#$&* course MTH 173 12/8 17:00I am sending this again along with the form confirmation as it has not appeared on my access page since it was submitted previously.
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Given Solution: `a** The question was about the motion of the particle. The graph of f(t) has at every point a slope of magnitude 1, except at the points where the slope changes. The graph of g(t) has at every point a slope of magnitude 1, except at the points where the slope changes. At points where the slope changes, it changes 'instantly', so at those points the slope is not defined. More formally at these points the quantity (f(t + `dt) - f(t) ) / `dt, whose limit defines the integral, has a different limit when `dt approaches zero through positive values than when it approaches 0 through negative values, so the limit is not defined. At all times t, except where the slope changes sign, | f ' (t) | = 1. Sometimes f ' (x) is positive, sometimes negative. At all times t, except where the slope changes sign, | g ' (t) | = 4. Sometimes g ' (x) is positive, sometimes negative. Thus at all times t, except where the slope changes sign, the speed of the particle is v = sqrt( 1^2 + 4^2 ) = sqrt(17), about 4.1. At clock time t = 0 the x coordinate is f(0) = 2 and the y coordinate is g(0) = 0. At clock time t =0 1/2 the x coordinate is f(1/2) = 1.5 and the y coordinate is g(1/2) = 1. So between t = 0 and t = 1/2, the particle moves from (2, 0) to (1.5, 1). At clock time t = 1 the x coordinate is f(1) = 1 and the y coordinate is g( 1 ) = 0 . So between t = 1/2 and t = 1, the particle moves from (1.5, 1) to (1, 0). At clock time t = 1.5 the x coordinate is f(1.5) = .5 and the y coordinate is g(1.5) = -1. So between t = 1 and t = 1.5, the particle moves from (1,0) to (.5,-1). At clock time t = 2 the x coordinate is f(2) = 0 and the y coordinate is g(2) = 0. So between t = 1.5 and t = 2, the particle moves from (.5, -1) to (0, 0). At clock time t = 2.5 the x coordinate is f(2.5) = .5 and the y coordinate is g(2.5) = 1. So between t = 2 and t = 2.5, the particle moves from (0, 0) to (.5, 1). During the next few half-second intervals the particle moves to (1, 0), (1.5, -1) and (2, 0) The first graph is of position vs. time, i.e., x vs. t for a particle moving on the x axis. The slope of the x position vs. time graph is the x component of 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 = 0, which occurs at t = 0 the slope becomes +1, indicating that the velocity of the particle instantaneously changes from -1 to +1 and the particle moves back off to the right. During the 4-second interval the x position therefore changes from x = 2 to x = 0 then back to x = 2. 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, and twice more between t = 2 and t = 4. At these points velocity goes from positive to negative or from negative to positive. So between t = 0 and t = 2, the y component of the position goes from 0 to 1 and back to 0, then to -1 and back up to 0. This occurs as x goes from 2 to 0, so the position of the particle zigzags from (2, 0) to (3/2, 1) to (1, 0) to (1/2, -1) to (0, 0). Between t = 2 and t = 4 the x coordinate moves back to the right, and the particle's path zigzags from (0, 0) to (1/2, 1) to (1, 0) to (3/2, -1) and back to the starting point at (2, 0). Since the x velocity is always 4 times the magnitude of the y velocity (except at the 'turning points' where velocity is not defined), the paths are straight lines. Another way of saying this is that the instantaneous speed, at all points where f ' and g ' are defined, is always v = sqrt( f ' ^2+ g ' ^2 ) = sqrt( 4^2 + 1^2) = sqrt(17) = 4.1, approx., as also specified earlier in this solution. The two paths form a shape on the graph that looks like two diamond shapes. The speed of the particle as it goes from point to point is always 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given ellipse centered (0,0), passing through (+-5, 0) and (0, +-7) The given ellipse is like a circle passing through (+-5,0) and (0,+-5), that is stretched along th y axis with a factor of (7/5). And since the farthest point on the x axis is (+-5,0), thus the x co-ordinate will oscillate between +-5, and thus x = 5cos(t) And since the farthest points on the y axis is (+-7,0), thus the y co-ordinate will oscillate between +-7, and thus y = 7sin(t) Thus the parameterization for the ellipse will be x = 5 cos(t) and y = 7sin(t), where t belongs to the interval [0,2pi) This is because the value of the function at 2pi will be the same as that at t = 0, that is the particle returns back to its original position when t = 2pi. