course mth 173 assignment #021DѼԸ\ē
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14:22:21 **** Query Problem 4.8.1 (3d edition 3.8.4). x graph v shape from (0,2) |slope|=1, y graph sawtooth period 2, |y|<=2, approx sine. Describe the motion of the particle described by the two graphs.
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RESPONSE --> The particle is moving back and forth from the point of observation, or the particle is moving in a circle on the same plane as the observer.
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14:25:46 ** The question was about the motion of the particle. The graphs are of position vs. time, i.e., x vs. t for a particle moving on the x axis. The slope of a position vs. time graph of a particle is the velocity of the particle. The first graph has slope -1 for negative values of t. So up to t = 0 the particle is moving to the left at velocity 1. Then when the particle reaches x = 2 the slope becomes +1, indicating that the velocity of the particle instantaneously changes from -1 to +1 at t = 0 and the particle moves back off to the right. On the second graph the velocity of the particle changes abruptly--instantaneously, in fact--when the graph reaches a sharp point, which it does twice between t = 0 and t = 2. At these points velocity goes from positive to negative or from negative to positive. Velocity is a maximum when the slope takes its greatest positive value as the graph passes upward through the x axis where the slope is probably 4 (I don't have the graph in front of me so that might be off, but if x goes from -2 to 2 as t changes by 1 the slope will be 4), and is a minimum when the slope takes its lowest negative value as the graph passes through the x axis going downward (slope -4?). So every 2 time units the particle will go from maximum positive velocity 4 to lowest negative velocity -4, as the ball goes from position x = -2 to x = +2 and back. **
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RESPONSE --> I didn't relize the second graph was part of the problem. Everything that you've said makes sense though.
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14:35:24 Query problem 4.8.21 (3d edition 3.8.16). Ellipse centered (0,0) thru (+-5, 0) and (0, +-7).Give your parameterization of the curve.
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RESPONSE --> x= cos(t) y=sin(t) 'pi/2< t < 4'pi / 2
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14:39:23 The standard parameterization of a unit circle (i.e., a circle of radius 1) is x = cos(t), y = sin(t), 0 <= t < 2 pi. An ellipse is essentially a circle elongated in two directions. To elongate the circle in such a way that its major and minor axes are the x and y axes we can simply multiply the x and y coordinates by the appropriate factors. An ellipse through the given points can therefore be parameterized as x = 5 cos (t), y = 7 sin (t), 0 <= t < 2 pi. To confirm the parameterization, at t = 0, pi/2, pi, 3 pi/2 and 2 pi we have the respective points (x, y): (5 cos(0), 7 sin(0) ) = (5, 0) (5 cos(pi/2), 7 sin(pi/2) ) = (0, 7) (5 cos(pi), 7 sin(pi) ) = (-5, 0) (5 cos(3 pi/2), 7 sin(3 pi/2) ) = (0, -7) (5 cos(pi), 7 sin(pi) ) = (5, 0). **
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RESPONSE --> I see how you parameterized the ellipse now, I was having trouble with the process and confused myself into making it harder than it really was. You put in the min / max figures corresponding to x and y to the unit circle formula as multipliers, which manipulates the formula to the figure that is given.
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14:46:35 Query 4.8.23 (was 3.8.18). x = t^3 - t, y = t^2, t = 2.What is the equation of the tangent line at t = 2 and how did you obtain it?
