A spherical soap bubble with a surface-tension of 0.005 lbf/ft is expanded from a diameter of 0.5 in to 3.0 in. How much work, in Btu, is required to expand this bubble?

Math 3040 - 1/27/16 February 7, 2016 1 Syllabus Notes ▯ There will be 1 in class prelim, some time near the drop deadline. ▯ There will be 1 in class ▯nal, date to be determined. ▯ There will be 1 paper, due near the end of the semester, to be writen in LaTeX. ▯ Finally, there will be up to 2 retries for all homeworks turned in, for a maximum of 100 2 How is Math di▯erent from other disciplines ▯ highly theoretical ▯ Mostly inductive, whereas other disciplines require more deduction. { However, discovering a new property in mathematics still requires some deduction ▯ Math permeates all sciences, whereas not all sciences permeate the other sciences ▯ Math is precise enough to rely only on formal logical proofs to derive ideas ▯ Once something is proven true, it remains that way forever. { it is very normal for math papers to have 25 year old references. Conversely, this is rare in scienti▯c disciplines, as hypotheses are often adjusted or disproved. { There exists some debate on what constitutes a valid proof, however - for example, some mathematicians don’t accept proof by contradic- tion ▯ Mathematics is less tangible - exists more conceptually 1 ▯ Math is a vehicle for other disciplines ▯ A question has one answer, but multiple paths to get there. ▯ In▯nities are an important concept in math, whereas other disciplines tend to deal with the ▯nite. 2 Math 3040 - 1/29/16 February 7, 2016 1 Sample Proof 1 Proposition: 4(1 ▯ 1+ 1▯ :::) = ▯ 3 5 7 Proof: d (arctanx) = 1 dt 1 + x2 1 = 1 ▯ (▯x ) = 1 + (▯x ) + (▯x ) + (▯x ) ::: = 1 ▯ x + x ▯ x ::: R By the fundamental theorem of calculus, x 12dt = arctan(x) ▯ arctan(0) = 0 1+t arctan(x). Thus, we can write: Z x t3 t5 t7 1 ▯ t + t ▯ t + t :::dt = [t ▯ + ▯ ::0] 0 3 5 7 3 5 7 x x x = x ▯ + ▯ ::: 3 5 7 = arctan(x) 1 1 1 1 ▯ Plugging in x = 1 gives 1 ▯ 3+ 5 ▯ 7 + 9::: =4 . Thus, ▯ = 4arctan(1) = 4(1 ▯ 1 + 1 ▯ ::::) 3 5 7 ▯ Well, this proof is ▯ne, but there are some things it cannot answer some questions. ▯ For example, how can we ▯nd the ▯rst 3 digits of pi using this proof How many terms will we need to calculate to ▯nd out the answer 1 ▯ Also consider the following proposition: Take any x 2 R. Then, there 1 1 exists a reordering of the terms 1;3; 5::: expressed as a1;a 2a 3:: such that ▯1a i x. This requires a deep understanding of the sample proof! 2 Integers ▯ There exist 2 approaches to ▯nd the properties of a set Z { We can try and derive each and every property, which is very di▯cult { Or, we can make some assumptions, called axioms, that we don’t need to prove ▯ We will take the Axiomatic approach. The axioms of the the integers are as follows: { Axiom 1.1: ▯ m + n = n + m (Commutativity of addition) ▯ (m + n) + p = m + (n + p) (Associativity of addition) ▯ m ▯ (n + p) = (m ▯ n) + (m ▯ p) (Distribution) ▯ m ▯ n = n ▯ m (Commutativity of multiplication) ▯ (m ▯ n) ▯ p = m ▯ (n ▯ p) (Associativity of Multiplication) { Axiom 1.2: There exists an element 0 2 Z such that for all n 2 Z, n + 0 = n. This is the Additive Identity. { Axiom 1.3: There exists an element 1 6= 0;1 2 Z such that for all m 2 Z, m ▯ 1 = m. This is the multiplicative identity. { Axiom 1.4: For all elements m 2 Z there exists a value ▯n such that n + ▯n = 0. This is the Additive inverse. { Axiom 1.5: Suppose. m;n;p 2 Z,and m = 6 0. If m ▯ n = m ▯ p, then n = p. This is Cancellation. Here is a sample proof using these axioms: 3 Sample Proof 2 Prop 1.11 v): Prove m(n + (p + q)) = (mn + mp) + mq. Proof: m(n + (p + q)) = m((n + p) + q) By associativity = m(n + p) + mq By distribution = (mn + mp) + mq By distribution Notice that we wrote out information for each and every step. Do we really need to write so much It would be tedious for larger proofs. 2 However, once you show that you know what you’re doing, you can always combine multiple small steps together, such as combining associativity and com- mutativity in one step. 4 What’s wrong with this proof Prop: (▯m)(▯n) = mn Proof: First, add (▯m ▯ n) to both sides. This gives (▯m)(▯n) + (▯m)n = mn + (▯m)n. By distributivity, we can reduce this to ▯m(▯n + n) = n(m + ▯m). By axiom 1.4, this reduces to ▯m ▯ 0 = n ▯ 0. By a previous proposition, this gives 0 = 0. ▯ This doesn’t actually prove anything, it just reduces a statement to a known fact. ▯ This is NOT the same as proving that the left side is, in fact, equal to the right side. ▯ in the case of this proof, you should use the axioms and previously proved statements to show that the left equals the right. 3