Math 67. Rumbos Fall Solutions to Review Problems for Final Exam. (a) Use the triangle inequality to derive the inequality

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1 Math 67. umbos Fall 8 Solutions to eview Problems for Final Exam. In this problem, u and v denote vectors in n. (a) Use the triangle inequality to derive the inequality Solution: Write v u v u for all u, v n. () u (u v) + v and applying the triangle inequality to obtain from which we get that u u v + v, u v v u. () Interchanging the roles for u and v in () we obtain from which we get Combining () and (3) yields which is (). v u u v. v u v u. (3) v u v u v u, (b) Use the inequality derived in the previous part to show that the function f : n given by f(v) v, for all v n, is continuous in n. Solution: Fix u n and apply the inequality in () to any v n to obtain that v u v u, or f(v) f(u) v u. (4) Next, apply the Squeeze Lemma to obtain from (4) that lim f(v) f(u), v u which shows that f is continuous at u for any u n.

2 Math 67. umbos Fall 8 (c) Prove that the function g : n given by g(v) sin( v ), for allv n, is continuous. Solution: Observe that g sin f, where f : n is as defined in part (b). Thus, g is the composition of two continuous functions, and is, therefore, continuous.. Define the scalar field f : n by f(v) v for all v n. (a) Show that f is differentiable in n and compute the linear map Df(u): n for all u n. What is the gradient of f at u for all u n? Solution: Let u n and compute f(u + w) u + w (u + w) (u + w) u u + u w + w u + w w u + u w + w, for w n, where we have used the symmetry of the dot product and the fact that v v v for all v n. We therefore have that f(u + w) f(u) + u w + w, for all u n and w n. (5) Writing and Df(u)w u w, for all u n and w n, (6) E u (w) w, for all u n and w n, (7) we see that (5) can be rewritten as f(u + w) f(u) + Df(u)w + E u (w), for all u n and w n, (8) where, according to (6), Df(u): n n defines a linear transformation, and, by virtue of (7), E u (w) w w, for w,

3 Math 67. umbos Fall 8 3 from which we get that E u (w) lim w w. Consequently, in view of (8), we conclude that f is differentiable at every u n, derivative at u given by (6). Since Df(u)w f(u) w, for all u and w in n, by comparing with (6), we see that f(u) u, for all u n. Alternate Solution: Alternatively, for u (x, x,..., x n ) n, we have that f(u) x + x + + x n; so that f x j (u) x j, for j,,..., n. Thus, all the partial derivatives of f are continuous on n ; that is, f is a C function. Consequently, f is differentiable on n. Furthermore, Df(u)w ( x x x n ) w, for all w n, which can be written as Df(u)w u w, for all w n. (9) It then follows that f(u) u for all u n. (b) Let v denote a unit vector in n. For a fixed vector u in n, define g : by g(t) u t v, for all t. Show that g is differentiable and compute g (t) for all t. Solution: Observe that g f σ, where σ : n is given by Thus, σ is a differentiable path with σ(t) u t v, for all t. () σ (t) v, for all t. () Thus, by the result from part (a), g is the composition of two differentiable functions. Consequently, by the Chain ule, g is differentiable with g (t) Df(σ(t))σ (t), for all t. ()

4 Math 67. umbos Fall 8 4 Thus, using (9) and (), we obtain from () that or g (t) σ(t) ( v), for all t, g (t) σ(t) v, for all t. (3) Thus, using (), we obtain from (3) that which leads to since v is a unit vector in n. g (t) (u t v) v, for all t, g (t) t u v, for all t, (4) (c) Let v be as in the previous part. For any u n, give the point on the line spanned by v which is the closest to u. Justify your answer. Solution: It follows from (4) that g (t) > for all t ; so that g has a global minimum when g (t). We then obtain from (4) that g(t) is the smallest possible when t u v. Consequently, the point on the line spanned by v which is the closest to u is (u v) v, or the orthogonal projection of u onto the direction of v. 3. Let I denote an open interval which contains the real number a. Assume that σ : I n is a C parametrization of a curve C in n. Define s: I as follows: s(t) arlength along the curve C from σ(a) to σ(t), (5) for all t I. (a) Give a formula, in terms of an integral, for computing s(t) for all t I. Answer: s(t) t a σ (τ) dτ, for all t I. (6)

