1. Suppose that the equation F (x, y, z) = 0 implicitly defines each of the three variables x, y, and z as functions of the other two:

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1 Final Solutions. Suppose that the equation F (x, y, z) implicitly defines each of the three variables x, y, and z as functions of the other two: z f(x, y), y g(x, z), x h(y, z). If F is differentiable and F x, F y, and F z are all nonzero, show that [] z x x z. Solution. From a Theorem on implicit differentiation done in class, we have that z x f x F x F z x h y F y F x z g z F z F y From these three equations, we have z x x z. 2. Can we have a differentiable scalar field f : R 2 R satisfying both of the following conditions? (a) The partial derivatives f x (, ) f y (, ), and (b) the directional derivative f (; i + j) 3. Solution. Since f x (, ) f y (, ), f(, ) f x (, ) i + f y (, )) j. But the fact that f is differentiable would imply that which contradicts condition (b). f (; i + j) f(, ) (i + j),

2 3. A cylinder whose equation is y f(x) is tangent to the surface z 2 + 2xz + y at all points common to the two surfaces. Find f(x). Solution. Let F (x, y, z) f(x) y, G(x, y, z) z 2 + 2xz + y, let S denote the set of point common to the two surfaces. Since these two surfaces are tangent to each other at each (x, y, z) S, we have that In other words, f(x, y, z) g(x, y, z), for (x, y, z) S. (f (x),, ) (2z,, 2z + 2x), for (x, y, z) S, where f (x) f x. Upon simplification, we have the ordinary differential equation f (x), for (x, y, z) S. 2z In S, we must have f(x) z 2 + 2xz. Solving for z in terms of x from this quadratic equaation, we obtain z x ± x 2 f(x). Therefore, f is solution to the ordinary differential equation f (x) dy dx x ± x 2 y. 4. Find three positive numbers whose sum is hundred and whose product is a maximum. Solution. Let x, y, z denote the three numbers. We need to maximize f(x, y, z) xyz, subject to the constraint g(x, y, z) x + y + z. Using the method of Lagrange s multipliers, we obtain the following system of equations (for some nonzero λ R) yz λ xz λ xy λ x + y + z. 2

3 From this system of equations, we obtain the equivalent system λx λy λz x + y + z. Since λ, we have that x y z, which upon substitution in x + y + z yields x y z 3. It is easy to see that this is a maximum, as f(98,, ) < f( 3, 3, 3 ). 5. Use Stokes Theorem to evaluate c (y+sin x) dx+(z2 +cos y) dy+x 3 dz, where C is the curve r(t) (sin t, cos t, sin 2t), t [, 2π]. Solution. Let F (x, y, z) Mi + Nj + P k (y + sin x) i + (z 2 + cos y) j + x 3 j. Clearly, the components of F : M y + sin x, N z 2 + cos y, P x 3 have continuous first partials everywhere in R 3. Furthermore, r(t) is a simple closed (r() r(2π)) and smooth curve that lies on the smooth surface z 2xy. Let S denote the part of the surface z 2xy bounded by r(t). Note that the projection of S onto the xy-plane is the unit disk D centered at origin. Also, C is traversed clockwise (when viewed from above) and S is oriented downward. For this S and C, the hypotheses of Stokes Theorem are satisfied. By the Stokes Theorem, we have that F dr ( F ) n dσ. C 2y S By a simple calculation, n i+ 2x j k 4x 2 +4y 2 + 4x 2 +4y 2 + 4x 2 +4y 2 + and F 2z i 3x 2 j k. Therefore, 8xy 2 + 6x 3 ( F ) n dσ 4x 2 + 4y 2 + da 4x 2 + 4y 2 + S π. D D 2π (8xy 2 + 6x 3 ) da (8r 3 cos θ sin 2 θ + 6r 3 cos 3 θ )r drdθ 6. Use the Gauss Divergence Theorem to evaluate S F n dσ, where F (x, y, z) z 2 x i + ( y3 3 + tan z) j + (x2 z + y 2 ) k 3

