Solutions to Homework 11

Transcription

1 Solutions to Homework 11 Read the statement of Proposition 5.4 of Chapter 3, Section 5. Write a summary of the proof. Comment on the following details: Does the proof work if g is piecewise C 1? Or did the proof assume that g C 1? The short answer is, yes, it does work, provided you either 1) think carefully about whether the fundamental theorem of calculus allows for finitely many discontinuities, or 2) justify that adding the lengths of the C 1 pieces, that make up the curve yields the same value as summing the corresponding integrals of the length of the velocity vector. The first option is perhaps easier to think about. Suppose that g(t) is a piecewise differentiable function, where g : [a, b] R. Does the fundamental theorem of calculus imply that b a g (t) dt = g(b) g(a)? It does. In fact, the fundamental theorem of caluclus applies if g (t) is integrable. The statement appears on p. 47. Of course, this perhaps avoids the question, since we now need to understand the proof of the fundamental theorem of calculus. You will see this in a course on real analysis. In the meantime, you might critique proofs of the fundamental theorem of calculus which appear in any calculus textbook that you have used previously. The second option sidesteps the question of whether the fundamental theorem of calculus applies to more general functions. The fundamental theorem of calculus is easy to prove if g (t) is assumed to be continuous. (This is the hypothesis as stated in calculus textbooks.) But to apply this simpler version of the thoerem, we need to justify that the the definition of the length of a parametrized curve g(t) is additive. The definition of the arc length l(g) of a continuous curve g : [a, b] R n is that it is equal to the supremum over all partitions P of [a, b] of the value of l(g, P) = k g(t i ) g(t i 1 ), i=1

2 where P = {a = t < < t k = b}. If g is C 1 on [a, b] except at a point c, where a < c < b, then consider the two curves x(t) and y(t) defined by restricting g(t) to [a, c] and [c, b], respectively. The claim is that l(g) = l(x) + l(y). While this is intuitively clear, it is somewhat delicate to prove rigorously. Suppose that P is a partition of [a, b]. Let P be the partition obtained by adding c. Then directly from the definition and the triangle inequality, it follows that l(g, P ) l(g, P). Restricting the partition P induces partitions P x of [a, c] and P y of [c, b]. Directly from the definition, l(p x, x) + l(p y, y) = l(p, g) On the other hand, if Q x and Q y are partitions of [a, c] and [c, b] respectively, then their union defines a partition Q of [a, b] with the property that l(q x, x) + l(q y, y) = l(q, g). From these two observations, it follows that l(g) = l(x) + l(y). Again, the above argument is somewhat delicate. What we have done is the following. Suppose that M i is the supremum of a set A i of real numbers, where i = 1, 2, 3. We have shown that if a 1 A 1, then there exist a 2 A 2 and a 3 A 3 such that a 2 + a 3 a 1. And we have shown that conversely that if a 2 A 2 and a 3 A 3, then there exists a 1 A 1 such that a 1 a 2 + a 3 (in fact equal). The conclusion is that since M 2 and M 3 are upper bounds of A 2 and A 3, respectively, M 2 + M 3 is an upper bound of A 1, and so M 2 + M 3 M 1 (since M 1 is less than or equal to every upper bound of A 1 ); conversely, since M 1 is an upper bound of A 1, M 1 M 2 is an upper bound of A 3, and since M 3 is the least upper bound of A 3, M 1 M 2 M 3, so that M 1 M 2 + M 3. Hence, M 1 = M 2 + M 3. (You will see similar arguments to the above when you study real analysis.) Clearly the above argument works if the parameter domain [a, b] is subdivided into finitely many subintervals, [a = c, c 1 ], [c 1, c 2 ],..., [c k 1, c k = b]. Call the curves obtained by restricting to these subintervals

3 x 1,..., x k. If each of these curves are C 1, then the fundamental theorem of calculus (with the more restrictive hypothesis that the integrand is continuous) can be applied to each. Thus, l(x i ) is equal to the integral of x (t) over [c i 1, c i ]. And the above argument shows that, like integrals, the function l is additive. Why is g(t + h) g(t) s(t + h) s(t)? The definition of s(t) is that it is equal to the length (meaning the value of the function l) of the restriction of g to [a, t]. (This is stated in the proof of 5.4.) So, by the above additive property for l, s(t + h) s(t) is the length of g restricted to [t, t + h]. On the other hand g(t + h) g(t) is the value of l(g, P), where P = {t < t + h} is the trivial partition of [t, t + h]. Since refining P cannot decrease the value of l(g, P), the inequality above holds true. Why is s(b) = l(g)? (Hint: There is a very simple answer. What is the definition of s(t)?) This is the definition of s(b). We ve said enough about Proposition 5.4. But I encourage you to read this and other textbooks critically since it is often the case that details are omitted from proofs. Usually this helps to keep matters less complicated and to still impart the important details of the proof, but occasionally this can lead to erroneous (wrong) statements. So, be skeptical and ask lots of questions. Read the statement of Theorem 6.4 of Chapter 7, Section 6. Use this theorem to solve exercises 3 and 4 in 7.6. These problems were solved in class on 4/23/12. But here is a recap: To determine the volume of the ellipsoid (x/a) 2 + (y/b) 2 + (z/c) 2 1 in R 3, let g : R 3 R 3 be the function u au g v = bv w cw Let be the unit ball, u 2 + v 2 + w 2 1 in R 3, where the coordinates are given by (u, v, w). Think of g as being given by three coordinate functions:

