#$&* course Mth 174 12/6 4 alculus IIAsst # 18
.............................................
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
......!!!!!!!!...................................
11:35:35
......!!!!!!!!...................................
**** Describe how the given slope field is consistent with your step-by-step results.
......!!!!!!!!...................................
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.
.........................................
11:35:53
......!!!!!!!!...................................
**** Is your approximation an overestimate or an underestimate, and what property of the slope field allows you to answer this question?
......!!!!!!!!...................................
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. **
.........................................
11:36:07
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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 using left Riemann sum we find `dy by saying `dy is approx. f`(x)*`dx, and f`(x) is evaluated at the beginning at each interval or `dx. When `dy is calculated then y is approx. y at beginning of interval + `dy(where `dy is symbol for change in y during interval). This is the same logic used when calculating using Eulers method.
.............................................
Given Solution: **** explain why Euler's Method gives the same result as the left Riemann sum for the integral
......!!!!!!!!...................................
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.
.........................................
11:36:36
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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: dB/dt + 2B = 50 = dB/dt = 50 2B = dB = (50 2B)*dt = dB/(50-2B) = dt dB/(50-2B) = dt, is in a form which we can solve Int(dB/(50-2B)) = Int(dt) = -½*ln|50-2B| = t + c ***Now solve for B**** ln|50-2B| = -2t + c(c absorbs the mult by -2) |50 2B| = e^(-2t+c) for c > 0 |50 2B| = c*e^(-2t) B = (c*e^(-2t))-25 for B<25 = (c*e^(-2t))+25 for B>25 So for B(1) = 100 now we solve for for B using second equation 100 = c*e^(-2*1) = c*e^(-2) c = 100/e^-2 = approx. 738.9056 = B(t) = 738.9056e^(-2*t), approx ??????This would not be considered a logistic equation would it???????????????
.............................................
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) ). **
.........................................
11:36:54
......!!!!!!!!...................................
**** What is the general solution to the differential equation?
......!!!!!!!!...................................
11:37:08 I'm not sure, I didn't find a general solution
.........................................
11:37:08
......!!!!!!!!...................................
**** Explain how you separated the variables for the problem.
......!!!!!!!!...................................
11:37:28 I just treated db and dt as normal variables and multiplied dt times the entire equation
.........................................
11:37:28
......!!!!!!!!...................................
**** What did you get when you integrated the separated equation?
......!!!!!!!!...................................
11:37:40 1/2B = 25t - B*t
.........................................
11:37:41
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I thought my solution looked not quite right, I left out the init 25 on right side of the equals operator. ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: t*dx/dt = (1+2ln t)tan x = dx/tan x=(1 + 2ln t)/(t/dt)=(1+2ln t)*dt/t=(1+2ln t)dt/t=(1 + 2 ln t)/t dt Int(dx/tan x) = Int((1 + 2ln t)/t dt) Int(dx/tan x) = -ln |cos x| Int((1 + 2ln t)/t dt) = Int(1/t dt) + Int(2ln t/t dt) = ln|t|+[2*Int(ln t/t dt)],using u sub=u= ln t, du = 1/t dt(getting rid of t in denom.) = 2*Int(u du) = 2*(u^2/2) = u^2, converting back to t var. = (ln t)^2, so right hand side ends up being . = ln|t| + ln|t|^2 + c, so our equation is Int(dx/tan x) = Int((1 + 2ln t)/t dt) = -ln |cos x| = ln|t| + ln|t|^2 + c !!!!Note, Because problem puts a limit that the we are only concerned with 1st quadrant only this keeps us from having to use any nonreals. In the 2nd and 3rd quad cos x is neg and taking the natural log of a neg number or zero is undefined and going to give a nonreal solution. ????Im guessing that is why it is mentioned in the problem, my text makes no reference except that assume x, y, t > 0???!!!!!! -cos x = t + t^2 + e^c, for c>0 = -cos x = t + t^2 + c cos x = -t t^2 -c x = cos^-1(-t t^2 c) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 ] ). **
.........................................
11:38:09
......!!!!!!!!...................................
**** What is the general solution to the differential equation?
......!!!!!!!!...................................
11:38:18 The same thing
.........................................
11:38:18
......!!!!!!!!...................................
**** Explain how you separated the variables for the problem.
......!!!!!!!!...................................
11:38:38 I multiplied dt by the entire equation and treated it as a normal variable
.........................................
11:38:38
......!!!!!!!!...................................
**** What did you get when you integrated the separated equation?
......!!!!!!!!...................................
11:39:05 ln|sin(x)| =(2 ln(t) + 1)^2/4 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??????The book has tan x listed in the short table of indef. intergrals as Int(tan x dx) = -ln |cos x| + C, is this not correct. Also could you reduce e^ln|t| and e^ln|t^2| to t and t^2 and then take inverse of the cos x fn to find the value of x ????????? ------------------------------------------------ Self-critique Rating:2 "