#$&* course Mth 279 6/26/13 around 5 pm Part I: The equation m x '' = - k x*********************************************
.............................................
Given Solution: If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t). Substituting this into the equation we have m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t Simplifying both sides we see that the equation is true. The same procedure can and should be used to show that the third equation is true, while the fourth is not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):####I believe I did it correctly, but I could only get it to simplify to 1 = 1, instead of m = k. ------------------------------------------------ Self-critique rating:####2 ********************************************* Question: `q002. An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2. After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2. Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality. Explain what is wrong with the reasoning given above. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: #### x' = 2 x + t, which when plugged in gives (x^2 + t^2 / 2)' = 2 x + t. When taking the derivative of the left side of the equation you cannot take it with respect to both t and x variables. Even though the two derivatives were correct if taken with respect to that variable, the equation as a whole would have a derivative that is taken with respect to one or the other but not both. Therefore it is incorrect, because the derivative was taken with respect to both variables instead of just one or the other. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):###ok ------------------------------------------------ Self-critique rating:####3 ********************************************* Question: `q003. The general solution to the equation m x '' = - k x is of the form x(t) = A cos(omega * t + theta_0), where A, omega and theta_0 are constants. (There are reasons for using the symbols omega and theta_0, but for right now just treat these symbols as you would any other constant like b or c). Find the general solution to the equation 5 x'' = - 2000 x: Substitute A cos(omega * t + theta_0) for x in the given equation. The value of one of the three constants A, omega and theta_0 is dictated by the numbers in the equation. Which is it and what is its value? One of the unspecified constants is theta_0. Suppose for example that theta_0 = 0. What is the remaining unspecified constant? Still assuming that theta_0 = 0, describe the graph of the solution function x(t). Repeat, this time assuming that theta_0 = 3 pi / 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: #### x = A cos(omega * t + theta_0), x' = -A * omega * sin(omega * t + theta_0), x'' = -A * (omega)^2 * cos(omega * t + theta_0). When plugging into the equation, m (-A * (omega)^2 * cos(omega * t + theta_0)) = -k (A cos(omega * t + theta_0)), dividing both sides by (-A cos(omega * t + theta_0)) gives, m * omega^2 = k. After dividing by m and taking the square root of both sides you get, omega = sqrt(k/m), meaning that omega is dictated by the numbers in the equation. In this specific case of having m = 5 and k = 2000, omega = sqrt(2000/5), which makes omega = 20. The remaining unspecified constant is A, because regardless of the value of A, omega will still be equal to sqrt(k/m). If theta_0 = 0, x(t) = A cos(20*t), The graph would be a cos wave graph withan amplitude of A and a period of (2*pi)/20 = pi/10. If theta_0 = 3 pi/2, x(t) = A cos (20 t + 3*pi/2), the graph would The graph would still have the same period but shifted to the left (3*pi)/40 units. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 did not go into as great detail as much as I could have on a few of the answers. ------------------------------------------------ Self-critique rating:####3 ********************************************* Question: `q004. In the preceding equation we found the general solution to the equation 5 x'' = - 2000 x. Assuming SI units, this solution applies to a simple harmonic oscillator of mass 5 kg, which when displaced to position x relative to equilibrium is subject to a net force F = - 2000 N / m * x. With these units, sqrt(k / m) has units of sqrt( (N / m) / kg), which reduce to radians / second. Our function x(t) describes the position of our oscillator relative to its equilibrium position. Evaluate the constants A and theta_0 for each of the following situations: The oscillator reaches a maximum displacement of .3 at clock time t = 0. The oscillator reaches a maximum displacement of .3 , and at clock time t = 0 its position is x = .15. The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0. The oscillator has a maximum velocity of 2, which occurs at clock time t = 0. (Hint: The velocity of the oscillator is given by the function x ' (t) ). 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ####The oscillator reaches a maximum displacement of .3 at clock time t = 0. A cos(theta_0) = .3, as stated above theta_0 = n*pi, with n being any positive or negative even number. Therefore, theta_0 = 0, making the equation, A(1) = .3, meaning A = .3. The oscillator reaches a maximum displacement of .3 , and at clock time t = 0 its position is x = .15. When plugging in the values, .3 cos(omega*0 + theta_0) = .15. Solving for that, you get .3 cos(theta_0) = .15. Then divide both sides by .3 to get cos(theta_0) = .5, which then goes to, theta_0 = cos^-1(.5). Meaning that theta_0 = pi/3. The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0. The value of A would remain the same as in the first example, meaning that A = .3. Plugging back into the original equation at t=0 .3cos(theta_0) = .3, and cos(theta_0) = 1. Therefore theta_0 = cos^-1(1), which equals 0. The oscillator has a maximum velocity of 2, which occurs at clock time t = 0. (Hint: The velocity of the oscillator is given by the function x ' (t) ) maximum velocity = A*omega = 2. We know from previous problems that omega = 20, so when plugging in, A*20 = 2, and A = .1. To find the maximum velocity, the value of sin(omega*t + theta_0) is set to zero. At t = 0, that just leaves sin(theta_0). Tha value is zero at n* pi, with n being any whole number, thus for simplicity we should set theta_0 = 0. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):####ok ------------------------------------------------ Self-critique rating:####3 ********************************************* Question: `q005. Describe the motion of the oscillator in each of the situations of the preceding problem. SI units for position and velocity are respectively meters and meters / second. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ####In the first situation, the pendulum starts at the position x=.3 meters, and is released. From there it will oscillate back and forth consistantly, with an amplitude of .3 m. In the second situation, the pendulum is begun at x = .15 meters, but will have a displacement of .3, thus it will be shifted to the left to begin with, because its maximum x-value will be .45 m, but will only go to x = -.15m, meaning that its central point is not at the origin. In the third situation, the pendulum will do as in the first situation, but will reach its maximum velocity of 2 m/s at x= 0 meters. THerefore the pendulum will be increasing in velocity as it approaches x = 0, and thn decrease until it reaches the next maximum displacement point, then begin again. In the fourth situation, it will do the same as the third, but the maximum displacement of the pendulum is only .1 meters instead of .3. confidence rating #$&*:#2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####I am not sure if I described them in the appropriate manner as intended by the instructor or if there was a different method to do such. ------------------------------------------------ Self-critique rating:####2 ********************************************* Question: Part II: Solutions of equations requiring only direct integration. `q006. Find the general solution of the equation x ' = 2 t + 4, and find the particular solution of this equation if we know that x ( 0 ) = 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: #### First we must integrate the solution to get x(t) = t^2 + 4t + C. At x(0) = C = 3, therefore, C = 3. Meaning that the general solution is x(t) = t^2 + 4t + 3. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):####I did the solution without checking to see if I made a careless mistake or not, which may be helpful just in case. ------------------------------------------------ Self-critique rating:###2 ********************************************* 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. When you plug in x'(0), you get x'(0) = 7 = C, so C=7. THus the general solution for x'(t) = t^2 -.5t + 7. x'(t) = t^2 -.5t + 7, x (t) = t^3/3 - .25 t^2 + 7t + C. At x(0) = C = 1, therefore C = 1. So the general solution for the equation is x (t) = t^3/3 - .25 t^2 + 7t + 1. Plugging back in the equations to check on my notes, I concluded that my integrals were indeed correct. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Integrating both sides we obtain x ' = t^2 - .5 t + c_1, where c_1 is an arbitrary constant. Integrating this equation we obtain x = t^3 / 3 - .25 t^2 + c_1 * t + c_2, where c_2 is an arbitrary constant. Our general solution is thus x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2. The condition x(0) = 1 becomes x(0) = 0^3 / 3 - .25 * 0^2 + c_1 * 0 + c_2 = 1 so that c_2 = 1. x ' (t) = t^2 - .5 t + c_1, so our second condition x ' (0) = 7 becomes x ' (0) = 0^2 - .5 * 0 + c_1 = 7 so that c_1 = 7. For these values of c_1 and c_2, our general solution x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2 becomes the particular solution x(t) = t^3 / 3 - .25 t^2 + 7 t + 1. You should check to be sure this solution satisfies both the given equation and the initial conditions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):####Ok ------------------------------------------------ Self-critique rating:#####3 ********************************************* Question: `q008. Use the particular solution from the preceding problem to find x and x ' when t = 3. Interpret your results if x(t) represents the position of an object at clock time t, assuming SI units. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: #### At t = 3, x (3) = 3^3/3 - .25* 3^2 + 7(3) + 1, which simplifies to x(3) = 9 - 2.25 +22 = 28.75. x'(3) = 3^2 -.5(3) + 7, which simplifies to x(3) = 9 - 1.5 + 7 = 14.5. An x vs. t grah would be at the point (3, 28.75), with a slope of 14.5. Where at t=3s, the position of the object would be at 28.75m and have a slope of 14.5m. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ####I could have better worded the way I described the point and its description, especially when talking about the slope. ------------------------------------------------ Self-critique rating:####3 ********************************************* 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: ##### Because x' is represented on the right side of the equation, it can be represented by any equation that includes x' in it. Therefore, the first, fourth, and last equations work, but not the second or third. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The right-hand side of the equation includes the function x ' but does not include the variable t or the function x. So the right-hand side can be represented by any function which includes among its variables x '. That function may also include x and/or t as a variable. The forms f(t) and f(x, t) fail to include x ', so cannot be used to represent this equation. All the other forms do include x ' as a variable, and may therefore be used to represent the equation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####I understand the concept, but should have mentioned that x and/or t could be included as variables in the equation to give my reasoning behind picking all three with x' in the function. ------------------------------------------------ Self-critique rating:####3 ********************************************* Question: `q010. If F_frict is zero, then the function x in the equation x '' = -F_frict / m - c / m * x ' represents the position of an object of mass m, on which the net force is - c * x '. Explain why the expression for the net force is -c * x '. Explain what happens to the net force as the object speeds up. