#$&* course Mth 174
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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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: OK ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With Euler’s method, we multiple the y’ value of the left endpoint of all the subintervals to find the change in y. We use dy = (y’)’dx. We do this for each interval and add all the dy changes to find the total change in y. We find the left Riemann sum for y’ by calculating y’ at the left subinterval endpoints and multiplying those numbers by ‘dx to obtain the area y’(‘dx). We use the rectangle areas and add them together to get the total area of the interval. 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): I should add that with both ways we are totaling the same y ' `dx results, obtaining identical final answers. ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We need to get dB and dt on opposite ends of the equation, so we get dB / (50 - 2B) = dt. After integration, we get -0.5 ln(50 - 2B) = t + c. After some rearranging and plugging in, we get B= 25 + 0.5 e^(-2(t+c)). If B(1)=100, we end up with c=( -1/2) ln(150) - 1. Also, B(t) = 25 + .5 e^(-2(t - 1/2 ln(150) - 1)) = 25 + 75 e^(-2 (t - 1) ). We now know that | 50 - 2B | = e^(-2c) * e^(-2 t). Therefore, solving for B gives us the equation B = 25 - C e^(-2t). We use the first solution because B(1)= 100, which is a value greater than 25. We have 100 = 25 + C e^(-2), and we solve for c to get c=75e^2. Therefore, the final solution is B = 25 + 75 e^-2 e^(-2t). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I could have simplified to get the solution B = 25 + 75 e^(-2(t - 1) ). ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I know that there is several steps involved and integration needed, but I am not sure how to begin this problem. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I now understand that we begin with x = arcsin(A*t^(ln(t) + 1)) and we must separate the variables to obtain cos x / sin x * dx = 1/t dt + 2 ln(t) / t * dt. Using integration and u-substitution, we come out with sin(x) = A t e^((ln(t))^2). ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!