4 x = au, y = bv, and z = cw. If is the unit ball u 2 + v 2 + w 2 1, then g() is the ellipsoid (x/a) 2 + (y/b) 2 + (z/c) 2 1. The transformation g effectively replaces u by x/a since x = au. The change of variables formula states that if g is one-to-one (this function is one-to-one since it is a linear transformation with non-zero determinant and, therefore, invertible) and C 1 (it is since linear transformations are continuously differentiable) and if Dg is invertible at each point of (this is so since Dg = g as g is linear), then f(y) dv y (f g)(x) det Dg(x) dv x g() The vector y refers to the coordinates (x, y, z) and x refers to the coordinates (u, v, w) for this problem. The notation (x, y, z)/ (u, v, w) is commonly used for the Jacobian determinant. In this notation, the change of variables formula reads f(x, y, z) dx dy dz = f(x(u, v, w), y(u, v, w), z(u, v, w)) (x, y, z) (u, v, w) du dv dw g() Since we are to compute the volume, f(y) = 1. The derivative of g is g and its determinant is abc. We may assume that a, b, c >. Thus, vol (g() = dv y = abc dv x = (abc) vol (). g() Therefore, the volume of the ellipsoid is (abc) times 4π/3. In exercise # 4 of 7.6, a triangle having vertices (, ), (1, ), and (, 1) is given. Call this region S. We are to compute the value of the integral over S of the function f(x, y) = exp( x y x+y ). I will use coordinate notation, rather than vector notation in what follows. In polar coordinates, g(r, θ) = (r cos θ, r sin θ) is the transformation. This, of course, should be thought of as consisting of the two coordinate functions x = r cos θ and y = r sin θ. In the notation of the change of variables theorem, S = g(), for some, as yet to be determined region. (This is the typical scenario: the region S

5 is expressed in coordinates (x, y), and you want to change to another coordinate system using a transformation g which expresses x and y as functions of the new coordinates. So, to express the region S in terms of the new coordinates, you need to compute g 1 (S).) The boundary of S consists of three lines and each can, in turn, be expressed in terms of r and θ. For example, the line joining (, 1) to (1, ) has equation x + y = 1, which transforms to r(cos θ + sin θ) = 1. The inequalities r (cos θ + sin θ) 1 and θ π/2 describe. The Jacobian determinant of g is r (which is the formal justification for why da = r dr dθ is correct). The change of variables formula implies that e x y cos θ sin θ x+y dx dy = e cos θ+sin θ r dr dθ S To simplify the above, use the following trick: cos θ + sin θ = 2( 1 2 cos θ sin θ) = 2 sin (θ + π 4 ), by the sum of angles formula for sine. Similarly, cos θ sin θ is equal to 2 cos (θ + π 4 ) by the sum of angles formula for cosine. The integral above becomes π/2 1 2 csc (θ+ π 4 ) e cot (θ+ π 4 ) r dr dθ = 1 4 π/2 e cot (θ+ π 4 ) csc 2 (θ + π 4 ) dθ = 1 4 (e e 1 ) To solve the same problem using the change of variables u = x y, v = x + y, first solve for x and y in terms of u and v (as this is useful in determining. We find that x = 1 2 (u + v) and that y = 1 2 (v u). Therefore, the lines which bound S transform to u + v = (corresponding to x = ), v u = (corresponding to y = ) and v = 1 (corresponding to x + y = 1). Thus, the region is another triangle. The Jacobian determinant of the transformation x = (u + v)/2, y = (v u)/2 is equal to 1/2. The change of variables formula implies that e x y 1 x+y dx dy = e uv 1 S 2 du dv

6 From a sketch of, it is clear that it is most sensible to integrate with respect to u first: 1 v v euv 1 du dv = 2 v(e e 1 ) dv = 1 4 (e e 1 ). Solve exercises #2, 4 (b,d), 5, 6 (b,c), 7 (a, b, c, e, f), 11 (a, c, d), and 14 of Chapter 8, Section 2. Solve exercises #1, 2, 3, 5, 6 (a, b, c) of Chapter 8, Section 3. Please refer to the additional handwritten solutions to homework # 11 for solutions to the problems in 8.2 and 8.3. The following problems are recommended, but will not be collected: Chapter 7, section 4, exercises #1, 3, 7, 12. The answers to each of the above problems is in the textbook. Chapter 7, section 5, exercises #1, 2, 5, 9. If you have questions about the problems on determinants listed above, please send your question via or stop by during office hours.