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Given parameters x = t^3 - t, y = t^2 and t = 2 The co-ordinate of the particle at t = 2 will be x(2) = 8 - 2 = 4 and y(2) = 4, thus the co-ordinate will be (6,4) dx/dt = 3t^2 - 1 and dy/dt = 2t thus the slope of the curve traced by the particle will be = dy/dt / dx/dt slope of the curve traced by the particle = 2t / (3t^2 - 1) the slope of the curve at t = 2 will be = 4/11 the tangent to the curve will pass through (6,4) and will have a slope of 4/11 the equation will be (y - 4)/(x - 6) = 4/11 11y - 44 = 4x - 24 11y = 4x + 20 y = (4x + 20)/11 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Derivatives are dx/dt = 3t^2 - 1, which at t = 2 is dx/dt = 11, and dy/dt = 2t, which at t=2 is 4. We have x = 6 + 11 t, which solved for t gives us t = (x - 6) / 11, and y = 4 + 4 t. Substituting t = (x-6)/11 into y = 4 + 4 t we get y = 4 + 4(x-6)/11 = 4/ll x + 20/11. Note that at t = 2 you get x = 6 so y = 4/11 * 6 + 20/11 = 44/11 = 4. Alternatively: The slope at t = 2 is dy/dx = dy/dt / (dx/dt) = 4 / 11. The equation of the line thru (6, 4) with slope 4/11 is y - 4 = 4/11 ( x - 6), which simplifies to y = 4/11 x + 20/11. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 4.8.12 (3d edition 3.8.22). x = cos(t^2), y = sin(t^2).What is the speed of the particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given parameters x = cos(t^2) and y = sin(t^2) v-x = speed in x direction = dx/dt = (cos(t^2))’ = -2tsin(t^2) v-y = speed in y direction = dy/dt = (sin(t^2))’ = 2tcos(t^2) thus the speed of the particle will be v = sqrt((v-x)^2 + (v-y)^2) v = sqrt(4t^2sin^2(t^2) + 4t^2cos^2(t^2)) v = sqrt(4t^2) = |2t| speed of the particle at t = 2 will be 2|t| confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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). Since sin^2(theta) + cos^2(theta) = 1 for any theta, this is so for theta = t^2 and the expression simplifies to 2 | t |. The speed at clock time t is 2 | t |. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qDoes the particle ever come to a stop? If so when? If not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The particle will come to stop when v-x = 0 and v-y = 0. Since v-x = -2tsin(t^2), v-x will be zero when t = 0 or when sin(t^2) = 0 sin(t^2) will be zero when t^2 = n(pi), where n is an integer v-y = 2tcos(t^2), v-y will be zero when t = 0 or when cos(t^2) = 0 cos(t^2) will be zero when t^2 = n(pi/2) where n is an integer Thus, we see that v-x and v-y will be 0 together when t = 0 Thus, the particle will stop at only one point, that is it will be at rest at t = 0. For all other t values, the particle will at least some finite v-x or v-y velocity and thus will keep moving and will not be at rest ever after. Thus we see that the particle starts from rest, that is 0 velocity at t = 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The particle isn't moving when v = 0. v = 2 | t | { sin^2(t^2) + cos^2(t^2) }^(1/2) is zero when t = 0 or when sin^2(t^2) + cos^2(t^2) = 0. However sin^2(t^2) + cos^2(t^2) = 1 for all values of t, so this does not occur. t = 0 gives x = cos(0) = 1 and y = sin(0) = 0, so it isn't moving at (1, 0). The particle is at rest at t = 0, and only at t = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.9.33 was 3.9.18 (3d edition 3.9.8) (was 4.8.20) square the local linearization of e^x at x=0 to obtain the approximate local linearization of e^(2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y = e^x The value of the function at x = 0 will be y(0) = e^0 = 1 The derivative of the function is y’ = e^x The derivative of the function at x = 0 will be y’(0) = 1 Thus the tangent to the curve at x = 0 will pass through the point (0,1) and will have a slope of 1 Equation of the tangent will be (y - 1)/(x - 0) = 1 Thus the equation of the tangent will be y = x + 1 Thus the approximate local linearization of e^x near x = 0 will be e^x = x + 1 squaring this approximation for the approximate local linearization of e^(2x) we get e^2x approximately = (x + 1)^2 = x^2 + 2x + 1 near x = 0, the value of x^2 becomes insignificant, making x^2 + 2x + 1 almost equal to 2x + 1 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT COMMENT I’m not sure I understood the question to be answered. INSTRUCTOR RESPONSE You were asked to square the local linearization The local linearization of e^x is 1 + x, which when squared is 1 + 2 x + x^2. The local linearization of e^(x^2) is 1 + 2 x. The squared local linearization of e^x is 1 + 2 x + x^2, the local linearization of e^(x^2) is 1 + 2 x. Near x = 0 the quantity x^2 becomes insignificant, so near x = 0 the two expressions are nearly identical. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Local linearization of e^x, is e^x = x + 1 When we multiply the local linearization with itself we get e^2x = (x + 1)^2 = x^2 + 2x + 1 The actual local linearization of e^2x is e^2x = 2x + 1 We see that the actual local linearization of e^2x differs from the local linearization of e^2x obtained by squaring the local linearization of e^x by a factor of x^2. Since near x = 0, the value of x^2 will be very small and will not be significant and thus can be ignored. This proves the consistency with the local linearization of e^(2x). However the first expression, that is the local linearization of e^x multiplied to itself is more accurate than the local linearization of e^(2x). This is because, e^(2x) is a concave up curve, whereas the local linearization of e^(2x) is a straight line quickly moving away from the graph. On the other side x^2 + 2x + 1 is also a concave up curve and will thus move away from the graph at a much slower rate as compared to the straight line. Since x^2 does not have a high considerable value for x values near x = 0, thus the value of the (x^2 + 2x + 1) will be equal to the value of (2x + 1), which supplements the accuracy of (x^2 + 2x + 1) 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). ** STUDENT COMMENT I see what I’ve been missing. The question is which tang line follows the actual curve more accurate and I see why this whould be the y =e^x local linearization squared. Would this not have something to do with adding the factor exponential growth to the local linearization, of a exponential Fn? INSTRUCTOR COMMENT: Good speculation. The exponential function, by definition, grows exponentially. However we cannot calculate the exact value of an exponential function at a given numerical value of x, except at x = 0. The point here is to find approximate values of the exponential function, first by a local linearization, then by adding a quadratic term to help us follow the curvature of the exponential graph. Later (2d semester) we see that we can keep adding high-power terms (e.g., multiples of x^3, then of x^4, etc.) we can approximate the effects of higher and higher derivatives to obtain functions that approximate the exponential better and better. Our approximation functions will be polynomials, which can be evaluated accurately, and we will obtain expressions for just how accurate we can expect the approximations obtained from these polynomials to be. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.9.21 (3d edition 3.9.12) T = 2 `pi `sqrt(L / g). How did you show that `dT = T / (2 L) * `dL? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given T = 2pi*(sqrt(L/g)) Here pi and g are constant, and thus T becomes a function of L Thus T = (2pi/sqrt(g))*(sqrt(L)) Differentiating both sides we get dT/dL = d((2pi/sqrt(g))*(sqrt(L)))/dL dT/dL = (2pi/sqrt(g))*(sqrt(L))’ Using chain rule of differentiation we get dT/dL = (2pi/sqrt(g))*(1/2)*(L^-1/2) dT/dL = pi/(sqrt(g)*sqrt(L))…………………………..(2) Since T = (2pi/sqrt(g))*(sqrt(L)) Thus pi/sqrt(g) = T/(2sqrt(L)) ………….……………..(1) Substituting (1) in (2) we get dT/dL = T/(2*sqrt(L)*sqrt(L)) = T/(2L) Thus dT = (T/(2L))*(dL) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In this problem you are referred to the equation T = 2 pi sqrt( g / L). and told that g remains constant while L varies. You are asked to find an expression for the resulting change in T. Within this context the given equation gives T as a function of L. The rate of change of T with respect to L is the derivative dT/dL. Roughly speaking, when you multiply dT / dL by a small change in L, you are multiplying the rate of change of T by the change in L, which gives you the approximate change in T. Another way of saying this is dT/dL * `dL = `dT, where `dT is the linear approximation of the change in T. The main thing we need to calculate is dT / dL: To start with, `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. ** STUDENT QUESTION I am having problems with problems that pertain to proving something equals something else were in the notes can I go back and review this concept? INSTRUCTOR RESPONSE Applications occur throughout this text, and there is no one approach to solving application problems. The best results come from a lot of practice, and from feedback you get on your solutions. This problem is an application of the differential. The only way to learn how to apply the differential is by first understanding what the differential means, in general. Then, to be a bit repetitive, I'll say again that understanding comes with extensive practice. I suggest that you give #22 another shot after looking over the solution below. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK- ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qIf we wish to estimate period to within 2%, within what % must we know L? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the previous question we obtained that dT = (T/(2L))*(dL) Rearranging this we get dT/T = (1/2)*(dL/L) period with 2% indicates that dT/T should be less than 0.02 if dT/T is less than 0.02, (1/2)*(dL/L) should also be less than 0.02 This indicates that dL/L should be less than 0.02*2 = 0.04. Thus if we wish to estimate period to within 2%, within 4% must we know L. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** `dT = T / (2 L) * `dL can be rearranged to the form `dT / T = 1/2 `dL / L. If we want `dT / T to be less than .02, then 1/2 `dL / L must be less than .02, so `dL / L must be less than .04. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 4.7.4 (3d edition 4.8.9) graphs similar to -x^3 and x^3 at a. What is the sign of lim{x->a} [ f(x)/ g(x) ]? How do you know that the limit exists and how do you know that the limit has the sign you say it does? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Whenever we have a function negative of the other, the ratio is always going to be equal to -1 and no matter what the limiting value is, the value is always going to be -1. Here it does not matter, whether the ratio turns out to be a 0/0, for x tending to a, because we have the same functions and what actually matters is what happens in the vicinity of that point. Thus even if the funciton is not defined at a, the ratio of the values of the function near x = a is going to be finite and equal to -1, since the functions are negative of each other. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). ** GOOD STUDENT COMMENT The limiting pt isn’t as important as what is adjacent to the limiting pt. It can’t be 0. INSTRUCTOR RESPONSE Good. To put it even more strongly: The actual point isn't important at all; only the neighborhood of the point (the region adjacent to the point, as you put it nicely) is relevant to the limit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given lim x tending to 0 (x / (sinx)^(1/3)) Let f(x) = x and g(x) = sin(x)^(1/3) Thus we have lim x tending to 0 f(x)/g(x) Since f(0) = 0 and g(0) = 0, L-hospital is applicable Using the rules of L-hospital we get lim x tending to 0 f(x)/g(x) = lim x tending to o f’(x)/g’(x) lim x tending to 0 1 / ((1/3)sin(x)^(-2/3)cos(x)) = 0 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 cos(x) * sin(x)^(-2/3), so f ' (x) / g ' (x) = 1 / (1/3 cos(x) * sin(x)^(-2/3) ) = 3 sin(x)^(2/3) / cos(x). 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. STUDENT QUESTION I am having trouble with defining the limiting value of the problem. I want to say it is less than zero but the answer says it is zero. Can you explain were I might have this wrong?????
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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. The derivative of the denominator sin(x)^(1/3) is cos(x) * 1/3 sin(x^(-2/3) ) = 1/3 * cos(x) / (sin(x))^(2/3). As x approaches zero sin(x) approaches zero, and so does the 2/3 power of sin(x). So the denominator of the derivative approaches zero, so the derivative cannot be evaluated at x = 0. Thus sin(x)^(1/3) does not have a local linearization at x = 0. Its derivative approaches infinity as x approaches 0. So 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: OK ********************************************* 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: No comment or any surprise or insights confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). ** "" Form Confirmation Thank you for submitting the following information: identifyingInfo: Submitting Assignment yourCourse: MTH 173 Contact_FirstName: Kevin Contact_LastName: Shah firstPendLength: query_21 email: krs2070@email.vccs.edu S1 12/2 2:30 response "" If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 021. `query 21 Question: `q**** Query Problem 4.8.3 was 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. Your solution: The first graph is the graph for the x component of the velocity. The graph of the x co-ordinate is a V shaped curve which has a slope of magnitude of 1 for all points except the points at which the slope changes. Thus the speed with which the x co-ordinate changes is thus 1. The y co-ordinate graph is a saw-tooth shape graph with a period of 