#$&* course MTH 279 1/20 5 Part I: The equation m x '' = - k x*********************************************
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Given Solution: If x = cos(t) then x ' = - sin(t) and x '' = - cos(t). Substituting the expressions for x and x '' into the equation we obtain m * (-cos(t)) = - k * cos(t). Dividing both sides by cos(t) we obtain m = k. If m = k, then the equation is satisfied. If m is not equal to k, it is not. If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t). Substituting this into the equation we have m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t Simplifying both sides we see that the equation is true. The same procedure can and should be used to show that the third equation is true. The question of the fourth equation is left to you. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I saw that if m = k then that was the only way that equation would be true. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2. After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2. Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality. Explain what is wrong with the reasoning given above. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x ' = 2 x + t 2 x + t = 2( x^2 + t^2 / 2 ) + t 2 x + t = 2x^2 + t^2 + t two sides aren't equal When integrating or deriving, you must keep in mind that you are doing it with respect to one variable. In the method above, the first part was integrated with respect to x and the second part was integrated with respect to t. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The given function is a solution to the equation, provided its derivative x ' satisfies x ' = 2 x + t. It would be tempting to say that the derivative of x^2 is 2 x, and the derivative of t^2 / 2 is t. The problem with this is that the derivative of x^2 was taken with respect to x and the derivative of t^2 / 2 with respect to t. We have to take both derivatives with respect to the same variable. Similarly we can't integrate the expression 2 x + t by integrating the first term with respect to x and the second with respect to t. Since in this context x ' represent the derivative of our solution function x with respect to t, the variable of integration therefore must be t. We will soon see a method for solving this equation, but at this point we simply cannot integrate our as-yet-unknown x(t) function with respect to t. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don't know if I plugged into that equation correctly. ------------------------------------------------ Self-critique rating: 3
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Given Solution: If x = A cos(omega * t + theta_0) then x ' = - omega A sin(omega * t + theta_0) and x '' = -omega^2 A cos(omega * t + theta_0). Our equation therefore becomes m * (-omega^2 A cos(omega * t + theta_0) ) = - k A cos(omega * t + theta_0). Rearranging we obtain -m omega^2 A cos(omega * t + theta_0) = -k A cos(omega * t + theta_0) so that -m omega^2 = - k and omega = sqrt(k/m). Thus the constant omega is determined by the equation. The constants A and theta_0 are not determined by the equation and can therefore take any values. No matter what values we choose for A and theta_0, the equation will be satisfied as long as omega = sqrt(k / m). Our second-order equation m x '' = - k x therefore has a general solution containing two arbitrary constants. In the present equation m = 5 and k = 2000, so that omega = sqrt(k / m) = sqrt(2000 / 5) = sqrt(400) = 20. Our solution x(t) = A cos(omega * t + theta_0) therefore becomes x(t) = A cos(20 t + theta_0). If theta_0 = 0 the function becomes x(t) = A cos( 20 t ). The graph of this function will be a 'cosine wave' with a 'peak' at the origin, and a period of pi / 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand that the points I specified would be at +A, not -A. I was looking at the equation x'' where the coeffiecient of A was negative, and not the solution equation that I should've been looking at. ------------------------------------------------ Self-critique rating: 3
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Given Solution: As seen in the preceding problem, a general solution to the equation is x = A cos(omega * t + theta_0), where omega = sqrt(k / m). For the current equation 5 x '' = -2000 x, this gives us omega = 20. In the current context omega = 20 radians / second. So x(t) = A cos( 20 rad / sec * t + theta_0 ). Maximum displacement occurs at critical values of t, values at which x ' (t) = 0. Taking the derivative of x(t) we obtain x ' (t) = - 20 rad / sec * A sin( 20 rad/sec * t + theta_0). The sine function is zero when its argument is an integer multiple of pi, i.e., when 20 rad/sec * t + theta_0 = n * pi, where n = 0, +-1, +-2, ... . A second-derivative test shows that whenever n is an even number, our x(t) function has a negative second derivative and therefore a maximum value. We can therefore pick any even number n and we will get a solution. If maximum displacement occurs at t = 0 then we have 20 rad / sec * 0 + theta_0 = n * pi so that theta_0 = n * pi, where n can be any positive or negative even number. We