Assignment 18 Query

course Mth 174

σ܁{󞁼assignment #018

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018. `query 18

Cal 2

12-06-2007

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20:27:53

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 -->

a) Here are my estimated values for x and y:

x=0, y=0

x=0.2, y=0

x=0.4, y=0.0016

x=0.6, y=0.0144

x=0.8, y=0.0576

x=1, y=0.16

So y(1) ~= 0.16

c) I believe this is an underestimate.

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20:28:05

what is your estimate of y(1)?

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RESPONSE -->

y(1) ~ 0.16

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22:18:31

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

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RESPONSE -->

For small values of x, y is very small, resulting in a very small positive slope. The y value is increasing as the slope increases, and moves to 1 as the x and y values move towards x=y.

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22:20:06

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 -->

This approximation is an underestimate because the slope field acts like a left Riemann sum. Basically the value of the slope multiplied over the interval is less than the actual value of the slope.

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22:28:41

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

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RESPONSE -->

Euler's Method and the left Riemann sum use the same basic process and even use the derivative.

Euler's Method finds the change in one variable by multiplying the slope by the change in the other variable, in essence making a box in which the area is found. This box starts from the left side of the interval and moves to the right.

In the left Riemann sum, the value of the derivative is taken at the left side of the interval and is multiplied by the distance of the interval.

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22:36:35

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 -->

Here's the process I followed to solve the equation:

dB/dt = 50 - 2B

dB/dt = 2(25 - B)

dB/ (25 - B) = 2 dt

ln | (25 - B) | = 2t + C

this would be - ln | 25 - B |; letting u = 25 - B you would get du = -dB.

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

B = 25 - e^(2t) e^C

Substituting B=100 when t=1:

100 = 25 - e^(2) e^C

75 = - e^(2) e^C

75e^(-2) = e^C

So the final equation becomes:

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

The - sign I mentioned above would give you

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

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22:36:41

what is your solution to the problem?

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RESPONSE -->

See above.

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22:37:46

What is the general solution to the differential equation?

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RESPONSE -->

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

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22:38:31

Explain how you separated the variables for the problem.

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RESPONSE -->

Basically I moved the different variables around using algebra until each type was on each side.

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22:39:11

What did you get when you integrated the separated equation?

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RESPONSE -->

I got the equation:

ln | 25 - B | = 2t + C

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23:02:04

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 -->

I got stuck on this one. Here's what I did:

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

ln| -ln | cos x | + C | = ln t (1 + 2ln t) - int( 2/t ln t )

Which then gets messier and messier as I continue to look for an antiderivative.

Any suggestions would be greatly appreciated.

** We separate variables, as you did.

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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23:02:12

what is your solution to the problem?

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RESPONSE -->

Didn't find one that made any sense.

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23:02:20

What is the general solution to the differential equation?

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RESPONSE -->

I don't know.

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23:02:57

Explain how you separated the variables for the problem.

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RESPONSE -->

Pretty much the same way as before, but this time it didn't seem to work right.

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23:03:04

What did you get when you integrated the separated equation?

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RESPONSE -->

Nothing useful.

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23:04:25

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

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RESPONSE -->

Interesting. A good extension of the first stuff introduced on differential equations, which builds on this, which builds on that, etc. Pretty cool stuff.

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Definitely powerful stuff.

You did well, though you got turned around a little when integrating that last equation. See my notes and let me know if you have questions.