Assignment 18

course Mth 174

}????????????assignment #018018. `query 18

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Cal 2

04-12-2007

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19:05:29

Query problem 11.3.4 (was 10.3.6) Euler y' = x^3-y^3, (0,0), `dx = .2, 5 steps

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19:13:07

what is your estimate of y(1)?

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RESPONSE -->

Starting from (0,0):

m = 0, delta y = 0.2*0 = 0, new point (0.2,0)

m = 0.2^3 = 0.008, delta y = 0.2*.008 = 0.0016, new point (0.4, 0.0016)

m = 0.4^3 - 0.0016^3 = 0.064, delta y = 0.2*0.064 = 0.0128, new point (0.6, 0.0144)

m = 0.6^3 - 0.0144^3 = 0.216, delta y = 0.2*0.216 = 0.0432, new point (0.8,0.0576)

m = 0.8^3 - 0.0576^3 = 0.512, delta y = 0.2*0.512 = 0.1024, new point (1.0, 0.16)

y(1) = 0.16

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19:18:37

Describe how the given slope field is consistent with your step-by-step results.

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Although difficult to discern due to the scale used on the slope field, the results of Euler's method appear to be consistent with the slope field as x increases from 0 to about 0.6, after which the slope field shows an increasing slope compared to the results obtained (the slope field shows an increasingly more positive slope than do the results obtained using Euler's method). Thus, Euler's method produces an underestimate.

Good. Compare with the following:

The slope field is rising and concave up, so that the slope at the left-hand endpoint will be less than the slope at any other point of a given interval. It follows that an approximation based on slope at the left-hand endpoint of each interval will be an underestimate.

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19:20:06

Is your approximation an overestimate or an underestimate, and what property of the slope field allows you to answer this question?

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The approximation is an underestimate, because the slope field shows an increasingly positive slope as x approaches 1, and the slope is much more positive than the slopes obtained by Euler's method.

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19:20:19

Query problem 11.3.10 (was 10.3.10) Euler and left Riemann sums, y' = f(x), y(0) = 0

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19:28:21

explain why Euler's Method gives the same result as the left Riemann sum for the integral

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RESPONSE -->

Since the initial point is (0,0) and dy/dx = f(x), the initial slope = f(x) = 0. For any delta x, delta y = 0 * delta x = 0, resulting in the next point being (delta x, 0). Since f(x) remains 0, the slope remains zero, and delta y remains 0 for all further points (f(x) = 0 for all x. Thus, the curve of f(x) is the x-axis.

Since f(x) = 0, all left Riemann sums = 0 * delta x = 0.

Therefore, the integral (0 to x) of f(t) dt = the integral (0 to x) of 0 dt = 0, which is the same result as the Euler method.

I don't think there is any assumption that f(x) remains 0. In any case, even if that is assumed in the given problem, it is not necessary. This will be true for any continuous function f(x):

** Euler's method multiplies the value of y ' at the left-hand endpoint of each subinterval to calculate the approximate change dy in y, using the approximation dy = y ' * `dx. This result is calculated for each interval and the changes dy are added to approximate the total change in y over the interval.

The left-hand Riemann sum for y ' is obtained by calculating y ' at the left-hand endpoint of each subinterval then multiplying that value by `dx to get the area y ' `dx of the corresponding rectangle. These areas are added to get the total area over the large interval.

So both ways we are totaling the same y ' `dx results, obtaining identical final answers. **

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19:33:19

Query problem 11.4.19 (3d edition 11.4.16) (was 10.4.10) dB/dt + 2B = 50, B(1) = 100

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19:40:56

what is your solution to the problem?

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RESPONSE -->

dB/dt + 2B = 50

dB/dt = 50 - 2B

dB/(50-2B) = dt

(-1/2)*dB/(B-25) = dt

(-1/2)*integral dB/(B-25) = integral dt

(-1/2) ln |B-25| = t + c

ln |B-25| = -2t + c

|B-25| = e^(-2t+c)

B-25 = Ae^(-2t) (A constant)

B = Ae^(-2t) + 25

Substituting: B(1) = 100

100 = Ae^(-2) + 25

A = 75/e^(-2) = 75e^2

Substituting:

B = 75e^2*e^(-2t) + 25

B = 75e^(2-2t) + 25

Excellent. See also the following. The form B = 75 e^(-2(t-1)) + 25 is used here (this form shows clearly how the function can be 'built up' from the basic function e^t using a vertical stretch, a horizontal compression, a vertical shift and a horizontal shift), but it equivalent to the form you give.

** We can separate variables.

We get dB/dt = 50 - 2 B, so that dB / (50 - 2B) = dt.

Integrating both sides we obtain

-.5 ln(50 - 2B) = t + c, where c is an arbitrary integration constant.

This rearranges to

ln | 50 - 2B | = -2 (t + c) so that
| 50 - 2B | = e^(-2(t+c)). Since B(1) = 100 we have, for t = 1, 50 - 2B =
-150 so that |50 - 2 B | = 2 B - 50. Thus
2B - 50 = e^(-2(t+c)) and
B = 25 + .5 * e^(-2(t+c)).

If B(1) = 100 we have

100 = 25 + .5 * e^(-2 ( 1 + c) ) so that
e^(-2 ( 1 + c) ) = 150 and
-2(1+c) = ln(150). Solving for c we find that
c = -1/2 * ln(150) - 1.

Thus

B(t) = 25 + .5 e^(-2(t - 1/2 ln(150) - 1))
= 25 + .5 e^(-2t + ln(150) + 2))
= 25 + .5 e^(-2t) * e^(ln(150) * e^2
= 25 + 75 e^2 e^(-2t)
= 25 + 75 e^(-2 (t – 1) ).

Note that this checks out:

B(1) = 25 + 75 e^2 e^(-2 * 1)
= 25 + 75 e^0 = 25 + 75 = 100.

Note also that starting with the expression

| 50 - 2B | = e^(-2(t+c)) we can write
| 50 - 2B | = e^(-2c) * e^(-2 t). Since e^(-2c) can be any positive number we replace this factor by C, with the provision that C > 0. This gives us

| 50 - 2B | = C e^(-2 t) so that

50 – 2 B = +- C e^(-2 t), giving us solutions

B = 25 + C e^(-2t) and
B = 25 – C e^(-2t).

The first solution gives us B values in excess of 25; the second gives B values less than 25.

Since B(1) = 100, the first form of the solution applies and we have

100 = 25 + C e^(-2), which is easily solved to give
C = 75 e^2.

The solution corresponding to the given initial condition is therefore

B = 25 + 75 e^-2 e^(-2t), which is simplified to give us

B = 25 + 75 e^(-2(t – 1) ). **

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19:42:20

What is the general solution to the differential equation?

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RESPONSE -->

The general solution is:

B = Ae^(-2t) + 25

(from previous response)

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19:42:30

Explain how you separated the variables for the problem.

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RESPONSE -->

See previous response.

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19:42:41

What did you get when you integrated the separated equation?

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RESPONSE -->

See previous response.

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19:44:32

Query problem 11.4.40 (3d edition 11.4.39) (was 10.4.30) t dx.dt = (1 + 2 ln t ) tan x, 1st quadrant

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19:59:28

what is your solution to the problem?

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RESPONSE -->

t*dx/dt = (1 + 2ln t)tan x

dx/tan x = t(1 + 2 ln t) dt

Integral dx/tan x = Integral t(1 + 2 ln t) dt

Integral dx/(sin x/cos x) = Integral t dt + 2*Integral ln t dt

Integral cos x/sin x dx = (t^2)/2 + 2(t*ln t - t) + c

u = sin x du = cos x dy

Integral (1/u) du = (t^2)/2 + 2t*ln t - 2t + c

ln |u| = (t^2)/2 + 2t*ln t - 2t + c

ln |sin x| = (t^2)/2 + 2t*ln t - 2t + c

sin x = e^(t^2)/2 + 2t*ln t - 2t + c)

sin x = e^[(t^2)/2 - 2t](e^c)[e^(2t*ln t)]

sin x = Ae^[(t^2)/2 - 2t]*(e^ln t)^(2t)

sin x = Ae^[(t^2)/2 - 2t]*t^(2t)

sin x = A*[t^(2t)]*(e^[(t^2)/2 - 2t]

x = arcsin A*t^(2t)*e^[(t^2)/2 - 2t]

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20:04:54

What is the general solution to the differential equation?

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Previous response was the general solution. Unless I'm missing something here, I don't believe that a specific solution is possible without an initial point.

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20:05:06

Explain how you separated the variables for the problem.

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RESPONSE -->

See previous response.

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20:05:19

What did you get when you integrated the separated equation?

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RESPONSE -->

See previous response.

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20:18:58

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

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It was very interesting to reverse the differentiation of implicit functions done in the first part of the book.

It's interesting that the solution to 11.4.45 (y' = x/y) which is x^2 - y^2 = 2C becomes the standard form of a hyperbola by dividing through by 2C.

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Assignment 18

Excellent work. See my note and let me know if you have questions.