Assignment 18

course MTH 174

}_j펚assignment #018

018. `query 18

Cal 2

11-16-2008

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20:13:02

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)

0 = (0^3 - 0^3)(1/5)

next y = 0, x = 1/5

0.008 = (0.2^3 - 0^3)(1/5)

next y = 0 + 0.0016, x = 2/5

0.0128 = (0.4^3 - 0(0.008)^3)(1/5)

next y = 0.0016 + 0.0128 = 0.0144, x = 3/5

0.0432 = (0.6^3 - 0.0144^3)(1/5)

next y = 0.0144 + 0.0432 = 0.576, x = 4/5

0.1024 = (0.8^3 - 0.576^3)(1/5)

next y = 0.1024 + 0.576 = 1.6, x = 5/5

So the estimate at of y(1) = 0.16

This is likely an underestimate because the graph is concave up.

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20:13:36

what is your estimate of y(1)?

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RESPONSE -->

y(1) = approx 0.16

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20:14:13

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

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RESPONSE -->

The slopes on the slope field are increasing as are the slopes in the results that I calculated.

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20:15:43

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 -->

The approximation is an underestimate due to the graph being concave up as is shown by the slopes of the graph as x gets larger from (0,0)

Good. More specifically:

** 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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21:22:27

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 -->

Approximating using Euler's method is the same as using the left Reimann sum.

Looking at it from Left Reimann Sum:

y0 = f(x0) * `dx

y1 = f(x1) * `dx

y2 = f(x2) * `dx

yn = f(xn-1) * `dx

You then add all these up to estimate the integral.

Euler's method is using the slope * `dx to get y

y1 = y0 + f(x0) * `dx

y2 = y1 + f(x1) * `dx = y0 + f(x0) * `dx + f(x1) * `dx

So adding all these up you have y0 + f(x0) * `dx + f(x1) * `dx ... yn-1+ f(xn-1) * `dx

When f(0) = 0, the y0 goes away and Euler's method is the same as adding up the parts of the left Reimann Sum.

Good. Another version of the same explanation:

** 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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21:22:46

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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21:45:12

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, B(1) = 100

dB/dt = 50 - 2B

dB/dt = 2(25 - B)

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

ln |25 - B| = 2t + C

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

25 - B = e^2t * e^C

Using B(1) = 100

25 - 100 = e^2 * e^C

-75 = e^2 * e^C

-75/e^2 = e^C

25 - B = e^2t * -75 * e^-2

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

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

Double checking equation:

B(1) = 75 * e^(2(1) - 2) + 25 = 100

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21:46:59

what is your solution to the problem?

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RESPONSE -->

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

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21:47:41

What is the general solution to the differential equation?

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RESPONSE -->

The general solution is

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

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21:48:41

Explain how you separated the variables for the problem.

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RESPONSE -->

I factored out 2 and paired it with `dt and divided both sides by (25 - B) to move the B over to be with dB.

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21:51:33

What did you get when you integrated the separated equation?

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RESPONSE -->

After integrating a solution to the problem is

25 - B = e^2t * e^C

-B = e^2t * e^C - 25

B = -e^C * e^2t + 25

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22:06:40

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 + 2 ln t) tan x

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

1/(sin x/cos x) = (1 + 2 ln t)/t dt

(cos x/sin x) = 1/t + (2 ln t)/t dt

Integrating both sides (u = ln t, du/dt = 1/t:

ln |sin x| = ln |t| + 2 u^2/2 + C

ln |sin x| = ln |t| + ln t * ln t + C

sin x = e^(ln |t| + ln t * ln t + C)

sin x = e^(ln |t|) * e^(ln t * ln t) * e^C

sin x = t * t^2 *e^C

e^(ln(t) * ln(t) ) = (e^ln(t))^ln(t) = t^(ln(t)), not t^2

sin x = t^3 *e^C, e^C = B

x = arcsin (Bt^3)

Good solution except for an apparent algebra error near the end (see my preceding note).

For comparison:

** 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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22:06:54

what is your solution to the problem?

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RESPONSE -->

x = arcsin (Bt^3)

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22:07:37

What is the general solution to the differential equation?

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RESPONSE -->

x = arcsin (Bt^3)

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22:09:14

Explain how you separated the variables for the problem.

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RESPONSE -->

I divided both sides tan x to move it to the other side of the equation with dx and multiplied both sides by dt to move the dt to be with 1 + 2 ln t.

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22:09:45

What did you get when you integrated the separated equation?

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RESPONSE -->

ln |sin x| = ln |t| + ln t * ln t + C

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22:09:52

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. &#