#$&* course MTH 279 Part I: The equation m x '' = - k x*********************************************
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Given Solution: If x = cos(t) then x ' = - sin(t) and x '' = - cos(t). Substituting the expressions for x and x '' into the equation we obtain m * (-cos(t)) = - k * cos(t). Dividing both sides by cos(t) we obtain m = k. If m = k, then the equation is satisfied. If m is not equal to k, it is not. If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t). Substituting this into the equation we have m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t Simplifying both sides we see that the equation is true. The same procedure can and should be used to show that the third equation is true. The question of the fourth equation is left to you. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2. After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2. Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality. Explain what is wrong with the reasoning given above. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Taking the integral of d/dx will stimply be x^2 + tx + C because one does not the take simple take the integral of t as well with an integration factor of dx, what they the person did was partial integration with relation to one variable at a time, but even at that, did a poor job. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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(omega)(sin(omega *t + theta_0)) x” = -A (omega^2) cos(omega*t+theta_0) m x '' = - k x -A (omega^2) cos(omega*t+theta_0) * m = -k * (A cos(omega * t + theta_0)) -m omega^2 = - k 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. m x '' = - k x m = 5 and k = 2000 omega = sqrt(k / m) = sqrt(2000 / 5) = sqrt(400) = 20. 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. Theta_0 = 3 pi /2 the function x(t) = A cos(20t +(3pi/2)). ). The graph of this function will be a 'cosine wave' with a peak at (-3pi/2, 0) , and a period of pi / 10. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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) ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -x(t) = A cos( 20 rad / sec * t + theta_0 ) 0.15 = 0.3 cos( 20 rad / sec * 0 + theta_0 ) 0.15 = 0.3 cos(theta_0 ) theta_0 = pi/3 -x ' (t) = - 20 rad / sec * A sin( 20 rad/sec * t + theta_0) 2 = - 20 rad / sec * 0.3 sin( 20 rad/sec * 0 + theta_0) 2 = - 20 rad / sec * 0.3 sin(theta_0) 2 = -6 sin(theta_0) sin(theta_0) = (-1/3) -x ' (t) = - 20 rad / sec * A sin( 20 rad/sec * t + theta_0) 2 = - 20 rad / sec * A sin( 20 rad/sec * 0 + theta_0) 2 = - 20 rad / sec * A sin(theta_0) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: x(t) = Acos(20t + theta_0) x'(t) = -20Asin(20t + theta_0) x(t) = 0.3 cos(theta_0) meters x'(t) = -6 sin(theta_0) meters/second x(t) = 0.3 cos(theta_0) = 0.15 meters x'(t) = -6 sin(theta_0) meters/second x(t) = 0.3 cos(theta_0) meters x'(t) = -6 sin(theta_0) = 2.0 meters/second x(t) = Acos(theta_0) meters x'(t) = -20Asin(theta_0) = 2.0 meters/second confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think understood what it was asking for, but I cannot be certain if I answered the question it wanted me to answer.
