course Mth 174 ڧ܂ۇassignment #003
......!!!!!!!!...................................
12:43:47 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
......!!!!!!!!...................................
RESPONSE --> (dy)(g) / t = (dy)((v + vd)/2) / t dy = change in distance g = acceleration (v + vd) / 2 = average velocity t = time This question confused me and I know you don't realy give credit to answer that I can't explain, but my answer was pretty much a guess.
.................................................
......!!!!!!!!...................................
12:44:01 how can you symbolically represent the give statement?
......!!!!!!!!...................................
RESPONSE --> This was in my last ""answer""
.................................................
......!!!!!!!!...................................
12:44:19 How can we show that the statement is true?
......!!!!!!!!...................................
RESPONSE --> Also in the ""answer"" two questions ago.
.................................................
......!!!!!!!!...................................
12:47:35 How can we use a graph to show that the statement is true?
......!!!!!!!!...................................
RESPONSE --> I do not understand the first answer, so I am not sure how to construct a graph. I would say that the x-axis is time. But that is as far as I got.
.................................................
......!!!!!!!!...................................
13:04:45 query problem 7.1.22 (3d edition #18) integral of `sqrt(cos(3t) ) * sin(3t)
......!!!!!!!!...................................
RESPONSE --> I have to admit that I am extremely confused at this substitution thing. My only attempt was that I found: w = -cos(3t) dw = -3sin(3t) * dt And this is not very much, but I just do not know what to do from here. Even the examples confused me. ** TYPICAL INCORRECT SOLUTION: (-2/3) (cos3t)^(3/2) INSTRUCTOR COMMENT: The derivative of cos(3t) is -3 sin(3t), wo the derivative of -2/3 (cos(3t))^(-3/2) is 3 sqrt(cos(3t) sin(3t). This solution didn't account for the Chain Rule. To perform the integral use substitution. Very often the first substitution you want to try involves the inner function of a composite, and that is the case here. Cos(3t) is the inner function of the composite sqrt(cos(3t)), so we try using this as our substitution: w = cos (3t) dw = -3 sin (3t) so that sin(3t) = -dw / 3. Thus our expression becomes w^(1/2) * (-dw / 3). The integral of -1/3 w^(3/2) with respect to w is -1/3 * (2/3) w^(3/2), treating w^(1/2) as a power function. This simplifies to -2/9 w^(3/2) or -2/9 * (cos(3t))^(3/2). The general antiderivative is -2/9 * (cos(3t))^(3/2) + c, where c is an arbitrary constant.**
.................................................
......!!!!!!!!...................................
13:04:57 what did you get for the integral and how did you reason out your result?
......!!!!!!!!...................................
RESPONSE --> I didn't get an integral
.................................................
......!!!!!!!!...................................
13:18:22 query problem 7.1.20 (3d edition #21) antiderivative of x^2 e^(x^3+1)
......!!!!!!!!...................................
RESPONSE --> This one I think I may have done a little better on. w = x^3 + 1 dw = 3x^2 * dx x = (w - 1) ^ (1/3) dx = 1 / [3 * (w -1) ^ (2/3)] [(w - 1) ^ (1/3)] ^ 2 * e ^ w
.................................................
......!!!!!!!!...................................
13:18:57 what is the antiderivative?
......!!!!!!!!...................................
RESPONSE --> I put it in my last answer. (3/5) * (w - 1) ^ (5/3) * e ^ w + C
.................................................
......!!!!!!!!...................................
13:19:21 What substitution would you use to find this antiderivative?
......!!!!!!!!...................................
RESPONSE --> I used x = (w - 1) ^ (1/3)
.................................................
......!!!!!!!!...................................
13:33:45 query problem 7.1.37 (3d edition #35) antiderivative of (t+1)^2 / t^2
......!!!!!!!!...................................
RESPONSE --> Well, my answer seems extremely wrong, but I still think I need some extra help with this substitution thing. I changed the equation to (t + 1)^2 * t^-2 w = t^-2 dw = -2t ^ -3 * dt t = w ^ (-1/2) dt = (-1/2) * w ^ (-3/2) Substituting t into the equation: (w ^ (-1/2) + 1) ^ 2 * (w ^ -1/2) ^ -2 (w ^ (-1/2) + 1) ^ 2 * w (w ^ (1/4) + 2w ^ (-1/2) + 1) * (w) w ^ (5/4) + 2w ^ (1/2) + w I am not sure if this equation is correct, but the antiderivative of w ^ (5/4) + 2w ^ (1/2) + w is: (4/9)w ^ (9/4) + (4/3)w ^ (3/2) + (1/2)w ^ 2
.................................................
......!!!!!!!!...................................
13:33:48 what is the antiderivative?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:34:11 What substitution would you use to find this antiderivative?
......!!!!!!!!...................................
RESPONSE --> I substituted t = w ^ (-1/2)
.................................................
......!!!!!!!!...................................
13:35:13 query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)
......!!!!!!!!...................................
RESPONSE --> This situation involves a composite of the power function 1 / z^2 and the linear function t + 7. The latter is the inner function of the composite. We therefore substitute u = t+7 to get du = dt; limits t = 1 and t = 3 become u = 8 and u = 10. So the integral becomes int( u^-2, u, 8, 10). Antiderivative can be -u^-1 or -1/u. So definite integral is -1/10 - (-1/8) = 1/8 - 1/10 = 1/40. **
.................................................
......!!!!!!!!...................................
13:35:15 What did you get for the definite integral?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:35:16 What antiderivative did you use?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:35:18 What is the value of your antiderivative at t = 1 and at t = 3?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:42:19 query 7.1.86. World population P(t) = 5.3 e^(0.014 t).
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:43:31 What were the populations in 1990 and 2000?
......!!!!!!!!...................................
RESPONSE --> The population in 1990: P = 5.3 * e ^ (0.014 * 0) = 5.3 The population in 2000: P = 5.3 * e ^ (.014 * 10) = 6.096451134
.................................................
......!!!!!!!!...................................
13:47:48 What is the average population between during the 1990's and how did you find it?
......!!!!!!!!...................................
RESPONSE --> I used the Fundamental Theorem using the intervals [0, 10]. I found the antiderivative of 5.3 * e ^ (.014 * t) to be 5.3 * e ^ (.014 * t). To find the average population: AntiD of P(10) - AntiD of P(0) = 6.09641134 + 5.3 = 11.39645113 11.39645113 / 2 = 5.698225567, which is the average population
.................................................
......!!!!!!!!...................................
13:48:57 What is the value of your antiderivative at t = 1 and at t = 3?
......!!!!!!!!...................................
RESPONSE --> AntiP at t = 1 is 5.374721832 AntiP at t = 3 is 5.527340737
.................................................
......!!!!!!!!...................................
13:49:21 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> The substitution thing was difficult for me to grasp, and still is. I need a lot more practice.
.................................................
......!!!!!!!!...................................
13:49:22 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE -->
.................................................