5 Math 67. umbos Fall 8 5 (b) Prove that s is differentiable on I and compute s (t) for all t I. Deduce that s is strictly increasing with increasing t. Solution: It follows from the assumption that σ is C, the Fundamental Theorem of Calculus, and (7), that s is differentiable and s (t) σ (t), for all t I. (7) Since we are also assuming that σ is a parametrization of a C curve, C, it follows that σ (t) for all t I. Consequently, we obtain from (7) that s (t) >, for all t I, which shows that s(t) is strictly increasing with increasing t. (c) Let l arclength of C, and suppose that γ : [, l] n is a a parametrization of C with the arclength parameter s defined in (5); so that, C {γ(s) s l}. Use the fact that σ(t) γ(s(t)), for all t [a, b], to show γ (s) is a unit vector that is tangent to the curve C at the point γ(s). Solution: Note that σ γ s is a composition of two differentiable functions, by the result of part (b). Consequently, by the Chain ule, σ (t) ds dt γ (s), for t (a, b). Thus, using (7), σ (t) σ (t) γ (s), for t (a, b). So, using the fact that σ (t) > for all t (a, b), γ (s) σ (t) σ (t), for t (a, b), which shows that γ (s) is a unit vector that is tangent to the curve C at the point γ(s).

6 Math 67. umbos Fall Let I denote an open interval of real numbers and f : I be a differentiable function. Let a, b I be such that a < b, and define C to the section of the graph of y f(x) from the point (a, f(a)) to the point (b, f(b)); that is, C {(x, y) y f(x) and a x b} (a) By providing an appropriate parametrization of C, compute the arclenth of C, l(c). Solution: Parametrize C by σ : [a, b] given by σ(t) (t, f(t)), for a t b. Then, so that Therefore, σ (t) (, f (t)), for a t b; σ (t) + [f (t)], for a t b. l(c) b a + [f (t)] dt. (8) (b) Let f(x) 5 x 3/, for x. Compute the exact arcength of y f(x) over the interval [, ]. Solution: We use the formula in (8) with f (t) 3t /, for t >. Thus, l(c) + 9t dt [ ] ( + 9t)3/ 7 ( ) 7 74.

7 Math 67. umbos Fall Let Φ: denote the map from the uv plane to the xy plane given by ( ) ( ) ( ) u u u Φ v v for all, v and let T be the oriented triangle [(, ), (, ), (, )] in the uv plane. (a) Show that Φ is ( differentiable ) and give a formula for its derivative, DΦ(u, v), u at every point in v. Solution: Write ( ) ( ) ( ) u f(u, v) u Φ for all, v g(u, v) v ( ) u where f(u, v) u and g(u, v) v for all. Observe that the v partial derivatives of f and g exist and are given by f f (u, v), u (u, v) v g g (u, v), (u, v) v. u v Note that the partial derivatives of f and g are continuous. Therefore, Φ is a C map. Hence, Φ is differentiable on and its derivative map at (u, v), for any (u, v), is given by multiplication by the Jacobian matrix ( ) DΦ(u, v) ; v that is, ( ) ( ) ( ) ( ) h h h DΦ(u, v) k v k vk ( ) h for all k. (b) Give the image,, of the triangle T under the map Φ, and sketch it in the xy plane. Solution: The image of T under Φ is the set Φ(T ) {(x, y) x u, y v, for some (u, v) } {(x, y) x, y x /4}. A sketch of Φ(T ) is shown in Figure.

8 Math 67. umbos Fall 8 8 y y x /4 x (c) Evaluate the integral Figure : Sketch of egion Φ(T ) dxdy, obtained in part (b). Solution: Compute by means of iterated integrals dxdy where is the region in the xy plane x /4 dy dx [ x 3 x 4 dx ] (d) Evaluate the integral T 3. det[dφ(u, v)] dudv, where det[dφ(u, v)] denotes the determinant of the Jacobian matrix of Φ obtained in part (a). Compare the result obtained here with that obtained in part (c). Solution: Compute det[dφ(u, v)] to get so that T det[dφ(u, v)] 4v. det[dφ(u, v)] dudv T 4 v dudv, where the region T, in the uv plane is sketched in Figure. Observe that,

9 Math 67. umbos Fall 8 9 v v u T u Figure : Sketch of egion T in that region, v, so that det[dφ(u, v)] dudv T Compute by means of iterated integrals det[dφ(u, v)] dudv T 4v dudv, T u u du 4v dvdu 3, which is the same result as that obtained in part (c). 6. Consider the iterated integral x x y dydx. (a) Identify the region of integration,, for this integral and sketch it. Solution: The region {(x, y) x y, x } is sketched in Figure 3. (b) Change the order of integration in the iterated integral and evaluate the double integral x y dxdy.