2 and |y| <= 2, thus the magnitude of slope of the saw-tooth graph will be 4 for all values of t except the points where the slope changes and thus the speed at which the y co-ordinate changes will be 4. Thus the speed of the particle will always be sqrt(16 + 1) = sqrt(17) = 4.123 approximately. At t = 0 the particle has a x coordinate of 2 and y coordinate 0, thus it starts from (2,0). At t = ½ the particle has a x coordinate of 1.5 and y coordinate 2, thus it’s coordinate is (1.5,2) At t = 1 the particle has a x coordinate of 1 and y coordinate 0, thus it’s coordinate is (1,0) At t = 1.5 the particle has a x coordinate of 0.5 and y coordinate -2, thus it’s coordinate is (0.5,-2) At t = 2 the particle has a x coordinate of 0 and y coordinate 0, thus it’s coordinate is (0,0) At t = 2.5 the particle has a x coordinate of 0.5and y coordinate 2, thus it’s coordinate is (0.5,2) At t = 3 the particle has a x coordinate of 1 and y coordinate 0, thus it’s coordinate is (1,0) At t = 3.5 the particle has a x coordinate of 1.5 and y coordinate -2, thus it’s coordinate is (1.5,-2) At t = 4 the particle has a x coordinate of 2 and y coordinate 0, thus it’s coordinate is (2,0) Thus we see that the graph has a slope of sqrt(17) for all x values, except the points at which the slope changes. It starts from (2,0) moves to (1.5,2), to (1,0), to (0.5, -2), to (0,0), to (0.5, 2), to (1, 0), to (1.5, -2), and finally to (2,0). The graph appears to be two diamond shape graphs attached by the point (1,0) along the positive x axis with y <= |2|. From the graph we thus see that the x co-ordinate has a negative slope and thus it decreases from x = 2 to x = 0, and the y co-ordinate has a positive slope from x = 0 to x = ½ and negative from x = ½ to x = 1.5 and again positive slope from x = 1.5 to x = 2. Thus the y coordinate changes from 0 to 2 and then decreases to - 2 and then again increases to 0. For the remaining time the x co-ordinates has a positive slope and thus increases and thus moves right from x = 0 to x = 2. The y co-ordinate behaves the same way it behaved for the previous x values. Thus the graph of the particle will also be a replica of the previous graph with the only difference that it will now be moving from left to right. Confidence Assessment: 3 Given Solution: `a** The question was about the motion of the particle. The graph of f(t) has at every point a slope of magnitude 1, except at the points where the slope changes. The graph of g(t) has at every point a slope of magnitude 1, except at the points where the slope changes. At points where the slope changes, it changes 'instantly', so at those points the slope is not defined. More formally at these points the quantity (f(t + `dt) - f(t) ) / `dt, whose limit defines the integral, has a different limit when `dt approaches zero through positive values than when it approaches 0 through negative values, so the limit is not defined. At all times t, except where the slope changes sign, | f ' (t) | = 1. Sometimes f ' (x) is positive, sometimes negative. At all times t, except where the slope changes sign, | g ' (t) | = 4. Sometimes g ' (x) is positive, sometimes negative. Thus at all times t, except where the slope changes sign, the speed of the particle is v = sqrt( 1^2 + 4^2 ) = sqrt(17), about 4.1. At clock time t = 0 the x coordinate is f(0) = 2 and the y coordinate is g(0) = 0. At clock time t =0 1/2 the x coordinate is f(1/2) = 1.5 and the y coordinate is g(1/2) = 1. So between t = 0 and t = 1/2, the particle moves from (2, 0) to (1.5, 1). At clock time t = 1 the x coordinate is f(1) = 1 and the y coordinate is g( 1 ) = 0 . So between t = 1/2 and t = 1, the particle moves from (1.5, 1) to (1, 0). At clock time t = 1.5 the x coordinate is f(1.5) = .5 and the y coordinate is g(1.5) = -1. So between t = 1 and t = 1.5, the particle moves from (1,0) to (.5,-1). At clock time t = 2 the x coordinate is f(2) = 0 and the y coordinate is g(2) = 0. So between t = 1.5 and t = 2, the particle moves from (.5, -1) to (0, 0). At clock time t = 2.5 the x coordinate is f(2.5) = .5 and the y coordinate is g(2.5) = 1. So between t = 2 and t = 2.5, the particle moves from (0, 0) to (.5, 1). During the next few half-second intervals the particle moves to (1, 0), (1.5, -1) and (2, 0) The first graph is of position vs. time, i.e., x vs. t for a particle moving on the x axis. The slope of the x position vs. time graph is the x component of 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 = 0, which occurs at t = 0 the slope becomes +1, indicating that the velocity of the particle instantaneously changes from -1 to +1 and the particle moves back off to the right. During the 4-second interval the x position therefore changes from x = 2 to x = 0 then back to x = 2. 