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MIT 18 01 - Calculus Practice Problems

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� 18.01 Calculus Jason Starr Fall 2005 Lecture 20. November 1, 2005 Practice Problems. Course Reader: 4C-2, 4C-6, 4D-1, 4D-4, 4D-8. 1. Average values. Given a function f (x) defined on some interval [a, b], what is the average value of f (x)? A reasonable first approximation is to choose a finite c ollection of points from [a, b] and compute the average value over those points. Break [a, b] into a union of n subintervals of ∗ k in the kth interval. For the length Δx ∗ k (b − a)/n. finitely many values yFrom each interval, choose a point; say x = ∗), the average value is, k = f (x nAverage ≈ yn k=1 1 ∗ .k18.01 Calculus Jason Starr Fall 2005 Multiplying and dividing by Δx gives, n1 � Average ≈ y∗Δx. nΔx k k=1 Now nΔx equals n(a − b)/n, which is a − b. So the average value is, n1 � Average ≈ y∗Δx. b − a k k=1 The sum is a Riemann sum. To get better approximations to the true average, increase the number of points n at which f (x) is “sampled”. In the limit, this gives the true average, n1 � � bAverage = lim y∗Δx = f (x)dx/(b − a).ab − a n→∞ k=1 k Example. Under ideal conditions, a wire-producing machine produces wire of uniform radius r0. Because of small vibrations in the machine, the actual radius of the wire varies as a function of the length, r(x) = r0 + A cos(ωx). The quantity A is much smaller than r0. What is the average radius of the wire? Because the variation is periodic, the average value over any number of periods equals the average value of one period. In other words, compute the average for the interval 0 ≤ x ≤ 2π/ω. The length of this interval is 2π/ω. Thus the average value is, � 2π/ω 1 Average = r0 + A cos(ωx)dx. (2π/ω) 0 Using the Fundamental Theorem of Calculus, this equals, 1 2π/ω (2π/ω) (r0x + (A/ω) sin(ωx)|0 . This evaluates to, 1 r0(2π/ω) = r0. (2π/ω) Thus, although the radius varies and does not usually equal its ideal value r0, the average value is indeed, Average = r0. 2. Average values: non-uniform distribution. It often happens that the average value of f (x) is desired in a s ituation where the values f (x) are not all uniformly likely. Typically, the probability that x has value in the range from x0 to x0 + Δx is approximately, Prob(x0 ≤ x ≤ x0 + Δx) ≈ ρ(x0)Δx,� � 18.01 Calculus Jason Starr Fall 2005 for some nonnegative continuous function ρ(x). The function ρ(x) is called a probability distribution. Assuming this approximation becomes arbitrarily good as the length Δx approaches zero, the exact probability that x has value in the range x0 to x1 is, x1 Prob(x0 ≤ x ≤ x1) = ρ(x)dx. x0 In particular, because x must take value somewhere in the interval [a, b], the total probability is 1. In other words, � b ρ(x)dx = 1. a This is called the normalization condition. The average value is computed as before. But this time, each value y∗ = f (x∗ k ) is weighted by the k approximate probability that x takes value in the kth interval, ρ(x∗ k )Δx. This gives, nAverage ≈ f (xk )∗ρ(xk )∗Δx. k=1 In the limit as n goes to ∞, this gives the exact average, � b Average = f (x)ρ(x)dx. a It must be noted, the probability distribution ρ(x) often does not satisfy the normalization condi-tion. In this case, the formula above is wrong. But it is easily correct, ( � b a f (x)ρ(x)dx)/( � b a ρ(x)dx).Average = Example. A particle is fired through a slit and strikes a screen on the other side. Measuring the position on the screen so that the origin is the closest point on the screen to the slit, the probability distribution is empirically observed to be, ρ(x) = Ce−x2/2σ2 , where σ is a constant determining the “width” of the probability distribution, and C is an unde- termined normalization constant. What is the average distance of the particle from the center of the screen? Assume the particle lies in an interval [−R, R], where R is very large. Remark. This differs from the formula given in lecture, which was Ce−x2/2σ for a particular choice of σ. The formula given here is more “standard”. I apologize for any confusion. The distance function is, � f (x) = |x| = −x, x, x < 0 x ≥ 018.01 Calculus Jason Starr Fall 2005 According to the formula, the average value is, � R � R ( f (x)ρ(x)dx)/( ρ(x)dx). −R −R The numerator is, � R |x|Ce−x2/2σ2 dx. −R It is easiest to c ompute this by breaking it into a sum of 2 integrals, � 0 � R (−x)Ce−x2/2σ2 dx + (+x)Ce−x2/2σ2 dx. 0−R Make the substitution u = −x2/2σ2 , du = (−x/σ2)dx to reduce this to, � 0 � −R2/2σ2 � 0 Ce u(σ2du) + Ce u(−σ2du) = 2 Cσ2 e udu. −R2/2σ2 0 −R2/2σ2 Using the Fundamental Theorem of Calculus, this equals, u02Cσ2 (e −R2/2Σ2 = 2Cσ2(1 − e−R2/2Σ2 ).| As R becomes large, the quantity e−R2/2Σ2 becomes vanishingly small. Thus, in the limit as R tends to ∞, the numerator equals, � R lim |x|Ce−x2/2σ2 dx = 2Cσ2 . R→∞ −R Unfortunately, this is not an answer, because the normalization constant C is unknown. The normalization c ondition is that, � R C lim e−x2/2σ2 dx = 1. R→∞ −R Simplify this by making the substitution, u = x/σ, du = dx/σ, and Q = R/σ to get, � R/σ � Q C lim e−u2/2σdu = Cσ lim e−u2/2du. R→∞ −R/σ Q→∞ −Q Notice the limit, � Q lim e−u2/2du, Q→∞ −Q� 18.01 Calculus Jason Starr Fall 2005 does not depend on σ. It is simply some number. Denoting this number by 1/C1, the normalization condition is, Cσ/C1 = 1. The solution is that C = C1/σ. Plugging this into the formula above, the average distance is, Average distance = 2C1σ, where, Q 1/C1 = lim e−u2/2du. Q→∞ −Q There is a beautiful argument that, C1 = 1/√2π. Unfortunately, we cannot yet prove this. Taking it as true gives the final answer, Average distance = 3. Volumes of solids of revolution: the shell method. An alternative to the disk and washer method is the shell method. A shell is the region between 2 cylinders of the same height. If the average radius of the cylinders is r, if the width of the region is w and if the height of the cylinders is h, then the approximate volume of the shell is, 2√2π.σ/Volume ≈= Circumference × height × width = 2πrhw. Take the plane region bounded by x = a, x = b, the x-axis and the curve y = f (x). Revolve


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MIT 18 01 - Calculus Practice Problems

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