course Mth 174 ???????????}??assignment #003???g??x??m??????Physics II
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17:07:08 query 6.5 #8 Galileo: time for unif accel object to traverse dist is same as if vel was ave of init and final; put into symbols and show why true
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17:20:39 how can you symbolically represent the give statement?
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RESPONSE --> S (sub 0) = starting distance S (sub x) = distance traversed from t=0 to t=x V (sub 0) = velocity at t=0 = 0 V (sub x) = velocity at t = x V (sub mean) = mean velocity between t = 0 and t = x t (sub x) = time elapsed at t = x a = uniform acceleration Distance calculated using equations of motion: S (sub x) = (a*t (sub x)^2)/2 + C, but C=0 since S (sub 0) =0, so S (sub x) = (a*t^2)/2 Distance calculated using Galileo's rule: S (sub x) = v (sub mean) * t (sub x) Solving for t and setting the two calculations equal (as stated by Galileo: t (sub x) = Sqrt [(2*S (sub x))/a] = S (sub x)/v (sub mean)
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17:37:45 How can we show that the statement is true?
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RESPONSE --> Using the equivalent equations for calculating distance rather than time: V (sub mean) = [((V (sub 0) + V (sub x))/2] * t (sub x) Equations of motion: S (sub x) = (a * t (sub x)^2)/2 + v (sub 0) * t (sub x) + S (sub 0) V (sub x) = a * t (sub x) + V (sub 0) We check to see if S (sub x) calculated both ways (equation of motion and Galileo's equation) are equivalent. Note: All ""="" below should have a ? above them except the final one, which should have check mark above it. (a * t (sub x)^2)/2 + v (sub 0) * t (sub x) + S (sub 0) = [(v (sub 0) + v (sub x))/2] * t (sub x) + S (sub 0) Substituting for v (sub x) on the righthand side: (a * t (sub x)^2)/2 + v (sub 0) * t (sub x) + S (sub 0) = [v (sub 0) + a * t (sub x) + V (sub 0)]/2 + s (sub 0) Simplifying the righthand side: (a * t (sub x)^2)/2 + v (sub 0) * t (sub x) + S (sub 0) = (a * t (sub x)^2)/2 + v (sub 0) * t (sub x) + S (sub 0) Thus, the two equations are equivalent.
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18:48:07 How can we use a graph to show that the statement is true?
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RESPONSE --> If we graph v against t, for the uniform acceleration situation, we get a linear increase in velocity from v (sub 0) to v (sub x). We can then easily calculate the triangular area under the graph: (1/2)*(v (sub (x) - v (sub 0)) *((t (sub x) - t (sub 0). This area is the total distance traversed. If we graph v (sub mean) against t, we get a rectangular area whose area we can also calculate: v (sub mean) * (t (sub x) - t (sub 0)). This is also the distance traversed. Since the distances and thus the areas must be equal, we can set the two area calculations as the two sides of an equality. As v (sub mean) = (v (sub x) - v(sub 0)) / 2, the two time intervals must necessarily be equal.
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18:49:54 query problem 7.1.22 (3d edition #18) integral of `sqrt(cos(3t) ) * sin(3t)
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19:03:52 what did you get for the integral and how did you reason out your result?
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RESPONSE --> Since sin 3t is a constant multiple (* -1/3) of the derivative of cos 3t, I substituted u = cos 3t, thus du = -3 sin 3t dt Dividing by -3: -du/3 = sin t dt Substituting in the orginial integral: Integral of u^(1/2) * -du/3 = -(1/3) Integral of u^(1/2) du = -(1/3) * (2/3) * u^(3/2) + C = -(2/9) * u(3/2) + C Substituting back: = -(2/9) * (cos 3t)^(3/2) + C
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19:04:18 query problem 7.1.20 (3d edition #21) antiderivative of x^2 e^(x^3+1)
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19:08:44 what is the antiderivative?
