Assignment 21

course Mth 173

????????????assignment #020

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Calculus I

04-12-2008

?JwC??????K??assignment #021

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Calculus I

04-12-2008

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11:22:14

**** 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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When t = 0, the particle is at (0, 1). For the first unit of time the particle moves from x = 0 to x = 1 and from y = 1 to y = 0. It then drops one unit x for each unit t until t = 3 and x = -1; from t = 1 to t = 3 the particle goes from y = 0 to y = -1 and then y = 0. The particle again reaches x = 0 and y = 1 when t = 4.

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11:22:16

** 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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11:22:53

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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When x = cos(t) and y = sin(t) we have (cos(t))^2 + (sin(t))^2 = 1.

If the ellipse goes through (+-5, 0) and (0, +-7) and is centered at the origin we have the parameters for this ellipse being x = 5 cos(t) and y = 7 sin(t)

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11:22:59

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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11:25:00

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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I am using your response to this question as an example for my notes, and I will practice on the other textbook questions.

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11:33:54

** 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.

The tangent line to the graph of x vs. t through the t = 2 point (2, 6) would have slope 11; the tangent line would therefore be x = -16 + 11 t.

The tangent line to the y vs. t graph at the t = 2 point (2, 4) would have slope 4; the tangent line would therefore be y = -4 + 4 t.

We construct a graph of the y vs. x tangent line from the x vs. t and y vs. t tangent lines. The resulting linear function can be found by eliminating t from the equations of the x vs. t and y vs. t tangent lines.

Solving x = -16 + 11 t for t we get t = (x + 16) / 11.

Substituting this into y = -4 + 4 t we get

y = -4 + 4 ( x + 16) / 11 = 4 t + 20 / 11.

It isn't clear to me where the x = 6 + 11 t and y = 4 + 4 t equations came from. They yield the correct y vs. x solution, but I can't imagine how I ever came up with those equations. It is possible that the problem was changed in a new edition of the text and in editing I simply overlooked this detail.

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.

Note the alternative solution, which stands as originally given:

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 do not understand exactly how you came to the line ""We have x = 6 + 11 t"" nor do I see how you cam to ""y = 4 + 4 t""

Although I do not know how to get those equations, I do understand solving for t so that you can substitute t = (x - 6) / 11 into the y equation.

I don't get it either. See my note above.

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11:53:45

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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To find the speed of the particle I found the distance it travelled: 'sqrt [(x') ^ 2 + (y') ^ 2] per one unit of time.

To find x': f(x) = cos(x), g(x) = t ^ 2

f'(x) = -sin(x), g'(x) = 2t

g'(x) * f'(g(x)) = 2t * -sin(t ^ 2)

To find y': f(x) = sin(x), g(x) = t ^ 2

f'(x) = cos(x), g'(x) = 2t

g'(x) * f'(g(x)) = 2t * cos(t ^ 2)

'sqrt [(2t * -sin(t^2)) ^ 2 + (2t * cos(t ^ 2)) ^ 2]

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11:54:24

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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11:59:57

Does the particle ever come to a stop? If so when? If not why not?

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I set the velocity, 2 | t | { -sin^2(t^2) + cos^2(t^2) }^(1/2), equal to zero. The particle is stopped when t = 0.

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12:00:19

** 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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12:12:54

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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e ^ x when x = 0 is 1.

e ^ 2(x) when x = 0 is also 1.

I still do not fully understand local linearization, but I: found the derivative of e ^ 2x to be 2e ^ 2x. When x = 0 2e ^ 2x is 2. Then I didn't know what to do.

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12:14:50

** 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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I thought we were supposed to find the slope of the tangent line at e ^ 2x.

You noted in your solution that when x = 0, the derivative of the e^(2x) function is 2. When x = 0, we have e^(2x) = e^(2 * 0) = 1. So the tangent line is the line through (0, 1) with slope 2, which is 1 + 2 x. This is the local linearization of e^(2 x).

The local linearization of e^x is found similarly; the slope at x = 0 is 1 and the local linearization is 1 + x.

The square of 1 + x is 1 + 2 x + x^2. This is the same as 1 + 2x, the local linearization of e^(2x), except for the x^2.

When x is small, x^2 is very small, and the local linearization of e^(2x) is very close to the square of the local linearization of e^x. Since e^(2x) is the square of e^x, this matches the expected result.

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12:14:59

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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12:20:51

** 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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I didn't really get the first part right so I didn't try hard to do this one. But after reading your answers I understand.

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13:00:56

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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I did not come to any concrete solutions. But i did try some different things.

I first found the derivative of T: T = 2'pi (L/g) ^ 1/2

T' = (1/2) (l/g) ^ -1/2

T' = 1 / 2 'sqrt(L / g)

I then took the equation 'dT = T / (2 L) * `dL and rewrote it: 'dT / 'dL = T / (2 L)

I then set T' = 'dT / 'dL. I was not real sure where to go from there (I won't show you all the algebra steps) so I took 1 / 2 'sqrt(L / g) = T / (2 L) and solved for T, g, and L.

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13:01:22

** `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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13:10:07

If we wish to estimate length to within 2%, within what % must we know L?

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13:11:30

** 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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I evaluated it with 'dL = .02, not .02 L.

I should have known better though.

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13:12:24

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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13:13:08

** 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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13:17:36

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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13:17:39

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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13:17:41

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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13:17:42

** 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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13:18:14

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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I am sorry I didn't not leave any responses to the last few problems. I had an appointment to make, but I will go back and review those items for the test.

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13:18:16

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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13:18:19

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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I expect you'll get l'Hopital's rule OK. But if you aren't sure, feel free to submit a copy of that problem with the given solutions, your solution and any self-critiques or questions you wish to add. Just mark your insertions with &&&&.