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RESPONSE --> x= t^3 -t y= t^2 x'= 3t ^2 -1 y'= 2t at t=2 we get (6, 4) and x'=11 y'=4 so the parametric equations for the tangent line are: x= 6 + 11t y= 4+ 4t
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14:50:26 ** Derivatives are dx/dt = 3t^2 - 1, which at t = 2 is dx/dt = 11, and dy/dt = 2t, which at t=2 is 4. We have x = 6 + 11 t, which solved for t gives us t = (x - 6) / 11, and y = 4 + 4 t. Substituting t = (x-6)/11 into y = 4 + 4 t we get y = 4 + 4(x-6)/11 = 4/ll x + 20/11. Note that at t = 2 you get x = 6 so y = 4/11 * 6 + 20/11 = 44/11 = 4. Alternatively: The slope at t = 2 is dy/dx = dy/dt / (dx/dt) = 4 / 11. The equation of the line thru (6, 4) with slope 4/11 is y - 4 = 4/11 ( x - 6), which simplifies to y = 4/11 x + 20/11. **
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RESPONSE --> I thought that the parametric equations were the ones for the tangent line, but they are for the curve, correct? so by finding the seperate velocites of x and y we can then just use them as 'dy and 'dx and work the tangent line as we normally would.
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14:57:01 Query 4.8.12 (3d edition 3.8.22). x = cos(t^2), y = sin(t^2).What is the speed of the particle?
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RESPONSE --> x= cos(t^2) y= sin(t^2) using the chain rule we can derive the derivatives: f(z)= cos(z) g(t)= t^2 F(g(t)= f'g * g' = -sin(t^2) * 2t f(z)= sin(z) g(t)= t^2 f(g(t))' = cos(t^2) * 2t since the speed of the particle is equal to the slope= 'dy/'dx (cos(t^2) * 2t) / (-sin(t^2) * 2t) cos(t^2) / -sin(t^2)
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15:02:35 The velocities in the x and y directions are dx / dt and dy / dt. Since x = cos(t^2) we have dx/dt = -2(t) sin (t)^2. Since y = sin(t^2) we have dy/dt = 2(t) cos (t)^2. Speed is the magnitude of the resultant velocity speed = | v | = sqrt(vx^2 + vy^2) so we have speed = {[-2(t) sin (t)^2]^2 + [2(t) cos (t)^2]^2}^1/2. This simplifies to {-4t^2 sin^2(t^2) + 4 t^2 cos^2(t^2) } ^(1/2) or (4t^2)^(1/2) { -sin^2(t^2) + cos^2(t^2) }^(1/2) or 2 | t | { -sin^2(t^2) + cos^2(t^2) }^(1/2). **
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RESPONSE --> I didn't bring my exponants down when finding the derivative of the equation. And I didn't quite get the slope / speed relationship right. Speed is the magnitude of the resultant velocity speed = | v | = sqrt(vx^2 + vy^2) so we have This statement is like the pythagoreum theorum so where is the hypotenuse graphically, is it the graph itself, so it's slope?
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15:06:03 Does the particle ever come to a stop? If so when? If not why not?
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RESPONSE --> Yes whenever the velocity changes from positive to negative the particle must ""stop"" and change direction
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15:06:41 ** The particle isn't moving when v = 0. v = 2 | t | { -sin^2(t^2) + cos^2(t^2) }^(1/2) is zero when t = 0 or when -sin^2(t^2) + cos^2(t^2) = 0. t = 0 gives x = cos(0) = 1 and y = sin(0) = 0, so it isn't moving at (1, 0). More generally: -sin^2(t^2) + cos^2(t^2) = 0 when sin(t^2) = cos(t^2). Since sin(z) = cos(z) when z = `pi/4 or when x = 5 `pi / 4, and in general when z = (4n + 1) `pi / 4, n = 0, 1, 2, 3, ... sin^2(t^2) = cos^2(t^2) when t^2 = `pi/4 or 5 `pi / 4 or (4n+1)`pi/4, i.e., when t = +- `sqrt( `pi/4), +- `sqrt(5 `pi / 4), or +-`sqrt(4n+1)`pi/4 for n = 0, 1, 2, 3, ... . **
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RESPONSE --> okay
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15:29:07 Query problem 3.9.18 (3d edition 3.9.8) (was 4.8.20) square the local linearization of e^x at x=0 to obtain the approximate local linearization of e^(2x)
