Assignment 18

course Mth 174

„ɬçã•š·þ»¼”xWë…cÙÉ{êúâÐ쇫assignment #018

018. `query 18

Cal 2

06-25-2009

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14:44:00

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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RESPONSE -->

When x = 1.0, y = .9109. I found this by starting with point (0,0) and plugged in the x and y values in the equation and multiplied that by (1/5). I then added 1 to this number to give be the new y value.

x= 0

y= 0

so 0^3 -0^3 * 1/5 = 0

x= .1

y = 0

so .1^3 - 0^3 * 1/5 = .0004

The new y value is 1 + .0004 = 1.0004

x = .2

y=1.0004

so .2^3 - 1.0004^3 = -.986

The new y value is 1 = -.986 = .8014

I repeated these steps until I reached x = 1.0

Your basic procedure is OK, but you used 10 steps rather than 5, while still multiplying by the increment 0.2.

** 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 = 0.2*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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14:44:42

what is your estimate of y(1)?

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RESPONSE -->

when x = 1.0, y = .9109

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14:50:50

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

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RESPONSE -->

I made a graph and plugged in each x and y value and then drew a line corresponding to the slopes between each point. This graph was consistent with the slope field given because both are concave up and increasing.

Good. Consider also the following:

** 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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14:51:39

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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RESPONSE -->

My approximation is an underestimation because the curve is concave up.

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14:54:11

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

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RESPONSE -->

Using Euler's method gives the same results as using left Riemann sums because both find the area under the curves, and when the graph is concave down the left Riemann sums and Euler's method both underestimate, and when the graph is concave up they both overestimate.

** 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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14:54:16

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

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RESPONSE -->

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15:21:48

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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RESPONSE -->

dB/dt= 2B - 50

dB/dt = 2 (B - 25)

dB/(B - 25) = 2dt

integral of dB/(B - 25) = integral of 2dt =

ln(B - 25) = 2t

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

B - 25 = e^c e^2t

B - 25 = B e^2t

B is a function of t. You wouldn't substitute B for the constant expression e^c.

B = 25 + B e^2t

= 25 + 100 e^2*1

=125e^2

** 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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15:21:50

what is your solution to the problem?

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RESPONSE -->

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15:22:23

What is the general solution to the differential equation?

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RESPONSE -->

B = B e^(2t + c)

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15:23:29

Explain how you separated the variables for the problem.

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RESPONSE -->

I subtracted - 2B from the right side and factored out a 2. I then put dB/ B - 25 = 2 dt and integrated.

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15:23:44

What did you get when you integrated the separated equation?

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RESPONSE -->

ln(B - 25) = 2t

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15:31:33

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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RESPONSE -->

t dx/dt = ( 1+ 2ln(t)) tanx

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

* t should be / t

dx/ tanx = t ( 1 + 2/t)

dx/tanx= t + 2

integral of dx/tanx = integral of t + 2

ln(tanx) =t^2/2 + 2t

tanx = e^(t^2/2 + 2t + c)

tanx = e^c e^(t^2/2 + 2t)

tanx = A e^(t^2/2 + 2t)

tanx = B e^(t/2/2 + 2t)

Right ideas, but you had an algebra error early in your solution.

** 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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15:31:40

what is your solution to the problem?

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RESPONSE -->

tanx = B e^(t/2/2 + 2t)

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15:32:00

What is the general solution to the differential equation?

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RESPONSE -->

tanx = e^(t^2/2 + 2t + c)

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15:32:21

Explain how you separated the variables for the problem.

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RESPONSE -->

This is shown previously

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15:32:30

What did you get when you integrated the separated equation?

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RESPONSE -->

Also shown previously

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15:32:43

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

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RESPONSE -->

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&#Your work looks good. See my notes. Let me know if you have any questions. &#