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MATH 2720 Winter 2012 Assignment 4 Questions 1, 2, 6 and 8 were marked. √ 1. Show that the curve c(t) = (t2 , 2t−1, t), t > 0, is a flow line of the vector field F (x, y, z) = (y+1, 2, 1/2z). Solution. The curve is a flow line if c0 (t) = F (c(t)). We have c0 (t) = (2t, 2, t−1/2 /2) and √ F (c(t)) = F (t2 , 2t − 1, t) 1 = (2t − 1) + 1, 2, √ 2( t) 1 = 2t, 2, √ 2 t = c0 (t), so c is a flow line of F . 2. Let c(t) be a flow line of a gradient field F = −∇V . Prove that V (c(t)) is a decreasing function of t. Solution. We obviously have to consider a function V : Rn → R, since c : R → Rn and we do not know what it means for a function to be decreasing except when it is scalar valued. So V (c(t)) is decreasing iff d V (c(t)) < 0. dt From the Chain Rule, d V (c(t)) = DV (c(t))Dc(t) dt = ∇V (c(t)) • c0 (t), because of the dimensionality of the functions involved. By assumption, ∇V = −F , so d V (c(t)) = −F (c(t)) • c0 (t). dt Since c is a flow line of F , there holds that c0 (t) = F (c(t)), so the previous equation takes the form d V (c(t)) = −c0 (t) • c0 (t). dt Recall that for any vector u, u • u = kuk2 . Thus, d V (c(t)) = −kc0 (t)k. dt For any u, kuk ≥ 0 with kuk = 0 only when u = 0. It follows that V is decreasing except where c is not smooth. Remarks. 1. This property is true if c is piecewise-smooth. Indeed, in more advanced calculus/analysis, it is shown that the definition of a decreasing function can be made more general; f : R → R is decreasing if f 0 (x) < 0 except maybe at a denumerable set of points. 2. This property plays a fundamental role in the study of some differential equations. A system of differential equations such that x0 (t) = F (x(t)) with F : Rn → Rn a vector field satisfying the property that ∃V such that ∇V = −F is called a gradient system. Gradient systems of differential equations enjoy a number of interesting properties, two of the most striking of which being that nontrivial solution curves x(t) are orthogonal to the level curves of V and that such systems do not admit nontrivial periodic solutions. 3. Let f : R3 → R. Prove that curl ∇f = 0. Solution. 4. Let F (x, y, z) = 2xyez~i + ez x2~j + (x2 yez + z 2 )~k. Here is a method that will allow us to find a function f : R3 → R such that F = ∇f . 4.a. Write that F = ∇f , that is, ∂f ∂f ∂f 2 z 2 ~ z 2~ z~ 2xye i + e x j + (x ye + z )k = , , . (1) ∂x ∂y ∂z Choose one term, say, ∂f /∂x. Compute the antiderivative of ∂f /∂x = 2xyez . You should obtain something of the form f (x, y, z) = φ(x, y, z) + K, with φ to be determined. 4.b. Explain why, in the expression you just found, K should in fact be a function of y and z, i.e., the antiderivative that you just computed should be f (x, y, z) = φ(x, y, z) + K(y, z), with K(y, z) unknown. [Hint: differentiate the expression with respect to x.] 4.c. Now differentiate the function f you found in 4.a with respect to y. Using (1) and the same type of reasoning as in 4.a and 4.b, show that this means that you can find an antiderivative of ∂K(y, z)/∂y of the form K(y, z) = ψ(x, y)+L(z), with L(z) an unknown function of z only. Substitute this K into the expression f (x, y, z) = φ(x, y, z)+K(y, z) found previously, giving you a function f (x, y, z) = φ(x, y, z)+K(y, z)+L(z). 4.d. Now differentiate the function f you found in 4.c with respect to z and obtain an expression for L0 (z). Taking the antiderivative of this and substituting into the function found in 4.c, you obtain finally the f such that F = ∇f . Check that this is indeed the case. Solution. 5. A sphere of mass m, radius a and uniform density has potential u and gravitational force F at a distance r from the centre (0, 0, 0) given by u= 3m mr2 − , 2a 2a3 m u= , r m F = − 3r (r ≤ a) a m F = − 3r (r > a), r where r = krk and r = x~i + y~j + z~k. 5.a. Check that F = ∇u on the inside and outside of the sphere. 5.b. Check that u satisfies Poisson’s equation uxx + uyy + uzz = K, with K a constant, inside the sphere. 5.c. Check that u satisfies Laplace’s equation uxx + uyy + uzz = 0 outside the sphere. Solution. 6. Evaluate the double integral ZZ |y| cos R πx dA 4 over the region R = [0, 2] × [−1, 0]. Solution. From Fubini’s theorem, ZZ πx dA = |y| cos 4 R 2Z 0 Z |y| cos 0 −1 πx dy dx. 4 Since the integrand takes the form f (x, y) = g(x)h(y) on a rectangular domain, we have ZZ πx |y| cos dA = 4 R 2 Z 0 Z πx cos dx 4 2 Z 0 |y| dy −1 Z 0 πx dx (−y) dy 4 0 −1 0 4 πx 2 −y 2 = sin π 4 0 2 −1 4 −1 = π 2 2 =− π = cos Remark. To check your answer, you could have used the following Maple command: int(int(abs(y)*cos(Pi*x/4),y=-1..0),x=0..2) 7. Let f be continuous on R = [a, b] × [c, d]. For x ∈ [a, b] and y ∈ [c, d], define Z xZ y F (x, y) = f (u, v) dv du. a c Using the fundamental theorem of calculus and Fubini’s theorem, show that ∂2F ∂2F = = f (x, y). ∂x∂y ∂y∂x Solution. 8. Let D be the region bounded by the positive x and y axes and the line 3x + 4y = 10. Compute ZZ x2 + y 2 dA. D Solution. The region is as shown in Figure 1. Let us describe it as a y-simple region. The equation of the Figure 1: The region D in Exercise 8. line as a function of x is y = 5/2 − 3x/4. It intercepts the x-axis at x = 10/3 and thus 5 3 10 2 D = (x, y) ∈ R ; 0 ≤ x ≤ , 0 ≤ y ≤ − x . 3 2 4 As a consequence, ZZ 2 2 Z x + y dA = D 10 3 0 Z 5 − 34 x 2 x2 + y 2 dy dx 0 5−3x y3 2 4 2 = x y+ dx 3 0 0 Z 10 3 3 1 5 3 3 2 5 = x − x + − x dx 2 4 3 2 4 0 " # 10 3 4 1 5 3 4 3 5 3 x − x + − x = 6 16 9 2 4 0 4 3 10 4 1 5 3 10 5 10 3 − + − . = 6 3 16 3 9 2 4 3 Z 10 3 Remarks. 1. There was no need to evaluate this result any further. But just in case you want to check, the result was 15625/1296. 2. In Maple, int(int(x^2+y^2,y =0..5/2-(3/4)*x),x = 0..10/3). If you want the antiderivative obtained after integrating with respect to x, do not specify the bounds for x: int(int(x^2+y^2, y=0..5/2-(3/4)*x),x)