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RESPONSE --> y= e^x y'=e^x x=0 y= 1 y'= 1 y= mx +b 1= 1(0) +b 1=b Ytan= x +1 (x+1)(x+1) = x^2 + 2x +2 y= e^(2x) y'= e^(2x) * 2 =2e^(2x) x=0 y=1 y'=2 (the same as if we had plugged it into the Ytan^2 equation) YTan= mx+b 1=2(0) +b b=1 YTan= 2x +1
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15:33:24 ** The local linearization is the tangent line. The line tangent to y = e^x at x = 0 is the line with slope y ' = e^x evaluated at x = 0, or slope 1. The line passes through (0, e^0) = (0, 1). The local linearization, or the tangent line, is therefore (y-1) = 1 ( x - 0) or y = x + 1. The line tangent to y = e^(2x) is y = 2x + 1. Thus near x = 0, since (e^x)^2 = e^(2x), we might expect to have (x + 1)^2 = 2x + 1. This is not exactly so, because (x + 1)^2 = x^2 + 2x + 1, not just 2x+1. However, near x = 0 we see that x^2 becomes insignificant compared to x (e.g., .001^ 2 = .000001), so for sufficiently small x we see that x^2 + 2x + 1 is as close as we wish to 2x + 1. **
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RESPONSE --> when I got my Ytan^2 I said in the answer(x + 1)^2 = x^2 + 2x + 2 but this was a typo, I had written down the correct answer. As long as x is very near 0 this equation is true.
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15:35:19 What do you get when you multiply the local linearization of e^x by itself, and in what sense is it consistent with the local linearization of e^(2x)? Which of the two expressions for e^(2x) is more accurate and why?
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RESPONSE --> you get: (x + 1)^2 = x^2 + 2x + 1 the actual tangent line for e(2x) is closer because it is the tangent line and it will be fairly close even when the x^2 has thrown off the other equation.
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15:37:02 ** The local linearization of e^(2x) is y = 2x + 1. The square of the local linearization 1 + x of e^x is y = (x + 1)^2 = x^2 + 2x + 1 . The two functions differ by the x^2 term. Near x = 0 the two graphs are very close, since if x is near 0 the value of x^2 will be very small. As we move away from x = 0 the x^2 term becomes more significant, giving the graph of the latter a slightly upward concavity, which for awhile nicely matches the upward concavity of y = e^(2x). The linear function cannot do this, so the square of the local linearization of e^x more closely fits the e^(2x) curve than does the local linearization of e^(2x). **
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RESPONSE --> okay I was wrong, because it isn't linear it will stay with the graph rather than travel away from it as the tangent line does.
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15:58:30 Query problem 3.9.22 (3d edition 3.9.12) T = 2 `pi `sqrt(L / g). How did you show that `dT = T / (2 L) * `dL?
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RESPONSE --> I don't understand how T / 2L is derived from T = 2 `pi `sqrt(L / g) I have the derivative as T'= .5L ^(-.5) and doesn't 'dt= T' * 'dL ?
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16:01:09 ** `sqrt(L/g) can be written as `sqrt(L) / `sqrt(g). It's a good idea to make this separation because L is variable, g is not. So dT / dL = 2 `pi / `sqrt(g) * [ d(`sqrt(L) ) / dL ]. [ d(`sqrt(L) ) / dL ] is the derivative of `sqrt(L), or L^(1/2), with respect to L. So [ d(`sqrt(L) ) / dL ] = 1 / 2 L^(-1/2) = 1 / (2 `sqrt(L)). Thus dT / dL = 2 `pi / [ `sqrt(g) * 2 `sqrt(L) ] = `pi / [ `sqrt(g) `sqrt(L) ] . This is the same as T / (2 L), since T / (2 L) = 2 `pi `sqrt(L / g) / (2 L) = `pi / (`sqrt(g) `sqrt(L) ). Now since dT / dL = T / (2 L) we see that the differential is `dT = dT/dL * `dL or `dT = T / (2 L) * `dL. **
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RESPONSE --> so the derivative was found by inplicit means instead of doing it as we have been, also it made it easier to understand when the 'sqrt function was broken down.
