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Calculus II
Asst # 18
Question:
**** Query problem 11.3.4 (as 10.3.6) Euler y' = x^3-y^3, (0,0), `dx =.2, 5 steps
Your solution:
Confidence Assessment:
Given Solution:
**** what is your estimate of y(1)?
** 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.
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11:35:53
I'm not sure exactly what you are asking here
** 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?
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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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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
Your solution:
Confidence Assessment:
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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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?
Your solution:
Confidence Assessment:
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
I'm not sure, I didn't find a general solution
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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 just treated db and dt as normal variables and multiplied dt times the
entire equation
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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
1/2B = 25t - B*t
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11:37:41
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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?
Your solution:
Confidence Assessment:
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
The same thing
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11:38:18
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**** Explain how you separated the variables for the problem.
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11:38:38
I multiplied dt by the entire equation and treated it as a normal variable
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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
ln|sin(x)| =(2 ln(t) + 1)^2/4
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