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, and twice more between t = 2 and t = 4. At these points velocity goes from positive to negative or from negative to positive. So between t = 0 and t = 2, the y component of the position goes from 0 to 1 and back to 0, then to -1 and back up to 0. This occurs as x goes from 2 to 0, so the position of the particle zigzags from (2, 0) to (3/2, 1) to (1, 0) to (1/2, -1) to (0, 0). Between t = 2 and t = 4 the x coordinate moves back to the right, and the particle's path zigzags from (0, 0) to (1/2, 1) to (1, 0) to (3/2, -1) and back to the starting point at (2, 0). Since the x velocity is always 4 times the magnitude of the y velocity (except at the 'turning points' where velocity is not defined), the paths are straight lines. Another way of saying this is that the instantaneous speed, at all points where f ' and g ' are defined, is always v = sqrt( f ' ^2+ g ' ^2 ) = sqrt( 4^2 + 1^2) = sqrt(17) = 4.1, approx., as also specified earlier in this solution. The two paths form a shape on the graph that looks like two diamond shapes. The speed of the particle as it goes from point to point is always 4. ** Self-critique (if necessary): OK Self-critique Rating: OK 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. Your solution: Given ellipse centered (0,0), passing through (+-5, 0) and (0, +-7) The given ellipse is like a circle passing through (+-5,0) and (0,+-5), that is stretched along th y axis with a factor of (7/5). And since the farthest point on the x axis is (+-5,0), thus the x co-ordinate will oscillate between +-5, and thus x = 5cos(t) And since the farthest points on the y axis is (+-7,0), thus the y co-ordinate will oscillate between +-7, and thus y = 7sin(t) Thus the parameterization for the ellipse will be x = 5 cos(t) and y = 7sin(t), where t belongs to the interval [0,2pi) This is because the value of the function at 2pi will be the same as that at t = 0, that is the particle returns back to its original position when t = 2pi. Confidence Assessment: 3 Given Solution: 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): OK Self-critique Rating: OK 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? Your solution: Given parameters x = t^3 - t, y = t^2 and t = 2 The co-ordinate of the particle at t = 2 will be x(2) = 8 - 2 = 4 and y(2) = 4, thus the co-ordinate will be (6,4) dx/dt = 3t^2 - 1 and dy/dt = 2t thus the slope of the curve traced by the particle will be = dy/dt / dx/dt slope of the curve traced by the particle = 2t / (3t^2 - 1) the slope of the curve at t = 2 will be = 4/11 the tangent to the curve will pass through (6,4) and will have a slope of 4/11 the equation will be (y - 4)/(x - 6) = 4/11 11y - 44 = 4x - 24 11y = 4x + 20 y = (4x + 20)/11 Confidence Assessment: 3 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): OK Self-critique Rating: OK 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? Your solution: Given parameters x = cos(t^2) and y = sin(t^2) v-x = speed in x direction = dx/dt = (cos(t^2))’ = -2tsin(t^2) v-y = speed in y direction = dy/dt = (sin(t^2))’ = 2tcos(t^2) thus the speed of the particle will be v = sqrt((v-x)^2 + (v-y)^2) v = sqrt(4t^2sin^2(t^2) + 4t^2cos^2(t^2)) v = sqrt(4t^2) = |2t| speed of the particle at t = 2 will be 2|t| Confidence Assessment: 3 Given Solution: `a 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). Since sin^2(theta) + cos^2(theta) = 1 for any theta, this is so for theta = t^2 and the expression simplifies to 2 | t |. The speed at clock time t is 2 | t |. ** Self-critique (if necessary): OK Self-critique Rating: OK Question: `qDoes the particle ever come to a stop? If so when? If not why not? Your solution: The particle will come to stop when v-x = 0 and v-y = 0. Since v-x = -2tsin(t^2), v-x will be zero when t = 0 or when sin(t^2) = 0 sin(t^2) will be zero when t^2 = n(pi), where n is an integer v-y = 2tcos(t^2), v-y will be zero when t = 0 or when cos(t^2) = 0 cos(t^2) will be zero when t^2 = n(pi/2) where n is an integer Thus, we see that v-x and v-y will be 0 together when t = 0 Thus, the particle will stop at only one point, that is it will be at rest at t = 0. For all other t values, the particle will at least some finite v-x or v-y velocity and thus will keep moving and will not be at rest ever after. Thus we see that the particle starts from rest, that is 0 velocity at t = 0. Confidence Assessment: 3 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. However