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RESPONSE --> I substituted: u = x^3 + 1, thus du = 3x^2 dx, and du/3 = x^2 dx Substituting in the integral: Integral of e^u * du/3 = 1/3 Integral of e^u du = (1/3) e^u + C Substituting back, the antiderivative is: (1/3) * e^(x^3 + 1) + C
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19:08:58 What substitution would you use to find this antiderivative?
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RESPONSE --> See previous response.
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19:20:18 query problem 7.1.37 (3d edition #35) antiderivative of (t+1)^2 / t^2
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19:25:16 what is the antiderivative?
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RESPONSE --> Since this problem didn't appear to have a constant mulitple of the derivative of one of the functions, I multiplied out the numerator, thus Integral of (t^2 + 2t + 1)/t^2 dt Dividing each term by t^2 results in: Integral of 1 + 2/t + 1/t^2 Bringing the constant 2 in the second term outside the integral and finding the antiderivative of each term results in: t + 2 ln |t| - 1/t + C
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19:37:53 What substitution would you use to find this antiderivative?
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RESPONSE --> As noted previously I found this problem worked without substitution. I don't see any reasonable way to do a substitution that would work (and be easier than the way I did it).
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19:39:03 query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)
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19:53:01 What did you get for the definite integral?
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RESPONSE --> Substituting u = t + 7, du= 1 dt Thus: Integral of 1/u^2 du (evaluated from 1 to 3) = - (1/u) evaluated from1 to 3 = -1/(t + 7) evaluated from 1 to 3 = -1/(3 + 7) - (-1/(1 +7) = -1/10 + 1/8 = 1/40
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19:53:42 What antiderivative did you use?
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RESPONSE --> See previous response
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19:54:33 What is the value of your antiderivative at t = 1 and at t = 3?
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RESPONSE --> At t = 1, value is -1/8 At t = 3, value is -1/10
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19:55:51 query 7.1.86. World population P(t) = 5.3 e^(0.014 t).
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20:03:02 What were the populations in 1990 and 2000?
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RESPONSE --> For 1990: P = 5.3 * e^(0.014*0) = 5.3* e^0 = 5.3 * 1 = 5.3 billion For 2000: P = 5.3 * e^(0.014*10) = 5.3 * e^0.14 = 6.096 billion (approx.)
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20:25:07 What is the average population between during the 1990's and how did you find it?
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RESPONSE --> I used the average value of a function rule: 1/(b-a) * Integal from a to b of f(x) dx Let b = 10 years and a = 0 years (assumes equation measures population at beginning of year) Substituting in the average value rule: 1/(10 - 0) * Integral from 0 to 10 of 5.3 * e^(0.014t) dt Let u = 0.014t du = 0.014 dt 378.571 du = 5.3 dt Substituting: 1/10 * Integral from 0 to 10 of e^u * 378.571 du Moving constant outside integral: 37.8571 * Integral from 0 to 10 of e^u du = 37.8751 * e^u, evaluated from 0 to 10 Substituting back for u: = 37.8751 * e^0.014t evaluated from 0 to 10 = 37.8751 * (e^0.14 - e^0) = 37.8751 * (1.15 -1 ) = 5.69 billion people (approx.)
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20:42:03 What is the value of your antiderivative at t = 1 and at t = 3?
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RESPONSE --> I presume you meant at t =0 and t=10 The values of the antiderivative before multiplying by 1/10 to get the average: t = 10: 435.46 billion (approx.) t = 0 : 378.571 billion The value after multiplying by 1/10: t = 10 : 43.546 billion (approx.) t = 0 : 37.857 billion (approx.) If you actually want t = 1 and t = 3 Before multiplying: t = 1: 383.908 billion (approx.) t = 3: 394.820 billion (approx.) After multiplying by 1/10: t = 1: 38.391 billion (approx.) t = 3: 39.482 billion (approx.)
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20:44:56 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I found that making the substitutions (assuming substitutions were possible) was easier than I expected. I'm still amazed at the places where e and ln show up. Integration by substitution seems like a very valuable tool.
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