, where F(x) (x 2 y 2, 2xy) and is described by 4. Evaluate x(t) 2(cos (2t), sin (2t)), t π/2. F Nds (F, F 2 ) (dy, dx) so: I() 6 π/2 π/2 (x 2 y 2 )dy 2xydx π/2 (4 cos 2 2t 4 sin 2 2t)(4 cos 2t) 8 cos 2t sin 2t( 4 sin 2t) dt (6 cos 3 2t + 6 sin 2 2t cos 2t) dt (cos 2t)(cos 2 2t + sin 2 2t) dt 6 ( ) π/2 6 2 sin 2t 8( ) 8 I() F dx π/2 cos 2t dt where F(x) (x 2 + y, z, z 4 + 2x), and is described by x(t) (cos t, sin t, cos t + sin t), t 2π. If you do this calculation straightforwardly it is quite a bit of work but notice you

7 have a nice closed region so much will evaluate to zero. I() 2π ( ( cos 2 t + sin t)( sin t) + (cos t + sin t)(cos t) +(cos t + sin t) 4 ( sin t + cos t) + 2 cos t( sin t) + 2 cos 2 t ) dt Note: 2π cos 2 t sin t dt 2π ( sin2 t + cos 2 t) dt 2π sin t cos t dt 2π (cos t + sin t)4 ( sin t + cos t) dt. The only evaluation left is 2π cos 2 t dt 2π ( + cos 2t) dt 2π. This is partly because F (x 2,, z 4 ) + (y, z, 2x) and f 3 x3 + 5 x5 is a potential for (x 2,, z 4 ) and partly because z x + y along. 5. Evaluate I() ( F) N dσ where is the portion of the surface 36z 8 4x 2 9y 2 for which z 2, and the normal is outward, and where F (F, F 2, F 3 ) is given by F (x, y, z) 3x 2 yz + yz 2, F 2 (x, y, z) x 3 z + xz 3 + z 2, F 3 (x, y, z) xye z2. Use tokes Theorem: ( F) N dσ F dx where is the boundary: for z 2, 4x 2 + 9y 2 36 which is an ellipse with major axis of 3 in the x direction, minor axis of 2 in the y direction and on the plane z 2. F dx (3x 2 yz + yz 2 )dx + (x 3 z + xz 3 + z 2 )dy + (xye z2 )dz ince z 2 and thus dz this reduces to: F dx (6x 2 y + 4y)dx + (2x 3 + 8x + 4)dy ince the ellipse is a simple closed curve use Green s Theorem: ( (6x 2 y+4y)dx+(2x 3 +8x+4)dy E x (2x3 + 8x + 4) ) y (6x2 y + 4y) dx dy E (6x x 2 4) dx dy E 4 dx dy 4(A(E)) 4(6π) 24π. 6. Let be the portion of the sphere x 2 + y 2 + z 2 9 for which z, and the normal is upward, and let F (F, F 2, F 3 ) be given by F (x, y, z) 3x + y 2 e z2 + z 3, F 2 (x, y, z) 6y + 2x 2 z 2, F 3 (x, y, z) x 2 + z.

8 Use the Divergence Theorem to compute the flux Φ (F) (F N)dσ. areful on this one. The Divergence Theorem applies to the entire surface of the region bounded and this surface does not include the cap on the bottom of the hemisphere. o 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 H (F N) dσ H B B T FdV ( ( ) 4 dv 2) 3 π(3)3 8π On the cap, T, the outward normal is k, z and x 2 + y 2 9 so that: (F N) dσ F(x, y, ) ( k)dσ T T 2π 3 x 2 dx dy r 2 cos 2 θr dr dθ T 2π 8 4 Therefore Φ (F) 8π π. 8 4 cos2 θ dθ 2π ( cos 2θ ) dθ 8π uppose is any portion of the plane ax + by + cz d, c >, whose area is A and that is positively oriented. Let F (α, β, γ) be a constant vector field. ompute the flux Φ (F) (F N) dσ Recall, a normal to the plane ax + by + cz d is (a, b, c). A unit normal in this direction (and we want this direction since c > is N (a, b, c). a2 + b 2 + c2 Therefore: F Ndσ (a, b, c) (α, β, γ)dσ a2 + b 2 + c2

9 aα + bβ + cγ a2 + b 2 + c 2 dσ aα + bβ + cγ aα + bβ + cγ A() a2 + b 2 + c2 a2 + b 2 + c A 2 8. Let F (P, Q), where P (x, y) 8e x2 + 3xy 3x 2 y and Q(x, y) 8e y2 + 4xy + 3x 2 y, and let be the circle x 2 + y 2 2 described in the positive direction. Use Green s Theorem to evaluate the circulation Ψ (F) F dx F dx 2 2π 2π 2 2 ( + 4xy + 3x 2 y) ) + 3xy 3x 2 y) dx dy x (8ey2 y (8ex2 (4y + 6xy 3x + 3x 2 ) dx dy ( 4r sin θ + 6r 2 cos θ sin θ 3r cos θ + 3r 2 cos 2 θ ) r dr dθ 3r 3 cos 2 θ dθ 2 2π 3πr 3 dr 3 4 π( 2) 4 3π. 3r 3 ( cos 2θ ) dθ dr