sin^2(t^2) + cos^2(t^2) = 1 for all values of t, so this does not occur. t = 0 gives x = cos(0) = 1 and y = sin(0) = 0, so it isn't moving at (1, 0). The particle is at rest at t = 0, and only at t = 0. Self-critique (if necessary): OK Self-critique Rating: OK Question: `qQuery problem 3.9.33 was 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) Your solution: Given function y = e^x The value of the function at x = 0 will be y(0) = e^0 = 1 The derivative of the function is y’ = e^x The derivative of the function at x = 0 will be y’(0) = 1 Thus the tangent to the curve at x = 0 will pass through the point (0,1) and will have a slope of 1 Equation of the tangent will be (y - 1)/(x - 0) = 1 Thus the equation of the tangent will be y = x + 1 Thus the approximate local linearization of e^x near x = 0 will be e^x = x + 1 squaring this approximation for the approximate local linearization of e^(2x) we get e^2x approximately = (x + 1)^2 = x^2 + 2x + 1 near x = 0, the value of x^2 becomes insignificant, making x^2 + 2x + 1 almost equal to 2x + 1 Confidence Assessment: 3 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. ** STUDENT COMMENT I’m not sure I understood the question to be answered. INSTRUCTOR RESPONSE You were asked to square the local linearization The local linearization of e^x is 1 + x, which when squared is 1 + 2 x + x^2. The local linearization of e^(x^2) is 1 + 2 x. The squared local linearization of e^x is 1 + 2 x + x^2, the local linearization of e^(x^2) is 1 + 2 x. Near x = 0 the quantity x^2 becomes insignificant, so near x = 0 the two expressions are nearly identical. Self-critique (if necessary): OK Self-critique Rating: OK 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? Your solution: Local linearization of e^x, is e^x = x + 1 When we multiply the local linearization with itself we get e^2x = (x + 1)^2 = x^2 + 2x + 1 The actual local linearization of e^2x is e^2x = 2x + 1 We see that the actual local linearization of e^2x differs from the local linearization of e^2x obtained by squaring the local linearization of e^x by a factor of x^2. Since near x = 0, the value of x^2 will be very small and will not be significant and thus can be ignored. This proves the consistency with the local linearization of e^(2x). However the first expression, that is the local linearization of e^x multiplied to itself is more accurate than the local linearization of e^(2x). This is because, e^(2x) is a concave up curve, whereas the local linearization of e^(2x) is a straight line quickly moving away from the graph. On the other side x^2 + 2x + 1 is also a concave up curve and will thus move away from the graph at a much slower rate as compared to the straight line. Since x^2 does not have a high considerable value for x values near x = 0, thus the value of the (x^2 + 2x + 1) will be equal to the value of (2x + 1), which supplements the accuracy of (x^2 + 2x + 1) Confidence Assessment: 3 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). ** STUDENT COMMENT I see what I’ve been missing. The question is which tang line follows the actual curve more accurate and I see why this whould be the y =e^x local linearization squared. Would this not have something to do with adding the factor exponential growth to the local linearization, of a exponential Fn? INSTRUCTOR COMMENT: Good speculation. The exponential function, by definition, grows exponentially. However we cannot calculate the exact value of an exponential function at a given numerical value of x, except at x = 0. The point here is to find approximate values of the exponential function, first by a local linearization, then by adding a quadratic term to help us follow the curvature of the exponential graph. Later (2d semester) we see that we can keep adding high-power terms (e.g., multiples of x^3, then of x^4, etc.) we can approximate the effects of higher and higher derivatives to obtain functions that approximate the exponential better and better. Our approximation functions will be polynomials, which can be evaluated accurately, and we will obtain expressions for just how accurate we can expect the approximations obtained from these polynomials to be. Self-critique (if necessary): OK Self-critique Rating: OK Question: `qQuery problem 3.9.21 (3d edition 3.9.12) T = 2 `pi `sqrt(L / g). How did you show that `dT = T / (2 L) * `dL? Your solution: Given T = 2pi*(sqrt(L/g)) Here pi and g are constant, and thus T becomes a function of L Thus T = (2pi/sqrt(g))*(sqrt(L)) Differentiating both sides we get dT/dL = d((2pi/sqrt(g))*(sqrt(L)))/dL dT/dL = (2pi/sqrt(g))*(sqrt(L))’ Using chain rule of differentiation we get dT/dL = (2pi/sqrt(g))*(1/2)*(L^-1/2) dT/dL = pi/(sqrt(g)*sqrt(L))…………………………..