are free to choose any such value of n, so we make the simplest choice, n = 0. This results in theta_0 = 0. Now if x = .3 when t = 0 we have A cos(omega * 0 + theta_0) = .3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I remember using the first and second derivative tests to find critical numbers and that they can correspond to max and min's, determining concavity, etc. I didn't think to use that in this question because that knoweldge was overshadowed by the way I learned how to play with this particular equation in physics class. ???? In the 4th scenario, do you pick an arbitrary value for theta_0 that makes sin = 1 in order to come up with a value for A? ------------------------------------------------ Self-critique rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Part II: Solutions of equations requiring only direct integration. `q006. Find the general solution of the equation x ' = 2 t + 4, and find the particular solution of this equation if we know that x ( 0 ) = 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = t^2 + 4t + C x(0) = t^0 + 4*0 + C = 3 C = 3 x = t^2 + 4t + 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Integrating both sides we obtain x(t) = t^2 + 4 t + c, where c is an arbitrary constant. The condition x(0) = 3 becomes x(0) = 0^2 + 4 * 0 + c = 3, so that c = 3 and our particular solution is x(t) = t^2 + 4 t + 3. We check our solution. Substituting x(t) = t^2 + 4 t + 3 back into the original equation: (t^2 + 4 t + 3) ' = 2 t + 4 yields 2 t + 4 = 2 t + 4, verifying the general solution. The particular solution satisfies x(0) = 3: x(0) = 0^2 + 4 * 0 + 3 = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q007. Find the general solution of the equation x ' ' = 2 t - .5, and find the particular solution of this equation if we know that x ( 0 ) = 1, while x ' ( 0 ) = 7. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x'' = 2t - .5 x' = t^2 - .5t + C x = t^3/3 - t^2/4 + Ct + D x = t^3/3 - t^2/4 + 7t + 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Integrating both sides we obtain x ' = t^2 - .5 t + c_1, where c_1 is an arbitrary constant. Integrating this equation we obtain x = t^3 / 3 - .25 t^2 + c_1 * t + c_2, where c_2 is an arbitrary constant. Our general solution is thus x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2. The condition x(0) = 1 becomes x(0) = 0^3 / 3 - .25 * 0^2 + c_1 * 0 + c_2 = 1 so that c_2 = 1. x ' (t) = t^2 - .5 t + c_1, so our second condition x ' (0) = 7 becomes x ' (0) = 0^2 - .5 * 0 + c_1 = 7 so that c_1 = 7. For these values of c_1 and c_2, our general solution x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2 becomes the particular solution x(t) = t^3 / 3 - .25 t^2 + 7 t + 1. You should check to be sure this solution satisfies both the given equation and the initial conditions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. Use the particular solution from the preceding problem to find x and x ' when t = 3. Interpret your results if x(t) represents the position of an object at clock time t, assuming SI units. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 3^3/3 - 3^2/4 + 7*3 + 1 x = 28.75 m x' = 3^2 -3/3 + 7 x' = 14.5 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our solution was x(t) = t^3 / 3 - .25 t^2 + 7 t + 1. Thus x ' (t) = t^2 - .5 t + 7. When t = 3 we obtain x(3) = 3^3 / 3 - .25 * 3^2 + 7 * 3 + 1 = 28.75 and x ' (3) = 3^2 - .5 * 3 + 7 = 14.5. A graph of x vs. t would therefore contain the point (3, 28.75), and the slope of the tangent line at that point would be 14.5. x(t) would represent the position of an object. x(3) = 28.75 represents an object whose position with respect to the origin is 28.75 meters when the clock reads 3 seconds. x ' (t) would represent the velocity of the object. x ' (3) = 14.5 indicates that the object is moving at 14.5 meters / second when the clock reads 3 seconds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q009. The equation x '' = -F_frict / m - c / m * x ', where the derivative is understood to be with respect to t, is of at least one of the forms listed below. Which form(s) are appropriate to the equation? x '' = f(x, x') x '' = f(t) x '' = f(x, t) x '' = f(x', t) x '' = f(x, x ' t) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x'' = f(x, x', t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The right-hand side of the equation includes the function x ' but does not include the variable t or the function x. So the right-hand side can be represented by any function which includes among its variables x '. That function may also include x and/or t as a variable. The forms f(t) and f(x, t) fail to include x ', so cannot be used to represent this equation. All the other forms do include x ' as a variable, and may therefore be used to represent the equation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm usually bad with definitions, so I'd like to see if I'm understanding this in my words: You don't have to write t because it is implied. This same follows for x because you have x' in the equation. You must, however, include x' since it encompasses these terms and to do otherwise would leave out a vital piece of information.