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Given Solution: Integrating both sides we obtain x(t) = t^2 + 4 t + c, where c is an arbitrary constant. The condition x(0) = 3 becomes x(0) = 0^2 + 4 * 0 + c = 3, so that c = 3 and our particular solution is x(t) = t^2 + 4 t + 3. We check our solution. Substituting x(t) = t^2 + 4 t + 3 back into the original equation: (t^2 + 4 t + 3) ' = 2 t + 4 yields 2 t + 4 = 2 t + 4, verifying the general solution. The particular solution satisfies x(0) = 3: x(0) = 0^2 + 4 * 0 + 3 = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. Find the general solution of the equation x ' ' = 2 t - .5, and find the particular solution of this equation if we know that x ( 0 ) = 1, while x ' ( 0 ) = 7. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Integrate both sides x’ = t^2 - ½ t + C x’(0) = 0^2 - ½ (0) + C = 7 C = 7 x’ = t^2 - ½ t + 7 x = (t^3)/3 - ¼ t^2 +7t +D x(0) = (0^3)/3 - ¼ 0^2 +7 (0)+D = 1 D = 1 x = (t^3)/3 - ¼ t^2 +7t + 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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 = (t^3)/3 - ¼ t^2 +7t +1 x(3) = (3^3)/3 - ¼ 3^2 +7(3) +1 x(3) = 9 - 9/4 + 21 + 1 = 28.75 x’ = t^2 - ½ t + 7 x’ (3) = 3^2 - ½ (3) + 7 x’(3) = 9 - 3/2 +7 = 14.5 x(t) represents 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our solution was x(t) = t^3 / 3 - .25 t^2 + 7 t + 1. Thus x ' (t) = t^2 - .5 t + 7. When t = 3 we obtain x(3) = 3^3 / 3 - .25 * 3^2 + 7 * 3 + 1 = 28.75 and x ' (3) = 3^2 - .5 * 3 + 7 = 14.5. A graph of x vs. t would therefore contain the point (3, 28.75), and the slope of the tangent line at that point would be 14.5. x(t) would represent the position of an object. x(3) = 28.75 represents an object whose position with respect to the origin is 28.75 meters when the clock reads 3 seconds. x ' (t) would represent the velocity of the object. x ' (3) = 14.5 indicates that the object is moving at 14.5 meters / second when the clock reads 3 seconds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had to read the given answer to understand what the second question was looking for in terms of interpretation, but when I read it , I understood what it was asking for. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. The equation x '' = -F_frict / m - c / m * x ', where the derivative is understood to be with respect to t, is of at least one of the forms listed below. Which form(s) are appropriate to the equation? x '' = f(x, x') x '' = f(t) x '' = f(x, t) x '' = f(x', t) x '' = f(x, x ' t) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x '' = f(x, x') x '' = f(x', t) x '' = f(x, x ' t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: Net Force = ma = mx'' Multiply by m mx '' = -Friction Force - c * x ' Friction Force = 0 F_net = mx '' = - c * x ' since x' = velocity increases, Net Force also increases confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Newton's Second Law gives us the general equation m x '' = F_net so that x '' = F_net / m. It follows that x '' = -F_frict / m - c / m * x ' represents an object on which the net force is -F_frict - c x '. If F_frict = 0, then it follows that the net force is -c x '. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. We continue the preceding problem. If w(t) = x '(t), then what is w ' (t)? If x '' = - b / m * x ', then if we let w = x ', what is our equation in terms of the function w? Is it possible to integrate both sides of the resulting equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. take the derivative, x”(t) 2. Substitute : w ' = - b / m * w (w ' * m)/-b = w 3. One can technically integrate anything since it is simply adding the parts under a curve together, so yes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x ' = (2 (.3) - .5) * (.2 + 1) x’ = (.6 - .5)*(.2+1) = .12 .12 is the slope of the tangent line x - .3 = .12(t - .2) is the equation of the tangent line dt’ = .2 difference thus dx’ = .2 * .12 = .024 x + dx’ = new x 3 + .024 = .324 (.4, .324) is our new point x ' = (2 (.324) - .5) * (.4 + 1) = .148 * 1.4 = .207 .207 is the new slope x-.324 = .207(t-.4) is the new equation of the tangent line t = .6 then dt’ = .2 and dx’ = .207 * .2 = .041 .324 +.041 = .365 (.6, .365) The slope of the latter slopes are greater than the former so we see that it is increasing as the point t becomes larger , and so we can safely assume our points in comparison to the solution curve are most likely below said curve as we underestimated said points. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: -.5, -.625, -.75, -.875, -1 0, 0, 0, 0, 0 .5, .625, .75, .875, 1 1, 1.25, 1.5, 1.75, 2 1.5, 1.875, 2.25, 2.625, 3 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). -.5, 0, .75, 1.75, 3 for t = .2 and x = .3 My curve is increasing at an increasing rate My x value will be .396 when t = 1 for t = .5 and x = .7 My curve is increasing at an increasing rate and faster than the former My x value will be .8035 when t = 1 The latter graph increases much faster in the x direction than the former. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I used the slope formula for finding x on my graph, I am not sure this was what was to be done.