10 Math 67. umbos Fall 8 y y x x Figure 3: Sketch of egion Solution: Compute x y dxdy y x y dxdy [ x y ] y y y dy. dy Next, make the change of variables u y to obtain that x y dxdy u du u du 7. What is the region over which you integrate when evaluating the iterated integral x x dy dx? x + y ewrite this as an iterated integral first with respect to x, then with respect to y. Evaluate this integral. Which order of integration is easier?

11 Math 67. umbos Fall 8 y y x x Figure 4: Sketch of egion Solution: The region {(x, y) y x, x } is sketched in Figure 4. Interchanging the order of integration, we obtain that x x + y dxdy x dxdy. (9) y x + y The iterated integral in (9) is easier to evaluate; in fact, x x + y dxdy x x + y dxdy y [ x + y ] y dy [ 4 + y y] We therefore get that x x + y dxdy 4 + y dy Evaluating the second integral on the right hand side of () yields dy. y dy. () y dy 3. ()

12 Math 67. umbos Fall 8 The first integral on the right hand side of () leads to [ y 4 + y dy 4 + y + 4 y ln y which evaluates to 4 + y dy ( 5 + ln + ) 8 +. () 5 Substituting () and () into () we obtain x x + y dxdy 5 + ln ], ( + ) Let f : denote a twice differentiable real valued function and define u(x, t) f(x ct) for all (x, t), where c is a real constant. Verify that u t u c x. Solution: Apply the Chain ule to obtain u x f (x ct) x (x ct) f (x ct). Similarly, u x f (x ct), (3) u t f (x ct) t (x ct) cf (x ct), and u t c f (x ct). (4) Combining (3) and (4) we see that which was to be verified. u t c f (x ct) c u x,

13 Math 67. umbos Fall Let f : denote a twice differentiable real valued function and define u(x, y) f(r) where r x + y for all (x, y). (a) Define the vector field F (x, y) u(x, y). Express F in terms of f and r. Solution: Compute where, by the Chain ule, F (x, y) u(x, y) u x î + u ĵ, (5) u x f (r) r x (6) and In order to compute r x u f (r) r. (7) r and x, write r x + y, (8) and differentiate with respect to x on both sides of (8) to obtain from which we get Similarly, Substituting (9) into (6) yields r r x x, r x x, for (x, y) (, ). (9) r r y, for (x, y) (, ). (3) r u x f (r) r Similarly, substituting (3) into (7) yields u f (r) r x. (3) y. (3)

14 Math 67. umbos Fall 8 4 Next, substitute (3) and (3) into (5) to obtain F (x, y) f (r) (x î + y ĵ), (33) r (b) ecall that the divergence of a vector field F P î +Q ĵ is the scalar field given by divf P x + Q. Express the divergence of the gradient of u, in terms of f, f and r. The expression div( u) is called the Laplacian of u, and is denoted by u or u. Solution: From (33) we obtain that P (x, y) f (r) x and Q(x, y) f (r) y, r r so that, applying the Product ule, Chain ule and Quotient ule, P x f (r) + x d [ ] f (r) r r dr r x f (r) r + x rf (r) f (r) r where we have also used (9). Simplifying the expression in (34) yields P x f (r) r Similar calculations lead to Q f (r) r + x f (r) r + y f (r) r x r, Adding the results in (35) and (36), we then obtain that divf P x + Q f (r) r + r f (r) r (34) x f (r) r 3. (35) y f (r) r 3. (36) r f (r) r 3, where we have used (8). Simplifying the expression in (37), we get that divf f (r) + f (r). r (37)