4 and S is top half of the sphere x 2 + y 2 + z 2. Solution. Let S denote the unit disk in the xy-plane centered at origin. Then S S S is a piecewise smooth closed surface enclosing a region E R 3. Also, the components of F : M z 2 x, N y tan z, P x2 z + y 2 have continuous first partials in an open set containing D S (as z π/2 anywhere in D S ). Therefore, the hypotheses of the Gauss Divergence Theorem are satisfied. By the Gauss Divergence Theorem, we have that F.n dσ F.n dσ + F.n dσ S S S For S, we have n k and z, and so F.n dσ ( y 2 ) da S S 2π π 4 r 2 (sin 2 θ)r drdθ By a simple calculation F x 2 + y 2 + z 2, and we have Finally, S E F dv F.n dσ 2π π/2 2π 5 E E ρ 2 ρ 2 sin φ dρ dφ dθ F dv. F dv F.n dσ 3π S 2. ( ( 7. If F (x, y) y i + x j, show that C F dr 2π along any counterclockwise oriented simple closed curve that encloses the origin. x 2 +y 2 ) x 2 +y 2 ) Solution. Let C be an arbitrary simple closed curve that encloses the origin. Let C be a counterclockwise oriented circle with center the origin and radius a, where a is chosen small enough so that C lies inside C. Let D be the region bounded by C and C. The the positively oriented boundary of D is C ( C ). 4

5 By the Circulation-Curl form of the Green s Theorem, we have that ( N M dx + N dy + M dx + N dy C C D x m ) da [ y 2 x 2 (x 2 + y 2 ) 2 y2 x 2 ] (x 2 + y 2 ) 2 da. D In other words, M dx + N dy M dx + N dy. C C Using the polar coordinates x a cos θ and y a sin θ, we have that C M dx + N dy 2π 2π. ( a sin t)( a sin t) + (a cos t)(a cos t) a 2 cos 2 t + a 2 sin 2 t 8. If F : R 3 R 3 is a vector field and f, g : R 3 R are scalar fields, show that (a) ( F ). ( ) f g f f g (b) g, g. Solution. (a) Let F M i + N j + P k. Then ( P F N ) ( M i + z z P ) ( N j + x x M ) k, and hence ( F ) x ( P N ) + ( M z z P ) + ( N x z x M ) 2 P x 2 N x z + 2 M z 2 P x + 2 N z x 2 M z. dt 5

6 (b) By the definition of the gradient, we have that ( ) ( ) ( ) ( ) f f f f i + j + k g g x g y g z ( ) ( ) ( fx g fg x fy g fg y fz g fg z i + j + g(f xi + f y j + f z k) f(g x i + g y j + g z k) g f f g. 9. Solve the differential equation (y 3 + xy 2 + y) dx + (x 3 + x 2 y + x) dy. Solution. From the equation, we have P y 3y 2 + 2xy +, Q x 3x 2 + 2xy +. ) k Clearly, the equation is not exact. We use an integrating factor of the form h h(u), where u xy. Let F (u) P y Q x yq xp 3(y2 x 2 ) xy(x 2 y 2 ) 3 u. Then the integrating factor h(u) e 3 u du u 3 x 3 y 3. Multiplying the differential equation by the integrating factor, we obtain the following exact differential equation ( x 3 + x 2 y + ) ( x 3 y 2 dx + y 3 + xy 2 + ) x 2 y 3 dy. We denote these new coefficients of dx and dy by P and Q respectively. We choose x y so that the rectangle with vertices x, x, y, y lies entirely in the region R {(x, y) R 2 x and y }, where P, Q, and all of their first partial exist and are continuous. A solution to this exact differential equation is given by x ( f(x, y) x 3 + x 2 y + ) y ( x 3 y 2 dx + y 3 + y 2 + ) y 3 6 dy,

7 that is 2x 2 + xy + 2x 2 y 2 + 2y 2 + y + 2y 2 c, which upon simplification yields the solution y 2 + 2xy + 2x 2 + 2x 2 y + kx 2 y 2.. (Bonus) Show that e x2 dx π 2. Solution. Let I e x2 dx. Then ( ) ( ) I 2 e x2 dx e y2 dy e (x2 +y 2) dx dy. (This is due to the Fubini s Theorem.) Converting this double integral into polar coordinates, we have I 2 π/2 e r2 r dr dθ. We now use the substitution r 2 u, to obtain I 2 2 π 4. π/2 e u du dθ Therefore, I π 2. 7