Assignment 18

course Mth 174

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018. `query 18
Cal 2
08-04-2008

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21:54:09
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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22:15:40
what is your estimate of y(1)?

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To find this I went from P(0) to P(5). P(0): x = 0 and y = 0. We already know that the next x will be x(0) + deltax = 0 + 1/5 = 1/5. To find the next y value I needed to find delta y. deltay = slope * delta x. The slope is dy/dx = x^3 - y^3, which at this point is (0) ^ 3 - 0 = 0. So at P(1), x = 1/5 and y = 0.

To find P(2) I went through the same process. I will do this without all of the commentary. The next x value is (1/5) + (1/5) = 2/5. The slope is (1/5)^3 - (0)^3 = (1/125). delta y = (1/125) * (1/5) = 1/625. So at P(2), x = 2/5 and y = 1/625.

The next x value is 2/5 + 1/5 = 3/5. The slope is (2/5)^3 - (1/625)^3 = .06399 (approx). delta y = [(2/5)^3 - (1/625)^3] * (1/5) = .012799. The new y value is y + delta y = .014399 (approx.) So at P(3), x = 3/5 and y = .014399 (approx)

The next x value is 3/5 + 1/5 = 4/5. The slope is (3/5)^3 - .014399^3 = .21599. delta y = .21599 * 1/5 = .043199. The next y value is y + delta y = .014399 + .043199 = .057598 (approx) So at P(4), x = 4/5 and y = .057598 (approx).

The next x value is 1. The slope is (4/5)^3 - (.057598)^3 = .511808. delta y = .511808 * 1/5 = .102361 (approx). The final y value is .057598 + .102361 = .1599597 (approx)

Good. Just for comparison:

** The table below summarizes the calculation. x values starting at 0 and changing by `dx = .2 are as indicated in the x column. y value starts at 0.

The y ' is found by evaluating x^3 - y^3. `dy = y ' * `dx and the new y is the previous value of y plus the change `dy.

Starting from (0,0):

m = 0, delta y = 0.20 = 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 = 0.20.064 = 0.0128, new point (0.6, 0.0144)
m = 0.6^3 - 0.0144^3 = 0.216, delta y = 0.20.216 = 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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22:17:30

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

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I am not real sure other than at the points of x and y's that I came up with do seem to match the slope field graph, but I do not know of a way to concludethis exactly.

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22:18:36

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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This part of the slope field graph is concave up, so I would say that my answer is an underestimate.

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22:35:41

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

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23:06:19

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

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I do not really know why exactly. I thought I had an answer but when I started to type it into my work, it sounded rediculous and I could not prove anything.

I'd say your answer was better than you thought, but it's very good to have a critical eye for one's own work and to understand when it doesn't quite hold together.

** 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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23:20:26

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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23:31:33

what is your solution to the problem?

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My solution: I first set it up to where each side of the equation had one variable. dB/dt + 2B = 50. I set dB/dt = 50 - 2B. Then I divided by 2 to get (1/2)( dB / dt) = 25 - B. I rearranged it to (1 / (25 - B)) dB = 2 dt.

I then took the integral of both to be ln abs(25 - B) = 2t + C. Which is abs(25 - B) = e^2t+C = e^2t * e^C. Which equals abs(25 - B) = A * e^2t, where A = e^C. -> 25 - B = +-A * e^2t = -B = D * e^2t - 25, where D = +-A.

So, B = -D * e^2t + 25. If B(1) = 100, then 100 = -D * e^2 + 25 -> -75 = D*e^2 -> D = -75 / e^2. I used D as a substitute for +- A so I wouldn't get confused. But in the end it didn't help. This is as far as I got and couldn't come up with an exact answer.

** 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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23:31:36

What is the general solution to the differential equation?

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23:31:38

Explain how you separated the variables for the problem.

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23:31:40

What did you get when you integrated the separated equation?

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23:42:23

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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I did well at setting this up, but I didn't turn in an answer becuase I couldn't get past finding the integrals for each side. I had it set up (1 / tan(x)) dx = (1 + (2ln(t)/t)) dt. I spent a good while trying to find the integrals, but I guess I'll have to go back and practice up.

** We separate variables.

t dx/dt = (1 + 2 ln t) tan x is rearranged to give
dx / (tan x) = (1 + 2 ln t) / t * dt = 1/t dt + 2 ln(t) / t * dt or
cos x / sin x * dx = = 1/t dt + 2 ln(t) / t * dt .

Integrating both sides

we let u = sin(x) on the left, obtaining du / u with antiderivative ln u =
ln(sin(x))

we let u = ln(t) and dv = 1/t * dt on the right and use integration by parts
to get antiderivative ln(t)^2 - int(ln(t) / t). Solving int(ln(t) / t) =
ln(t)^2 - int(ln(t) / t) for ln((t) / t we get int(ln(t) / t) = ln(t)^2 / 2.

int(1/t * dt) = ln(t).

Our equation therefore becomes

ln(sin(x)) = ln(t) + 2 * ln(t)^2 / 2 + c so that
sin(x) = e^(ln(t) + ln(t)^2 + c) = e^(ln(t)) * e^(ln(t)^2) * e^c = A * t *
t^(ln(t)) = A * t^1 * t^(ln(t)) = A * t^(1 + ln(t))

so that

x = arcsin(A * t^(1 + ln(t)) ).

This makes sense for t > 0, which gives a real value of ln(t), as long as A *
(t^(1 + ln(t) ) < 1. **

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