Query 18

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course mth 174

12:40 6/5

......!!!!!!!!...................................Calculus II

Asst # 18

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Question:

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

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Your solution:

We are asked to use 5 steps of the Euler method for y’=x^3 -y^3 we are given that delta x is .2 and the starting point is (0,0)

So we begin:

y1= y0 +deltax(x0, y0)

0+.2*0=0 so new point is (.2, 0)

y2= y1 +deltax(x1, y1)

0+.2*.2^3=.0016 so new point is (.4, .0016)

y3= y2 +deltax(x2, y2)

.0016+.2(.4^3 - .0016^3) =.0144 so new point is (.6, .0144)

y4= y3 +deltax(x3, y3)

.0144 + .2(.6^3-.0144^3) = .0576 so new point is (.8, .0576)

y5= y4 +deltax(x4, y4)

.0576+.2(.8^3 - .0576^3) = .16 new point is (1, .16)

confidence rating #$&*: 3

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Given Solution:

**** what is your estimate of y(1)?

***&&& The points we get are after the given (0,0) are (.2, 0) (.4, .0144), (.8, .0576), (1, .16)

So y(1) is .16*********

** 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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11:35:35

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

***&&& the slope field is consistent with the step by step pints we get because the graph by the points found shows that there is little increases in the area under the slope for the first few approximations and then the area under the slope increases at an increasing rate. ********

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11:35:53

** 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 contributions of the first three intervals are small, then get progressively larger.

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11:35:53

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**** Is your approximation an overestimate or an underestimate, and what

property of the slope field allows you to answer this question?

***&&&& the slope of the graph is concave up so if you use a left approximation, it would be an underestimate. IF you use a right approximation it would be an overestimate.****

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11:36:07

An underestimate

** 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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11:36:07

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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Question:

**** Query problem 11.3.13 4th and 3d editions 11.3.10 (formerly 10.3.10) Euler and left Riemann sums, y' = f(x), y(0) = 0

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Your solution:

We want to know explain why both the Euler and left Reimann sum for y’=f(x) and y(0)=0 will give us the same answers using each method.

When you use the Euler’s method, you are multiplying the value of y’ at the left endpoint of each interval by delta x. So as you go further along all of the delta x’s for each interval are added to the change in y and you end up with the total change along the interval.

The left Reimann sum is found by finding the value of y’ at the left endpoint and multiplying it by delta x so here you are also adding all of the delta x’s and the changes in y to end up with the total change over the interval.

Both of these methods give you the same result.

confidence rating #$&*: 3

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Given Solution:

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

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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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11:36:36

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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Question:

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

**** what is your solution to the problem?

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Your solution:

We are given that dB/dt + 2B=50 and that B(1)=100

We can rearrange this equation to get dt by itself and dB separate and perform a separate integration.

So we have dB/dt+2B=50

dB/dt=(50-2B)dt

dt=dB/(50-2B)

now we can integrate

integral dt = integral of dB/(50-2B) for the second side u=(50-2B) du=.2db so -1/2du=dB

so now we have integral of dt = integral of -1/2 * integral of dB/u

so integrating we get

t+c=-1/2 ln of absolute value of (50-2B)

so we begin to simplify and get B by itself

ln of absolute value (50-2B) =-2(t+c)

absolute value of (50-2B) = e^-2(t+c)

we can break the second half apart getting

absolute value of (50-2B) = e^-2t to e^-2c we know that the c in the exponent will be positive because e raised to a negative number is positive so we can just substitute in C for e^-2c getting

absolute value of (50-2B) = Ce^-2t

50-2B= + or - Ce^-2t at this point we solve for the + and - versions and determine which to use since we know B(1) = 100

So -2B=CE-2t -50

B= (Ce^-2t -50)/-2

B=25+ce^-2t

t(1) = 25.1

So -2B=-CE-2t -50

B= -(Ce^-2t -50)/-2

B=25-ce^-2t

t(1) = 24.86

Since we know B(1) is 100 then we use the equation which gives us the largest value for t(1) which is B=25+ce^-2t

So we substituite in for B(1) getting

100=25+Ce^-2

Ce^-2=75

C=75/e^-2

C=75e^2

So go back to the equation B=25+ce^-2t and plug in C getting

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

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

B=25 +75 e^(-2(t-1))

To ensure this is accurate substitute in known value of B(1) and put in 1 for t getting

100=25+75e^0

100=100

confidence rating #$&*: 2

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Given 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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11:36:54

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**** What is the general solution to the differential equation?

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11:37:08

***&&&The general solution is 50-2B= + or - Ce^-2t*****

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11:37:08

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**** Explain how you separated the variables for the problem.

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11:37:28

***&&&I rearranged the equation to get dt on one side and dB on the other so that I could integrate each side of the equation separately I got

Dt=dB/(50-2B)****************

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11:37:28

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**** What did you get when you integrated the separated equation?

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11:37:40

&&&&**** t+c=-1/2 * ln of absolute value of (50-2B) which can be simplified int

50-2B= + or - Ce^-2t***************

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11:37:41

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Self-critique (if necessary):

### I did fairly well on this and understood almost every step but I do have a couple of questions. In this step after we have separated and integrated we have

integral dt = integral of dB/(50-2B) getting the integral below

t+c=-1/2 ln of absolute value of (50-2B)

I understand the integration but why do we have integration constant on the left and not the right????