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16:03:11 If we wish to estimate length to within 2%, within what % must we know L?
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RESPONSE --> If you are estimating length withen 2% then you will know L to within 2% when it is estimated, as L gets larger then so will the descrepancy.
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16:04:59 ** If `dL = .02 L then `dT = T / (2 L) * .02 L = .02 T / 2 = .01 T. This tells us that to estimate T to within 1% we need to know L to within 2%. **
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RESPONSE --> I think that there might have been a typo in the prevous question, but when your estimating L within 2% yo must know T to within 1% because of the fraction in 'dT
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16:17:00 Query problem 4.7.4 (3d edition 4.8.9) graphs similar to -x^3 and x^3 at a. What is the sign of lim{a->a} [ f(x)/ g(x) ]? How do you know that the limit exists and how do you know that the limit has the sign you say it does?
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RESPONSE --> using L' Hopitals rule (-x^3)/(x^3) = (-3x^2) / (3x^2) lim{a->a} since the negative dominates then a will be negative.
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16:19:15 ** If one graph is the negative of the other, as appears to be the case, then for any x we would have f(x) / g(x) = -1. So the limit would have to be -1. It doesn't matter that at x = 0 we have 0 / 0, because what happens AT the limiting point doesn't matter, only what happens NEAR the limiting point, where 'nearness' is unlimited by always finite (i.e., never 0). **
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RESPONSE --> so the (-3x^2) / (3x^2) would have basically simplified to -1 / 1 or -1 which is our limit.
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16:23:14 Query 4.7.8 (3d edition 3.10.8) lim{x -> 0} [ x / (sin x)^(1/3) ].What is the given limit and how did you obtain it?
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RESPONSE --> x / (sinx) ^1/3) = x / (1/3 cosx ^ -2/3) = 3x^1/3 / x = 3 / x^1/3 as x-> 0
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16:25:08 As x -> 0 both numerator f(x) = x and the denominator g(x) = sin(x)^(1/3) both approach 0 as a limit. So we use l'Hopital's Rule f ' (x) = 1 and g ' (x) = 1/3 sin(x)^(-2/3), so f ' (x) / g ' (x) = 1 / (1/3 sin(x)^(-2/3) ) = 3 sin(x)^(2/3). Since as x -> 0 we have sin(x) -> 0 the limiting value of f ' (x) / g ' (x) is 0. It follows from l'Hopital's Rule that the limiting value of f(x) / g(x) is also zero.
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RESPONSE --> The book used a problem similar to this to show the ""dominance"" of a function, I understand what it means by ""Dominance"" but I don't understand exactly what indicates it using I'Hopital's Rule or when to use that indication. PLease Explain.
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16:30:56 What are the local linearizations of x and sin(x)^(1/3) and how do they allow you to answer the preceding question?
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RESPONSE --> The local linearizations would vary with regard to x but they would allow you to more easily compare sin(x)^(1/3) to x by extending a point on the graph.
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16:31:54 ** The local linearization of the numerator is just y = x. The denominator doesn't have a local linearization at 0; rather it approaches infinite slope. This means that as x -> 0 the ratio of the denominator function to the numerator function increases without bound, making the values of numerator / denominator approach zero. **
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RESPONSE --> okay
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16:33:34 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I am understand how to ""use"" I'Hoptals rule but I don't understand how to read it. although, I think that I have the parametric equations semi-undercontroll.
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16:35:13 I am confused on l'Hopital`s rule: How do you know when it can or cannot be used to evaluate a fn? I understand that f(a)=g(a)=0 and g'(x) cannot equal zero, but what are the other limitations? ** Those are the only limitations. Check that these conditions hold, then you are free to look at the limiting ratio f '(a) / g ' (a) of the derivatives. For example, on #18 the conditions hold for (a) (both limits are zero) but not for (b) (numerator isn't 0) and not for (c) (denominator doesn't have a limit). **
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RESPONSE --> this helps some, but I still need help with dominance. But it's basically the ratio of one to the other as they near the limit. right?
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