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: #### -c * x' is the net force acting upon the object because it is the value of the drag that is acting on the object and would be the only net force acting on the object if there is no friction. As the object speeds up, the net force on the object increases, because x' is the velocity, and it is on top of the fraction. Therefore the value of the net force would increase. confidence rating #$&*:#2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ####I feel like I understand the concept but did not give a reason to back up my answer, such as showing where net force derives from. ------------------------------------------------ Self-critique rating:####2 ********************************************* 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: #### 1.) If w(t) = x '(t), then w' (t) = x'' (t). 2.) Since w = x' and w' = x'', the equation can go to w'(t) = -b/m * w(t). Then integrate to get w(t) = integral( -b/m * w(t)). 3.) No, we do not know enough about the right side to make an appropriate integration at this point in time. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ####OK ------------------------------------------------ Self-critique rating:####3 ********************************************* Question: Part III: Direction fields and approximate solutions `q012. Consider the equation x ' = (2 x - .5) * (t + 1). Suppose that x = .3 when t = .2. If a solution curve passes through (t, x) = (.2, .3), then what is its slope at that point? What is the equation of its tangent line at this point? If we move along the tangent line from this point to the t = .4 point on the line, what will be the x coordinate of our new point? If a solution curve passes through this new point, then what will be the slope at the this point, and what will be the equation of the new tangent line? If we move along the new tangent line from this point to the t = .6 point, what will be the x coordinate of our new point? Is is possible that both points lie on the same solution curve? If not, does each tangent line lie above or below the solution curve, and how much error do you estimate in the t = .4 and t = .5 values you found? 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ####The slope at (.2, .3) would be x' = (2(.3) - .5) * (.2+1). Which with simplification goes to x'= (.1)*(1.2), which equals .12. thus the slope at that point is .12. The tangent line equation is x - .3 = .12(t-.2), which when simplified gives the tangent line x = .12t + .24. If we go along the tangent line and solve for x at t=.4, our x- moves by two change in slopes so it would be .3 +(.2 * .12) = .324, giving us the point (.4 , .324). At the new point x' = (2 * .324 - .5) * (.4 + 1), which when simplified down gives x' = .207. If we move to t = .6, we must do the same thing as before, but with the new slope, so .324 + (.2 * .207), which equals .365. This gives us the points (.6, .365). Since the slopes change between the two, with the second slope being much greater than the first slope calculated, it means that the solution curve increases as we go to the right. Since ours is based off the assumption of a straight line, we can estimate that our answers are a little below the actual curve. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):####Ok ------------------------------------------------ Self-critique rating:####3 ********************************************* 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: ##### Third row : .5, .625, .75, .875, 1. Fourth row : 1, 1.25, 1.5, 1.75, 2. Values of x' in order: -.5, 0, .75, 1.75, 3. The curve is increasing at an increasing rate, becasue as it increases on the t- plane, the slope of the t values continually increase and as it increases on the x-plane, the slopes continue to increases. Based off my curve, at t = 1, the x-value for the curve (.2, .3) would be at approximately 1. Based off the curve, at t= 1, the x-value would be 6.4. The curve is increasing at an increasing rate, which is the same as the (.2, .3) curve, but it started at a higher point, making the x-value greater at t=1. confidence rating #$&*:#2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####I am not sure my description of the curves are right. ------------------------------------------------ Self-critique rating:####2 ********************************************* 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? 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ####The graph is increasing at an increasing rate, starting at just above zero and nearly approaching one, but the slope increases the further right the graph gets. At t= 1, the value of the function is .03220 e^(3) + .25 = .8968, which is just below my predicted value of one for the answer. Therefore, my curve was slightly off, but fairly close. At t = .4, the value of the function is .3341, which is slightly greater than my .324 estimate, but it is fairly close. At t = .6, the value of the function is .4032, which is a bit larger than the .365 that was predicted earlier, but because we were basing the estimation off a straight line, instead of the curve. confidence rating #$&*:#2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####ok ------------------------------------------------ Self-critique rating:####2 ********************************************* 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: #### 1.) (2 x - .5) * (t + 1) = 1, 2x - .5 = 1/(t+1), x = (1/(2t+2)) +.25, 2.) (2 x - .5) * (t + 1) = 0, 2x - .5 = 0 x = .25. It is on the vertical straight line of x = .25. 3.) (2 x - .5) * (t + 1) = 2 2x - .5 = 2/(t+1) x = (1/(t+1)) + .25 4.) (2 x - .5) * (t + 1) = -1 2x - .5 = -1/(t+1) x = (-1/2(t+1)) +.25. My sketch is fairly consistent with the previously sketch of the directional field, but is off on a few points by small margins. Based on my sketched curve, my estimate for the point (.2, .25) is ( 1, .25), because at x = .25, there is a constant horizontal line. confidence rating #$&*:#3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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:####3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!