@&

You get integration constants on both sides, but the constant on the right can be subtracted from both sides, leaving just a constant (equal to the difference of the two original constants) on the left.

Since the constants were arbitrary in the first place, meaning that either one of the could take any value whatsoever, the same is true of their difference--it could be constant value whatsoever.

So having arbitrary constants on both side of an equation is no different than having a single arbitrary constant on only the one side we might want to choose.

In this case we choose the side with t, simply because of the way we solve teh resulting equation. (If there was an arbitrary constant on the side with ln | 50 - 2 B |, we would want to subtract it from both sides before exponentiating).

*@

Next question when we have B= (Ce^-2t -50)/-2 how does this simplify into

B=25+Ce^-2t and not B=25+(1/2)Ce^-2t??????

@&

C would be an arbitrary constant (i.e., it could be any value whatsoever).

So 1/2 C, which would also take any value, is still an arbitrary constant. The 1/2 is pointless and would require extra work, so we just leave it off.

Thus the C we get after multiplying through by 1/2 is not the same as the C we had before, but it's still an arbitrary constant.

We say that arbitrary constants 'absorb' multiplication by nonzero constants, meaning simply that when we multiply an arbitrary constant by a nonzero constant we still get an arbitrary constant.

As implicit in my preceding not, arbitrary constants also 'absorb' addition, or subtraction of other arbitrary constant (or for that matter multiplication or division by other arbitrary constants).

*@

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Self-critique Rating:3

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Question:

**** Query problem 11.4.40 was 11.4.39 (was 10.4.30) t dx/dt = (1 + 2 ln t ) tan x,

1st quadrant

**** what is your solution to the problem?

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Your solution:

We begin with tdx/dt=(1+2lnt)tanx we need to first separate getting dx an dt on opposite sides of the equation so we can do a separate integration.

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

tdx/tanx=dt * (1+2lnt)

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

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

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

so now we can do separate integrations

integral of dx/tanx = integral 1/t + 2 times the integral (ln(t))/t

for the left side we can rewrite dx/tanx as cos(x)/sin(x) dx letting u = sin(x) du = cos(x) we rewrite as integral of 1/u du which is ln(u) which is ln(sinx)

for the left side we have integral (1/t) + 2*integral ln(t)/t

for the first we get ln(t) and for the second we let u =ln(t) du = 1/t so we have 2*(1/2)u^2 which is ln(t^2) so now we put all together and add our integration constant to the left

all together after integration we have

ln(sin(x))= ln(t) +ln(t^2) +c

sin(x)=e^(lnt+lnt^2+c)

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

we can substitute a constant A in for e^c because this will be a positive number for all values of c that are positive so we have

sin(x)= Ate^lnt^2)

x=arcsin(Ate^ln(t^2))

confidence rating #$&*: 3

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Given Solution:

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

** 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

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

ln(sin(x))

Thus our antiderivative of the left-hand side is ln(sin(x)).

On the right-hand side 1/t dt + 2 ln(t) / t * dt we first find an antiderivative of ln(t) / t

Letting u = ln(t) we have du = 1/t dt so our integral becomes int(u^2 du) = u^2 / 2 = (ln(t))^2 / 2.

Thus integrating the term 2 ln(t) / t we get (ln(t))^2.

The other term on the right-hand side is easily integrated: int(1/t * dt) = ln(t).

Our equation therefore becomes

ln(sin(x)) = ln(t) + ln(t)^2 + c so that

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

= e^(ln(t)) * e^(ln(t)^2) * e^c

= A t e^(ln(t))^2.

where A = e^c is an arbitrary positive constant (A = e^c, which can be any positive number)

so that

x = arcsin(A t e^(ln(t)^2)

This makes sense for t > 0, which gives a real value of ln(t), as long as t e^(ln(t)^2 < = 1 (the domain of the arcsin function is the interval [ -1, 1 ] ).

**

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11:38:09

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**** What is the general solution to the differential equation?

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11:38:18

****&& x=arcsin(Ate^ln(t^2))*****

11:38:18

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**** Explain how you separated the variables for the problem.

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11:38:38

&&*** I made algebraic arrangements to get dx on one side and dt on the other so that separate integrals could be performed.

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11:38:38

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**** What did you get when you integrated the separated equation?

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11:39:05

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

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Self-critique (if necessary):

#####ok except for sin(x) = e^ln(t) * e^ln(t)^2 * e^c

we can substitute a constant A in for e^c because this will be a positive number for all values of c that are positive (I understand all of the preceding)

When I separate I get

sin(x)= Ae^ln(t)*e^ln(t^2)

sin(x)=Ae^ln(t)^3 but the given solution is sin(x)= Ate^lnt^2

Why does 1 of the t’s come down from the e power????

@&

My notation in the given solution was careless and misleading.

The four lines you are asking about should read

ln(sin(x)) = ln(t) + ( ln(t) )^2 + c so that

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

= e^(ln(t)) * e^((ln(t))^2) * e^c

= A t e^((ln(t))^2).

(ln(t))^2 is different that ln(t^2). Had it been the latter, your simplification would have worked.

However there is no way to simplify the product of two logarithms. ln(t^2) = 2 ln(t), but ln(a) * ln(b) cannot be simplified (except as ln( a ^(ln(b) ), which leads nowhere). So (ln(t))^2 = ln(t) * ln(t) = ln(t ^ (ln(t) ), but this is not a simplification.

*@

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Self-critique Rating:3

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#*&!

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