#$&* course Mth 279 2/11 Q_A_QuestionsThere are three parts to this set of questions.
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Given Solution: If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t). Substituting this into the equation we have m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t Simplifying both sides we see that the equation is true. The same procedure can and should be used to show that the third equation is true, while the fourth is not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2. After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2. Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality. Explain what is wrong with the reasoning given above. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We cannot yet solve an equation like this in the form x'(t) = f(x,t), because to solve this we must integrate both sides of the equation. When we integrate one side, it must be with respect to the same variable. That means that in an equation like this one, we cannot simply integrate the side with x and t with respect to different variables. They must be the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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(t) = A cos(omega * t + theta_0) x'(t) = -A(omega) sin(omega * t + theta_0) x''(t) = -A(omega)^2 cos(omega * t + theta_0) m x'' = - k x m (-A(omega)^2 cos(omega * t + theta_0)) = - k (A cos(omega * t + theta_0)) -m(omega)^2 Acos(omega * t + theta_0) = - k Acos(omega * t + theta_0) -m(omega)^2 = - k omega = sqrt(k/m) omega = sqrt(2000/5) = 20 The constants A, and theta_0 cannot be determined by the equation, and could be any value. Solution x(t) = A cos(omega * t + theta_0) becomes -> x(t) = A cos(20 t + theta_0) If theta_0 = 0 the function becomes x(t) = A cos( 20 t ), which would be a graph of the cosine curve with amplitude of A. theta_0 = 3 pi / 2 would give a similar graph but only vary apon the position of the wave. confidence 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): ------------------------------------------------ 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) ). 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 , and at clock time t = 0 its position is x = .15. 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 The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0 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) The oscillator has a maximum velocity of 2, which occurs at clock time t = 0 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) A and theta_0 could be any values as long as the equation is true, if theta_0 = 0 then A would be undefined, In this particular problem, if either A or theta_0 are zero then the equation is undefined. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Part II: Solutions of equations requiring only direct integration. `q006. Find the general solution of the equation x ' = 2 t + 4, and find the particular solution of this equation if we know that x ( 0 ) = 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x ' (t) = 2 t + 4 x ' (t) = 2 (t + 2) First I chose to simplify the right side, factoring out constants. Then Integrate both sides, left with respect to x, and right with respect to t. x(t) = 2((t^2/2) + 2t) + c x(t) = t^2 + 4t + c Simplified, redistrubuting the constants makes it more reconizable. x(0) = 3 x(0) = 0^2 + 4(0) + c = 3 c = 3 x(t) = t^2 + 4t + 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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: x''(t) = 2t - 0.5 First the integration of both sides once yeilds, x'(t) = t^2 - 0.5t + c Since x'( 0 ) = 7 x'(0) = 0^2 - 0.5(0) + c = 7 c = 7 x'(t) = t^2 - 0.5t + 7 Then the integration of both sides for the second time finally yeilds a solution for our second order differential equation. x(t) = (1/3)t^3 - (1/4)t^2 + 7t + c Since x( 0 ) = 1 x(0) = (1/3)(0)^3 - (1/4)(0)^2 + 7(0) + c = 1 c = 1 Our equation and final solution becomes: x(t) = (1/3)t^3 - (1/4)t^2 + 7t + 1 And if we take the derivative of this twice, we can find as a check that it does equal the original problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Integrating both sides we obtain x ' = t^2 - .5 t + c_1, where c_1 is an arbitrary constant. Integrating this equation we obtain x = t^3 / 3 - .25 t^2 + c_1 * t + c_2, where c_2 is an arbitrary constant. Our general solution is thus x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2. The condition x(0) = 1 becomes x(0) = 0^3 / 3 - .25 * 0^2 + c_1 * 0 + c_2 = 1 so that c_2 = 1. x ' (t) = t^2 - .5 t + c_1, so our second condition x ' (0) = 7 becomes x ' (0) = 0^2 - .5 * 0 + c_1 = 7 so that c_1 = 7. For these values of c_1 and c_2, our general solution x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2 becomes the particular solution x(t) = t^3 / 3 - .25 t^2 + 7 t + 1. You should check to be sure this solution satisfies both the given equation and the initial conditions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Use the particular solution from the preceding problem to find x and x ' when t = 3. Interpret your results if x(t) represents the position of an object at clock time t, assuming SI units. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x(t) = (1/3)t^3 - (1/4)t^2 + 7t + 1 finding x(t), when t =3 yeilds x(3) = (1/3)(3)^3 - (1/4)(3)^2 + 7(3) + 1 x(3) = 28.75 meters finding x'(t) when t = 3 yeilds x'(t) = t^2 - 0.5t + 7 x'(3) = (3)^2 - 0.5(3) + 7 x'(3) = 14.5 meters/second confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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): ------------------------------------------------ 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) other forms fail to include x' confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The right-hand side of the equation includes the function x ' but does not include the variable t or the function x. So the right-hand side can be represented by any function which includes among its variables x '. That function may also include x and/or t as a variable. The forms f(t) and f(x, t) fail to include x ', so cannot be used to represent this equation. All the other forms do include x ' as a variable, and may therefore be used to represent the equation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. If F_frict is zero, then the function x in the equation x '' = -F_frict / m - c / m * x ' represents the position of an object of mass m, on which the net force is - c * x '. Explain why the expression for the net force is -c * x '. Explain what happens to the net force as the object speeds up. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F_net = ma = mx'' Therefore if we put the original equation in this form by multiplying through by m, we obtain: mx '' = -F_frict - c * x ' And since F_frict = 0, it becomes: F_net = mx '' = - c * x ' x' = velocity so as it increases F_net also must increase. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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: If w(t) = x'(t) then w'(t) = x''(t) If x '' = - b / m * x ' and w = x' then x '' = - b / m * w w'(t) = - b / m * w and w(t) = - b / m * integration of (w(t)) There is not enough about w(t) t0 solve it confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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? 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: This problem appears to be worked already above. There is no given solution, so I think that maybe the solution above was supposed to go down there. I wrote down most of what was above, and there is nothing left to answer. At this point my own solution to the above would be a rehashing of the above. The good news though, is now I know how to do one of these problems of this general form. If given a different equation and numbers, then perhaps I could use this one as an example. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: The values of x ' at the five points (0, 0), (1/4, 1/4), (1/2, 1/2), (3/4, 3/4) and (1, 1) are -0.5, 0, 0.75, 1.75, and 3 As a curve passes through point (0.2, 0.3) From the above problem, q012, we gather that x' = 0.12 and its equation of the tangent is x= 0.12t + 0.24 and when we use this equation to see what x will be when t is 1, x = -0.12 x' = 1.35 x-0.7 = 1.35 (t - 0.5) x = 1.35t + 0.2 from this we can gather that when t =1, x = 1.15 Compared to the earlier problem, the slope has changed drastically. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: Again in this problem, there is alot of work already done like in q012. The general solution is x = A e^(t^2 + 2 t) + .25 and particular solution for (t, x) -> (0.2, 0.3) is x(t) = .05 / e^(.44) * e^(t^2 + 2 t) + .25 using the above particular solution, for x(t) for t = 1, we get x(t) = 0.8968 and t = 0.4 -> x(t) = 0.0841 and t = 0.6 -> x(t) = 0.1532 These values are significantly different than in the previous problems, and makes me wonder if I did something wrong. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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? (2x - 0.5)(t + 1) = 1 -> x = (1/2(t + 1) + 0.25) This is a version of a familiar curve x=1/t vertically compressed by a factor of 2, shifted to the left by 1 and 0.25 vertically Asympototes are at t = -1, and x = 0.25 at t = 0, the point (0, 0.75) is seen at t = 1, the point (1, 0.5) is shown x ' = 0? (2x - 0.5)(t + 1) = 0 -> x = 0.25 this is a horizontal line, which is also the horizontal asymptote of the previous solution. x ' = 2? (2x - 0.5)(t + 1) = 2 -> x = (1/(t + 1) + 0.25) This is a version of a familiar curve x=1/t shifted to the left by 1, and 0.25 vertically Asympototes are at t = -1, and x = 0.25 at t = 0, the point (0, 1.25) is seen at t = 1, the point (1, 0.75) is shown x ' = -1? (2x - 0.5)(t + 1) = -1 -> x = (-1/2(t + 1) + 0.25) This is a version of a familiar curve x=1/t but it is also reflected across the x axis vertically compressed by a factor of 2, shifted to the left by 1 and 0.25 vertically Asympototes are at t = -1, and x = 0.25 at t = 0, the point (0, -1/12) is seen at t = 1, the point (1, 0) is shown How consistent is your sketch with your previous sketch of the direction field? Accurate. 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. Therefore t = 1 would still give an x of 0.25 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: x ' = 1 when (2 x - .5) * (t + 1) = 1. Solving for x we obtain x = 1/2 ( 1 / (t + 1) + .5) = 1 / (2(t + 1)) + .25. The resulting curve is just the familiar curve x = 1 / t, vertically compressed by factor 2 then shifted -1 unit in the horizontal and .25 unit in the vertical direction, so its asymptotes are the lines t = -1 and x = .25. The t = 0 and t = 1 points are (0, .75) and (1, .5). Similarly we find the curves corresponding to the other values of x ': For x ' = 0 we get the horizontal line x = .25. Note that this line is the horizontal asymptote to the curve obtained in the preceding step. For x ' = 2 we get the curve 1 / (t + 1) + .25, a curve with asymptotes at t = -1 and x = .25, including points (0, 1.25) and (1, .75). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: These questions can be challenging if you are rusty on first-year calculus and trigonometric functions. Fortunately we won't be using these functions a lot near the beginning of the course, but they become very important later, and if you're rusty, or if you never really mastered these functions, you want to get a good start now. If you are rusty, try to answer the following without the use of a calculator: Quick refresher on trigonometric functions: What are the maximum and minimum possible values of cos(theta)? 1, and -1 #$&* For what values of theta does cos(theta) take its maximum value, and for what values of theta is cos(theta) equal to zero? 2n pi, n = any integer pi /2 #$&* 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? 1, 2 pi