15 Math 67. umbos Fall 8 5. Let f(x, y) 4x 7y for all (x, y), and g(x, y) x + y. (a) Sketch the graph of the set C g () {(x, y) g(x, y) }. Solution: The curve C is the graph of the equation x / + y, which is sketched in Figure 5. y / x Figure 5: Sketch of ellipse (b) Show that at the points where f has an extremum on C, the gradient of f is parallel to the gradient of g. Solution: Let σ : [, π] denote the C parametrization of C given by ( ) σ(t) cos t, sin t, for all t [, π]. We then have that Differentiating on both sides of (38) yields that g(σ(t)), for all t. (38) g(σ(t)) σ (t), for all t,

16 Math 67. umbos Fall 8 6 where we have applied the Chain ule, which shows that g(x, y) is perpendicular to the tangent vector to C at (x, y). Next, suppose that f(σ(t)) has a critical point at t o. Then, the derivative of f(σ(t)) at t o is ; that is, f(σ(t o )) σ (t ), where we have applied the Chain ule. It then follows that f(x o, y o ) is perpendicular to the tangent vector to C at a critical point (x o, y o ). Hence, f(x o, y o ) must be parallel to g(x o, y o ). (c) Find the largest and the smallest value of f on C. Solution: By the result of part (b), at a critical point, (x, y), of f on C, it must be the case that for some non zero real number λ, where and g(x, y) λ f(x, y), (39) f(x, y) 4 î 7 ĵ, (4) g(x, y) 4x î + y ĵ. (4) Substituting (4) and (4) into (39) yields the pair of equations and x λ (4) y 7λ. (43) Substituting the expressions for x and y in (4) and (43), respectively, into the equation of the ellipse x + y, yields that from which we get that 57 4 λ, λ ± (44)

17 Math 67. umbos Fall 8 7 The values for λ in (44), together with (4) and (43), yield the critical points ( ) ( , 7 and ) , (45) 57 Evaluating the function f at each of the critical points in (45) we obtain that ( ) f 57, 7 ( 57 and f ) , Consequently, the largest value of f on C is 57 and the smallest value is 57.. Let ω be the differential form in 3 given by ω x dx + y dy + z dz. (a) Compute the differential, dω, of ω. Solution: Compute dω dx dx + dy dy + dz dz. (b) If possible, find a differential form, f, such that ω df. Solution: Let f(x, y, z) x + y + z. Then, df x dx + y dy + z dz ω. (c) Let C be parametrized by a C connecting P o (,, ) to P (,, ). Compute the line integral ω. C Solution: Apply the Fundamental Theorem of Calculus, ω df f(p ) f(p o ), C C where f is as given in part (b). Consequently, ω f(,, ) f(,, ) 3 3. C

18 Math 67. umbos Fall 8 8 (d) Let C denote any simple closed curve in 3. Evaluate the line integral ω. C Solution: ω, since C is closed and ω is exact. C. Let f denote a differential form in 3 and ω a a differential form in 3. (a) Verify that d(df). Solution: Compute ( f d(df) d x so that d dx + f ( ) f dx + d x ) f dy + z dz ( ) f dy + d ( ) f dz z f x dx dx + f x dy dx + f z x + f x dx dy + f dy dy + f z d(df) dz dx + f x z dx dz + f z dy dz + f z ( ) f z f dy dz ( z ) f + z x f dz dx ( x z ) f + x f dx dy. x dz dy dz dz, (46) It follows from (46) and the fact that the mixed second partial derivatives of a C function are equal that d(df). (b) Verify that d(dω). Solution: Since ω is a differential form, we may write ω f dx + f dy + f 3 dz, where f, f and f 3 are differential forms. Thus, since the operator d is linear d(dω) d(df ) dx + d(df ) dy + d(df 3 ) dz,

19 Math 67. umbos Fall 8 9 which is since d(df j ), for j,, 3, by the result from part (a). Hence, d(dω), which was to be shown. 3. Let f and g denote differential forms in 3, and ω and η a differential forms in 3. Derive the following identities (a) d(fg) g df + f dg. Solution: Compute d(fg) (fg) x (fg) dx + ( f g x + g f ) dx x ( + f g + g f ) dy dy + (fg) z ( + f g z + g f ) dz, z dz (47) where we have used the Product ule. earranging terms in (47) we obtain that ( ) f f f d(f g) g dx + dy + x z dz f g df + f dg. ( g g dx + x ) g dy + z dz (b) d(fω) df ω + f dω. Solution: Write ω g dx + g dy + g 3 dz, where g, g and g 3 are differential forms. We then have that fω fg dx + fg dy + fg 3 dz, so that d(fω) d(fg ) dx + d(fg ) dy + d(fg 3 ) dz. (48)