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Given Solution: Newton's Second Law gives us the general equation m x '' = F_net so that x '' = F_net / m. It follows that x '' = -F_frict / m - c / m * x ' represents an object on which the net force is -F_frict - c x '. If F_frict = 0, then it follows that the net force is -c x '. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q011. We continue the preceding problem. If w(t) = x '(t), then what is w ' (t)? If x '' = - b / m * x ', then if we let w = x ', what is our equation in terms of the function w? Is it possible to integrate both sides of the resulting equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: w' (t) = x'' (t) w' = -b/m * w I should think we would be able to integrate both sides of the equation, since both w and w' are both functions of t. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If w(t) = x ' (t) then w ' (t) = (x ' (t) ) ' = x '' ( t ). If x '' = - b / m * x ', then if w = x ' it follows that x '' = w ', so our equation becomes w ' (t) = - b / m * w (t) The derivative is with respect to t, so if we wish to integrate both sides we will get w(t) = integral ( - b / m * w(t) dt), The variable of integration is t, and we don't know enough about the function w(t) to perform the integration on the right-hand side. [ Optional Preview: There is a way around this, which provides a preview of a technique we will study soon. It isn't too hard to understand so here's a preview: w ' (t) means dw / dt, where w is understood to be a function of t. So our equation is dw/dt = -b / m * w. It turns out that in this context we can sort of treat dw and dt as algebraic quantities, so we can rearrange this equation to read dw / w = -b / m * dt. Integrating both sides we get integral (dw / w) = -b / m integral( dt ) so that ln | w | = -b / m * t + c. In exponential form this is w = e^(-b / m * t + c). There's more, but this is enough for now ... ]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see that there isn't enough information to perform the actual integration, even though they are both functions of t.
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Given Solution: At the point (.2, .3) in the (t, x) plane, our value of x ' is x ' = (2 * .3 - .5) * (.2 + 1) = .12, approximately. This therefore is the slope of any solution curve which passes through the point (.2, .3). The equation of the tangent plane is therefore x - .3 = .12 * (t - .2) so that x = .12 t - .24. If we move from the t = .2 point to the t = .4 point our t coordinate changes by `dt = .2, so that our x coordinate changes by `dx = (slope * `dt) = .12 * .2 = .024. Our new x coordinate will therefore be .3 + .024 = .324. This gives us the new point (.4, .324). At this point we have x ' = (2 * .324 - .5) * (.4 + 1) = .148 * 1.4 = .207. If we move to the t = .6 point our change in t is `dt = .2. At slope .207 this would imply a change in x of `dx = slope * `dt = .207 * .2 = .041. Our new x coordinate will therefore be .324 + .041 = .365. Our t = .6 point is therefore (.6, .365). From our two calculated slopes, the second of which is significantly greater than the first, it appears that in this region of the x-t plane, as we move to the right the slope of our solution curve in fact increases. Our estimates were based on the assumption that the slope remains constant over each t interval. We conclude that our estimates of the changes in x are probably a somewhat low, so that our calculated points lie a little below the solution curve. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q013. Consider once more the equation x ' = (2 x - .5) * (t + 1). Note on notation: The points on the grid (0, 0), (0, 1/4), (0, 1/2), (0, 3/4), (0, 1) (1/4, 0), (1/4, 1/4), (1/4, 1/2), (1/4, 3/4), (1/4, 1) (1/2, 0), (1/2, 1/4), (1/2, 1/2), (1/2, 3/4), (1/2, 1) (3/4, 0), (3/4, 1/4), (3/4, 1/2), (3/4, 3/4), (3/4, 1) (1, 0), (1, 1/4), (1, 1/2), (1, 3/4), (1, 1) can be specified succinctly in set notation as { (t, x) | t = 0, 1/4, ..., 1, x = 0, 1/4, ..., 1}. ( A more standard notation would be { (i / 4, j / 4) | 0 <= i <= 4, 0 <= j <= 4 } ) Find the value of x ' at every point of this grid and sketch the corresponding direction field. To get you started the values corresponding to the first, second and last rows of the grid are -.5, -.625, -.75, -.875, -1 0, 0, 0, 0, 0 ... ... 1.5, 1.875, 2.25, 2.625, 3 So you will only need to calculate the values for the third and fourth rows of the grid. List your values of x ' at the five points (0, 0), (1/4, 1/4), (1/2, 1/2), (3/4, 3/4) and (1, 1). Sketch the curve which passes through the point (t, x) = (.2, .3). Describe your curve. Is it increasing or decreasing, and is it doing so at an increasing or decreasing rate? According to your curve, what will be the value of x when t = 1? Sketch the curve which passes through the point (t, x) = (.5, .7). According to your curve, what will be the value of x when t = 1? Describe your curve and compare it with the curve you sketched through the point (.2, .3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x' = -.5, 0, .75, 1.75, 3 respectively It is increasing at an increasing rate. It looks like x = 1.125 when t = 1 On the second curve, it looks like x = 1.5 The second curve is also increasing at an increasing rate, but it looks like it is increasing faster than the first curve sketched. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q014. We're not yet done with the equation x ' = (2 x - .5) * (t + 1). x ' is the derivative of the x(t) function with respect to t, so this equation can be written as dx / dt = (2 x - .5) * (t + 1). Now, dx and dt are not algebraic quantities, so we can't multiply or divide both sides by dt or by dx. However let's pretend that they are algebraic quantities, and that we can. Note that dx is a single quantity, as is dt, and we can't divide the d's. Rearrange the equation so that expressions involving x are all on the left-hand side and expressions involving t all on the right-hand side. Put an integral sign in front of both sides. Do the integrals. Remember that an integration constant is involved. Solve the resulting equation for x to obtain your general solution. Evaluate the integration constant assuming that x(.2) = .3. Write out the resulting particular solution. Sketch the graph of this function for 0 <= t <= 1. Describe your graph. How does the value of your x(t) function at t = 1 compare to the value your predicted based on your previous sketch? How do your values of x(t) at t = .4 and t = .6 compare with the values you estimated previously? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dx / dt = (2 x - .5) * (t + 1) dx/ (2x - .5) = (t + 1) dt 1/2 ln( abs(2x - .5) = t^2/2 + t + C ln( abs( 2x - .5) = t^2 + 2t + C 2x - .5 = e^(t^2 + 2t + C) 2x - .5 = e^(t^2 + 2t)C x = e^(t^2 + 2t)K + .25 .3 = e^(.2^2 + 2*.2)K + .25 K = .05/e^.44 x = (.05/e^.44) * e^(t^2 + 2t) + .25 x(1) = 0.9 It looks like my approximation was a little high compared to the actual value. x(.4) = .33 x(.6) = .40 The previous approximations appear pretty good. They were .324 and .365, respectively. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The equation is easily rearranged into the form dx / (2 x - .5) = (t + 1) dt. Integrating the left-hand side we obtain 1/2 ln | 2 x - .5 | Integrating the right-hand side we obtain t^2 / 2 + 4 t + c, where the integration constant c is regarded as a combination of the integration constants from the two sides. Thus our equation becomes 1/2 ln | 2 x - 5 | = t^2 / 2 + t + c. Multiplying both sides by 2, then taking the exponential function of both sides we get exp( ln | 2 x - 5 | ) = exp( t^2 + 2 t + c ), where as before c is an arbitrary constant. Since the exponential and natural log are inverse functions the left-hand side becomes | 2 x + .5 |. The right-hand side can be written e^c * e^(t^2 + 8 t), where c is still an arbitrary constant. e^c can therefore be any positive number, and we replace e^c with A, understanding that A is a positive constant. Our equation becomes | 2 x - .5 | = A e^(t^2 + 2 t). For x > -.25, as is the case for our given value x = .3 when t = .2, we have 2 x - .5 = A e^(t^2 + 2 t) so that x = A e^(t^2 + 2 t) + .25. Using x = .3 and t = .2 we find the value of A: .05 = A e^(.2^2 + 2 * .2) so that A = .05 / e^(.44) = .03220, approx.. Our solution function is therefore x(t) = .05 / e^(.44) * e^(t^2 + 2 t) + .25, or approximately x(t) = .03220 e^(t^2 + 2 t) + .25 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q015. OK, this time we are really going to be done with this equation. Again, x ' = (2 x - .5) * (t + 1) Along what line or curve is x ' = 1? Along what line or curve is x ' = 0? Along what line or curve is x ' = 2? Along what line or curve is x ' = -1? Sketch these three lines and/or curves for 0 <= t <= 1. Along each of these lines x ' is constant. Along each sketch 'slope segments' with slopes equal to the corresponding value of x '. How consistent is your sketch with your previous sketch of the direction field? Sketch a solution curve through the point (.2, .25), and estimate the coordinates of the t = 1 point on this curve. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2 x - .5) * (t + 1) = 1 x = 1/[2(t + 1)] + .25 (2 x - .5) * (t + 1) = 0 x = .25 (2 x - .5) * (t + 1) = 2 x = 1/(t+1) + .25 (2 x - .5) * (t + 1) = -1 x = -1/ [2(t + 1)] + .25 The sketches seem very consistent with the direction field previously sketch. Comparing points they have in common, such as (0, .25), (0, .75) , (1, .25) , (1, .75), the slopes from the earlier computation are the same as the slopes from the curves. As for the last question, sketching a curve through (.2, .25), I don't know how you can sketch a curve through an asymptote. To my knowledge, the curve would be undefined at this point. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: x ' = 1 when (2 x - .5) * (t + 1) = 1. Solving for x we obtain x = 1/2 ( 1 / (t + 1) + .5) = 1 / (2(t + 1)) + .25. The resulting curve is just the familiar curve x = 1 / t, vertically compressed by factor 2 then shifted -1 unit in the horizontal and .25 unit in the vertical direction, so its asymptotes are the lines t = -1 and x = .25. The t = 0 and t = 1 points are (0, .75) and (1, .5). Similarly we find the curves corresponding to the other values of x ': For x ' = 0 we get the horizontal line x = .25. Note that this line is the horizontal asymptote to the curve obtained in the preceding step. For x ' = 2 we get the curve 1 / (t + 1) + .25, a curve with asymptotes at t = -1 and x = .25, including points (0, 1.25) and (1, .75). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: