Assignment 18

course Mth 174

Please explain problem 11.4.19

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018. `query 18

Cal 2

08-11-2009

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14:59:10

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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15:00:49

what is your estimate of y(1)?

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My estimate of y(1) is 0.

** 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

If you follow the slope field starting at x = 0 and move to the right until
you reach x = 1 your graph will at first remain almost flat, consistent with
the fact that the values in the Euler approximation don't reach .1 until x has
exceeded .6, then rise more and more quickly. The result y(1) = .448 is very
plausible.

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15:01:27

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

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The slope field shows that the slopes of the various points between x=0 and x=1 are mostly flat, deviating only slightly.

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15:02:29

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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It is an underestimate. This can be told by the fact that the slopes of the area in consideration are all positive, therefore the function would be concave up at this point, causing Euler method to yield an underestimate.

Good, but the fact that the slopes are positive doesn't imply that the slope field is concave up. The fact that slopes increase from left to right implies that the slope field is concave up.

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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15:58:05

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

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16:04:02

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

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Because if f(x) is concave up, then both the left endpoint method and Euler's method will give an underestimate. The same is true of a concave down function, except that the methods will result in overestimates.

Your statements are true and insightful, but they don't address the question.

** 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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16:25:03

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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16:25:26

what is your solution to the problem?

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

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16:32:04

What is the general solution to the differential equation?

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B=25+75(2-2t)+C

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16:41:04

Explain how you separated the variables for the problem.

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The only seperation that was really needed was to multiply both sides by dt.

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16:48:24

What did you get when you integrated the separated equation?

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B^2/2=25t+C

You have a correct result, but you don't appear to have a correct process in your solution.

** 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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17:39:42

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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17:40:20

what is your solution to the problem?

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sinx=te^((ln(t))^2+1/4)

** 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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17:40:50

What is the general solution to the differential equation?

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sinx=te^((ln(t))^2+1/4)

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17:41:39

Explain how you separated the variables for the problem.

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Multiplied both sides by dt, then divided by tan(x) and t.

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17:42:30

What did you get when you integrated the separated equation?

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lnAbs(sinx)=((2ln(t)+1)^2)/2+C

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17:42:34

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

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Assignment 18

&#See my notes and let me know if you have questions. &#