(2) Since T = (2pi/sqrt(g))*(sqrt(L)) Thus pi/sqrt(g) = T/(2sqrt(L)) ………….……………..(1) Substituting (1) in (2) we get dT/dL = T/(2*sqrt(L)*sqrt(L)) = T/(2L) Thus dT = (T/(2L))*(dL) Confidence Assessment: 3 Given Solution: `a** In this problem you are referred to the equation T = 2 pi sqrt( g / L). and told that g remains constant while L varies. You are asked to find an expression for the resulting change in T. Within this context the given equation gives T as a function of L. The rate of change of T with respect to L is the derivative dT/dL. Roughly speaking, when you multiply dT / dL by a small change in L, you are multiplying the rate of change of T by the change in L, which gives you the approximate change in T. Another way of saying this is dT/dL * `dL = `dT, where `dT is the linear approximation of the change in T. The main thing we need to calculate is dT / dL: To start with, `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. ** STUDENT QUESTION I am having problems with problems that pertain to proving something equals something else were in the notes can I go back and review this concept? INSTRUCTOR RESPONSE Applications occur throughout this text, and there is no one approach to solving application problems. The best results come from a lot of practice, and from feedback you get on your solutions. This problem is an application of the differential. The only way to learn how to apply the differential is by first understanding what the differential means, in general. Then, to be a bit repetitive, I'll say again that understanding comes with extensive practice. I suggest that you give #22 another shot after looking over the solution below. Self-critique (if necessary): OK- Self-critique Rating: OK Question: `qIf we wish to estimate period to within 2%, within what % must we know L? Your solution: From the previous question we obtained that dT = (T/(2L))*(dL) Rearranging this we get dT/T = (1/2)*(dL/L) period with 2% indicates that dT/T should be less than 0.02 if dT/T is less than 0.02, (1/2)*(dL/L) should also be less than 0.02 This indicates that dL/L should be less than 0.02*2 = 0.04. Thus if we wish to estimate period to within 2%, within 4% must we know L. Confidence Assessment: 3 Given Solution: `a** `dT = T / (2 L) * `dL can be rearranged to the form `dT / T = 1/2 `dL / L. If we want `dT / T to be less than .02, then 1/2 `dL / L must be less than .02, so `dL / L must be less than .04. Self-critique (if necessary): OK Self-critique Rating: OK 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{x->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? Your solution: Whenever we have a function negative of the other, the ratio is always going to be equal to -1 and no matter what the limiting value is, the value is always going to be -1. Here it does not matter, whether the ratio turns out to be a 0/0, for x tending to a, because we have the same functions and what actually matters is what happens in the vicinity of that point. Thus even if the funciton is not defined at a, the ratio of the values of the function near x = a is going to be finite and equal to -1, since the functions are negative of each other. Confidence Assessment: 3 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). ** GOOD STUDENT COMMENT The limiting pt isn’t as important as what is adjacent to the limiting pt. It can’t be 0. INSTRUCTOR RESPONSE Good. To put it even more strongly: The actual point isn't important at all; only the neighborhood of the point (the region adjacent to the point, as you put it nicely) is relevant to the limit. Self-critique (if necessary): OK Self-critique Rating: OK Question: `qQuery 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? Your solution: Given lim x tending to 0 (x / (sinx)^(1/3)) Let f(x) = x and g(x) = sin(x)^(1/3) Thus we have lim x tending to 0 f(x)/g(x) Since f(0) = 0 and g(0) = 0, L-hospital is applicable Using the rules of L-hospital we get lim x tending to 0 f(x)/g(x) = lim x tending to o f’(x)/g’(x) lim x tending to 0 1 / ((1/3)sin(x)^(-2/3)cos(x)) = 0 Confidence Assessment: 3 Given Solution: `a 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 cos(x) * sin(x)^(-2/3), so f ' (x) / g ' (x) = 1 / (1/3 cos(x) * sin(x)^(-2/3) ) = 3 sin(x)^(2/3) / cos(x). 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. STUDENT QUESTION I am having trouble with defining the limiting value of the problem. I want to say it is less than zero but the answer says it is zero. Can you explain were I might have this wrong?????