course Mth 173 4/5 11 Question: `q**** 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.
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Given Solution: `a** 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. ** 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. 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Looked at given solution to get problem/solution and solve. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery 4.8.23 (was 3.8.18). x = t^3 - t, y = t^2, t = 2.What is the equation of the tangent line at t = 2 and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x’ = 3t^2 – 1 y’ = 2t x’(2) = 11 y’(2) = 4 m = 4/11 x = (2)^3 – 2 = 6 x = 6, y = 4, m = 4/11 y = 4/11 + b *I’m not sure how to solve for “b”. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): x = 6 + 11 t t = (x - 6) / 11 and y = 4 + 4 t y = 4 + 4(x-6)/11 = 4/ll x + 20/11 (b) = 20/11 ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery 4.8.12 (3d edition 3.8.22). x = cos(t^2), y = sin(t^2).What is the speed of the particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x’ = -2t sin(t^2) y’ = 2t cos(t^2) sqrt (vx^2 + vy^2) ‘sqrt (-2t sin(t^2)^2) + (2t cos(t^2)^2) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Answer not simplified, ={-4t^2 sin^2(t^2) + 4 t^2 cos^2(t^2) } ^(1/2) = (4t^2)^(1/2) { -sin^2(t^2) + cos^2(t^2) }^(1/2) = [2 | t | { -sin^2(t^2) + cos^2(t^2) }^(1/2)] ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qDoes the particle ever come to a stop? If so when? If not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, at (0, 0) because it is stopped at cos(0) and sin(0). confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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, ... . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): UGHHH!! I rushed it and forgot that cos(0) = 1. It comes to a complete stop at (1, 0). ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* Question: `qQuery problem 3.9.18 (3d edition 3.9.8) (was 4.8.20) square the local linearization of e^x at x=0 to obtain the approximate local linearization of e^(2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = e^x y’ = e^x (0, 1) y = (1)x + 1 y = e^(2x) y’ = 2x [2x] confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): y’ = e^2x = 2x + 1 ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* Question: `qWhat do you get when you multiply the local linearization of e^x by itself, and in what sense is it consistent with the local linearization of e^(2x)? Which of the two expressions for e^(2x) is more accurate and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: e^2x = 2x + 1 e^x of x + 1 = x^2 + 2x + 1 e^2x is more accurate the further away you get from 0. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery problem 3.9.22 (3d edition 3.9.12) T = 2 `pi `sqrt(L / g). How did you show that `dT = T / (2 L) * `dL? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2’pi(sqrt(L) / sqrt(g)) I do not know how to show this. confidence rating #$&* 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** `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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): `dT = dT/dL * `dL or `dT = T / (2 L) * `dL. ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* Question: `qIf we wish to estimate length to within 2%, within what % must we know L? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2% = .02 .02L/2 = .01T confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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%. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery problem 4.7.4 (3d edition 4.8.9) graphs similar to -x^3 and x^3 at a. What is the sign of lim{a->a} [ f(x)/ g(x) ]? How do you know that the limit exists and how do you know that the limit has the sign you say it does? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The limit is -1. Not sure if (0) is a limiting pt. confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The limiting pt isn’t as important as what is adjacent to the limiting pt. It can’t be 0.
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Given Solution: `aAs 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): = 1 / (1/3 sin(x)^(-2/3) ) .. in this step sin(x) is raised to the negative 2/3, not positive as I have it. ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* Question: `qWhat are the local linearizations of x and sin(x)^(1/3) and how do they allow you to answer the preceding question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: They are proportional in the fact that the higher the denominator is, the close the pt is to 0. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. I am confused on l'Hopital`s rule: How do you know when it can or cannot be used to evaluate a fn? I understand that f(a)=g(a)=0 and g'(x) cannot equal zero, but what are the other limitations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I understand that f(a)=g(a)=0 and g'(x) cannot equal zero, but what are the other limitations? confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* "