AP Calculus (Mr. Surowski) Homework from Chapter 7 (and some of Chapter 8) Lesson 30 Integral as accumulation (7.):, 3, 5, 8 0, 7, 20 22, 25 (to do quadratic regression on our (TI-84 calculators, refer to footnote # on the sllabus for Chapter 6; obviousl ou ll need to adapt this in the obvious wa from the discussion of logistic regression), 30. Here s a miniproject. Assume that ou are to drive a vehicle along a straight road for 00 miles. Assume also that ou know our velocit as a function of distance where (i) The graph of the velocit against distance is elliptical (see the graph to the right); (ii) v(0) = v(00) = 0; (iii) the maimum velocit is 60 mph. From this information, show that (A) The distance s along the road as a function of time is given b s(t) = 50+50 sin( 6 5 t π 2 ), 0 t 5π 6. (B) The time required to make this trip is 5π 6 ( 2.6) hours. Let s make it more interesting. Suppose that ou are a driver delivering goods for a compan and that the costs incurred to the compan are the operation of the vehicle and, of course, our pa. Assume that the cost of operating the vehicle is (30 + v/2)/00 dollars/mile and that ou get paid $2/hour. (C) Show that if ou drive according to the elliptical velocit described above, then the total cost to the compan for the 00 mile trip is 30 + 35π/2 $85. Finall, let s make it reall interesting! Suppose that the velocit curve follows the same sort of elliptical curve as a function of distance s, but that this time, the maimum velocit reached is a mph (rather than 60 mph). (D) Show that the minimum cost to the compan will occur when a = 40 3 mph in which case the total cost to the compan is 30 + 0π 3 ( 84.4) dollars.
Homework from Chapter 7 (and some of Chapter 8), cont d (7.2): 0, 3, 4, 8 24, 37, 4, 42, 45. (A) Miniproject. Consider the function = n, n 0, and consider the interval [a, b] on the -ais, where 0 < a < b. On the -ais is the corresponding interval [c, d], 0 < c < d, as indicated in the figures below: Lesson 3 Planar areas d c S R n = n>0 d c = n<0 n a b a b In both cases illustrated above we have a region R below the graph of = n, above the -ais, and bounded b the lines = a and = b. Correspondingl, we have a region S to the left of the graph of = n, to the right of the -ais and bounded b the lines = c and = d. Find the ratio area S as a function of n. area R Lesson 32 Volumes of revolution (disks and washers) (7.3): 2 (all parts), 3 6, 7, 8, 0, 3 8, 30, 3. (A) Compute the volume of the region that is common the the clinder of radius 2 centered on the -ais and the clinder of radius 2 centered on the -ais.
Homework from Chapter 7 (and some of Chapter 8), cont d Quiz over Lessons 32 34 Lesson 33 More volumes of revolution (shells) surface area Lesson 34 Arc lengths Lesson 35 Miscellaneous applications (7.3): 33 38, 47, 48, 50, 55-58. (A) Think about the region in problem 33. If ou were to compute the volume of the solid of revolution b revolving about the -ais, which method would ou use, disks (washers) or shells? What about revolving about the line =? (B) Think about the region in problem 34. If ou were to compute the volume of the solid of revolution b revolving about the line = 2, which method would ou use, disks (washers) or shells. What about revolving about the -ais? (This is more subtle!) (C) Consider infinite surface of revolution formed b revolving the graph of = /, about the -ais, 0 <. Show that this surface has finite volume but infinite surface area. Doesn t this mean that the surface can t be painted with a finite amount of paint? And et, let s pour enough paint into this infinitel long horn (which has finite volume) and dump it out. Haven t we painted this surface? What s wrong here?!! (7.4): 5, 7 0; 6, 8, 20. (7.5): 6, 7, 7, 9, 20 2. (A) (Work and Kinetic Energ) In AP Phsics, students learn (b being told ) that if we have one-dimensional motion of an object with mass m having force F applied to it, then the total work done on this object is the change in kinetic energ: Work = ke 2 ke. Without using calculus, this can onl be proved for an applied force of constant value. Using integration b parts, we can see wh it s alwas true. Follow these steps: (i) If the object starts at position = (corresponding to time t = t ) and ends at position = 2 (corresponding at time t = t 2 ), show that the work can be written as where v = d dt. W = 2 F d = m t2 (ii) Use integration b parts to conclude that t2 v (t)v(t) dt = v 2 t t t t2 2 t t v (t)v(t) dt, v (t)v(t) dt (iii) Now what? (You re almost done!)
Lesson 36 Improper Integrals Lesson 37 Trig substitutions Homework from Chapter 7 (and some of Chapter 8), cont d (8.4): 4, 4, 9 23, 27 29, 30, 3 34, 39, 42. Also consider these problems: (A) Determine which values of p guarantee that the improper integral d converges. p d (B) Show that for all values of p, the improper integral 0 diverges. p (C) Determine which values of p guarantee that the improper integral p+2 + d converges. p+ (D) Graph the function =, > 0. Net, draw the rectangles 2 corresponding to a right Riemann sum, where each rectangle has right side at =, 2, 3,.... Which is on the basis of our drawing, which d is larger, or? Based on this what can ou sa about the 2 n2 n=2 infinite series. (We ll see much more of this in Chapter 9.) n2 n= See problems 0 on the Integration Practice document Review for Chapter 7 Test Chapter 7 Test Application to Statistics. Here, I would once again like to consider some of the topics in statistics from the more mature point of view of calculus. Thus we assume that X is a (continuous) random variable, taking on values in the real numbers. Associated with X is its densit function f, which is useful inasmuch as it determines the distribution of X. For instance, the probabilit that we observe a X b is given b the integral P (a X b) = b f()d. The familiar fact that P ( < X < ) = is epressed b an improper integral: a f()d =. The mean and variance of the random variable X are given b the improper integrals:
µ = E(X) = f()d, σ 2 = Var(X) = ( µ) 2 f()d. The normal random variable X with mean µ and variance σ 2 has densit function Note that this implies that f() = σ 2π σ ( µ) 2 2π e 2σ 2. e ( µ)2 2σ 2 d =. () Eercise. Show that using the substitution u = ( µ)/ 2σ reduces the proof of equation () to that of showing that e u2 du = π. (2) (Notice that equation (2) cannot be proved b simple techniques since e u2 elementar antiderivative.) doesn t have an Eercise 2. Assuming the validit of either equations () or (2), show that (Hint: recall that σ 2π (odd function) d = 0.) Eercise 3. Likewise, show that σ 2π (Hint: this will involve an integration b parts!) e ( µ)2 2σ 2 d = µ. ( µ) 2 e ( µ)2 2σ 2 d = σ 2. Eercise 4. Consider the uniform densit function { if 0, f() = 0 otherwise. Compute the mean and variance of the associated random variable X. Eercise 5. Consider the densit function { 2 if 0, f() = 0 otherwise.
Compute the mean and variance of the associated random variable X.