20 Math 67. umbos Fall 8 Applying the result from part (a), we obtain from (48) that d(fω) (fdg + g df) dx + (fdg + g df) dy +(fdg 3 + g 3 df) dz f dg dx + g df dx +f dg dy + g df dy (49) +f dg 3 dz + g 3 df dz f(dg dx + dg dy + dg 3 dz) +df g dx + df g dy + df g 3 dz, where we have used the bi linearity of the wedge product. linearity again, we obtain from (49) that Using bi which was to be shown. d(fω) f dω + df (g dx + g dy + g 3 dz) f dω + df ω, (c) d(ω η) dω η ω dη. Solution: Write ω f dx + f dy + f 3 dz, where f, f and f 3 are differential forms, and compute so that ω η (f dx + f dy + f 3 dz) η f dx η + f dy η + f 3 dz η, d(ω η) d(f dx η) + d(f dy η) + d(f 3 dz η). (5) Using the result from part (b), we obtain from (5) that d(ω η) f d(dx η) + df dx η +f d(dy η) + df dy η (5) +f 3 d(dz η) + df 3 dz η,

21 Math 67. umbos Fall 8 where we have used the associativity of the wedge product. Next, write η g dx + g dy + g 3 dz, where g, g and g 3 are differential forms. Then, dx η g dx dy + g 3 dx dz, so that d(dx η) dg dx dy + dg 3 dx dz g z dz dx dy + g 3 ( g z g ) 3 dx dy dz, dy dx dz (5) where we have used the anti commutativity of the wedge product. On the other hand, note that dη dg dx + dg dy + dg 3 dz ( g x dx + g dy + g ) z dz dx ( g + x dx + g dy + g ) z dz dy ( g3 + x dx + g 3 dy + g ) 3 z dz dz,

22 Math 67. umbos Fall 8 so that, using the anti commutativity of the wedge product, dη g dy dx + g z + g x dx dy + g z dz dx + g 3 x dx dz + g 3 g 3 dy dz + g z + g 3 x dx dz + g z dz dy dz dy dy dz dz dx + g x dx dy + g ( g3 g ) z dy dz ( g + z g ) 3 x dz dx dy dx ( g + x g ) dx dy, so that dη ( g3 g ) z dy dz ( g + z g ) 3 x dz dx ( g + x g ) dx dy, (53) Taking the wedge product with dx on the left of (53) yields ( g3 dx dη g ) dx dy dz, (54) z

23 Math 67. umbos Fall 8 3 Comparison of (5) and (54) yields the identity d(dx η) dx dη (55) Similar calculations using (53) yield the additional identities d(dy η) dy dη (56) and d(dz η) dz dη. (57) Next, substitute the identities in (55), (56) and (57) into (5) to obtain d(ω η) f (dx dη) + df dx η f (dy dη) + df dy η f 3 (dz dη) + df 3 dz η, which leads to d(ω η) (f dx + f dy + f 3 dz) dη) +(df dx + df dy + df 3 dz) η, (58) by virtue of the bi linearity of the wedge product. We therefore obtain from (58) that d(ω η) ω dη + dω η, which was to be shown. 4. Let denote the square, {(x, y) x, y }, and denote the boundary of oriented in the counterclockwise sense. Evaluate the line integral (y + x 3 ) dx + x 4 dy.

24 Math 67. umbos Fall 8 4 Solution: Apply the Fundamental Theorem of Calculus to get (y + x 3 ) dx + x 4 dy d[(y + x 3 ) dx + x 4 dy] so that (y + x 3 ) dx + x 4 dy y dy + 3x dx) dx + 4x 3 dx dy y dy dx + 4x 3 dx dy (4x 3 y) dx dy, (4x 3 y) dxdy, (59) since is oriented in the counterclockwise sense. Evaluating the double integral in (59) we obtain that (y + x 3 ) dx + x 4 dy (4x 3 y) dxdy [ x 4 xy ] dy ( y)dy [ y y ].