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: x ' = 1 (2 x - .5) * (t + 1) = 1 x = 1/2 ( 1 / (t + 1) + .5) x = 1 / (2*(t + 1)) + .25 The curve is x = 1 / t, compressed by 2 times and then moved -1 unit in the x direction and .25 units in the y direction Its asymptotes are t = -1 and x = .25 and when t = 0 and t = 1 the points are (0, .75) and (1, .5) respectively x ' = 0 we get the horizontal line x = .25 x ' = 2 we get the curve 1 / (t + 1) + .25 which is a curve with asymptotes at t = -1 and x = .25 and having points at (0, 1.25) and (1, .75) respectively. x’ = -1 we get the curve (2x - 0.5)(t + 1) = -1 x = (-1/2(t + 1) + 0.25) compressed on y axis by a factor of 2, moved to the left by 1 and 0.25 vertically Asympototes are at t = -1, and x = 0.25 t = 0, the point (0, -1/12) t = 1, the point (1, 0) How consistent is your sketch with your previous sketch of the direction field? There are some variations, but the most part, it is fairly close Sketch a solution curve through the point (.2, .25), and estimate the coordinates of the t = 1 point on this curve. If I plug in this point for x ' = (2 x - .5) * (t + 1) x' = 0 which is accurate because x= 0.25 is a horizontal line and t = 1 would still give an x of 0.25 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: x ' = 1 when (2 x - .5) * (t + 1) = 1. Solving for x we obtain x = 1/2 ( 1 / (t + 1) + .5) = 1 / (2(t + 1)) + .25. The resulting curve is just the familiar curve x = 1 / t, vertically compressed by factor 2 then shifted -1 unit in the horizontal and .25 unit in the vertical direction, so its asymptotes are the lines t = -1 and x = .25. The t = 0 and t = 1 points are (0, .75) and (1, .5). Similarly we find the curves corresponding to the other values of x ': For x ' = 0 we get the horizontal line x = .25. Note that this line is the horizontal asymptote to the curve obtained in the preceding step. For x ' = 2 we get the curve 1 / (t + 1) + .25, a curve with asymptotes at t = -1 and x = .25, including points (0, 1.25) and (1, .75). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: These questions can be challenging if you are rusty on first-year calculus and trigonometric functions. Fortunately we won't be using these functions a lot near the beginning of the course, but they become very important later, and if you're rusty, or if you never really mastered these functions, you want to get a good start now. If you are rusty, try to answer the following without the use of a calculator: Quick refresher on trigonometric functions: What are the maximum and minimum possible values of cos(theta)? -1 and 1 #$&* For what values of theta does cos(theta) take its maximum value, and for what values of theta is cos(theta) equal to zero? maximum at theta = 0 and 2 pi and 0 when theta is pi/2 or 3pi/2 and any of these can be multiplied by a positive whole number integer “k” and be the same #$&* For what values of t does y = cos(5 t) take its maximum value? For what t values does this function take its minimum value? it takes its maximum value when t = k*pi when k is a positive even number or 0, it takes its minimum value when k is a positive odd number #$&* What is the maximum possible value of y = 4 cos(83 t)? y = 4 #$&* Find at least one value of t for which y = cos(2 t + pi) takes the value zero. t = - pi/4 #$&* Find at least one value of t for which y = 3 cos(2 t + pi/4) takes its maximum value. t = 7 pi / 8 #$&* Find at least one value of t for which the derivative of y = cos(t) is zero. t = pi #$&* Find at least one value of t for which the derivative of y = cos(t) is zero, and the second derivative is negative. t = 0 #$&* Find the critical points of y = 4 sin(2 pi t + pi/2), then using a second-derivative test find the values of t at which this function takes its maximum and minimum values. y’ =8 pi (cos(2pi(t) + pi/2)) y’ = 0 when t = 0, ½, 1, 3/2, 2, … Global max at all whole numbers because it is increasing at it approaches the number and declines afterward Global minimum at all halfed numbers because it declines as it approaches the number and increases afterward Feel free to submit a copy of these questions with your answers.