#$&* course Mth 174 Calculus IIAsst # 16
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Given Solution: We first shift the interval (- pi , pi) to the interval (0, 1). To shift the interval (-pi, pi) to (-1/2, 1/2) you would replace x by 2 pi x. To then shift the interval to (0, 1) you would in addition substitute x-1/2 for x. The function sin(k x) would become sin(2 pi k ( x - 1/2) ) = sin(2 pi kx - k pi). For even k this would be just sin(2 pi k x); for odd k this would be -sin(2 pi k x). (recall that replacing x with x - b shifts the graph b units in the x direction) The integral of the function itself over the interval is 1, so you would have a0 = 1/2. The integral of x * sin(2 pi k x) from 0 to 1 is easily found by integration by parts to be 1 / (2 pi k) (details of the integration: let u = x, dv = sin(2 pi k x) dx; v = -1 / (2 pi k) cos(2 pi k x), so the integral of v du is a multiple of sin(2 pi k x) and hence yields 0 at both limits; the u v term is -x cos(2 pi k x) / (2 pi k), which yields 0 at the left limit and -1 / (2 pi k) at the right limit). It follows that the first five terms of the series would be 1/2, -1/(2 pi), -1/(4 pi), -1 / (6 pi) and -1 / (8 pi). This yields the Fourier polynomial 1/2 - (1/pi)sin(2pix) - 1/(2pi)sin(4pix) - 1/(3pi)sin(6pix) - 1/(4pi)sin(8pix). The graph below depicts this function on the approximate interval (-1, 2) in red, as well as the function y = x (in blue). On the interval (0, 1) the two functions agree very well. You can see that the Fourier series repeats with period 1. . **** What substitution did you use to compensate for the fact that the period of the function is not 2 `pi?
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17:19:41 student solution: f has period b = 1, so I let x = bt/2'pi = t/2'pi &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: **** Query problem 10.5.25 (4th edition 10.5.24 was 9.5.24) integral of cos^2(mx) from -`pi to `pi is `pi **** which formula from the table did you use to establish your result and what substitution did you use?
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21:32:53 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: int(cos^2(mx)) = 1/2 cos x sin x/m + 1/2 x + C int(cos^2(mx) dx, -`pi, `pi) = cos`pi sin`pi/2m + `pi/2 - cos(-`pi)sin(-`pi)/2m +`pi/2 = `pi 0/2m - 0/2m = 0 thus, int( cos^2(mx) dx, -`pi, `pi) = `pi confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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21:32:53 ......!!!!!!!! STUDENT QUESTION on Problem #18: """"Justify the formula b(sub k) = (1/pi)integral f(x)sin(kx) dx (-pi to pi)..."""" I was able to complete this problem using the methodology on p.511-512. But I don't totally understand their justification for reducing the right side to one term (section above first boxed formula on p. 512. sin(kx) cos(mx) is the product of an even and an odd function, so is itself an odd function. Its integral over any region symmetric with respect to the origin must be 0. Note that cos(kx) goes through k complete cycles between x = -pi and x = pi, while cos(mx) goes through m complete cycles. Assuming k > m (which can be assumed without loss of generality since the values of k and m can be interchanged without changing the integrand), cos(kx) and cos(mx) will go 'in and out of phase' k - m times between x = -pi and x = pi, each time on an interval of length 2 pi / (k - m). This is easy to see by graphing two cosine functions for different integer values of k and m. Partition the interval [-2 pi, 2 pi] into subintervals each having length 2 pi / (k - m). On each such subinterval the product cos(kx) * cos(m x) is antisymmetric with respect to the midpoint of the interval and hence contributes 0 to the integral. Since the interval [-pi, pi] is partitioned into such subintervals, the integral from -pi to pi is 0. This can be easily but somewhat messily 'nailed down' with algebraic-trigonometric manipulations, which however don't add much to this geometric explanation. The function depicted below is a superposition of sine and cosine functions. The function is defined on the interval -pi < x < pi, which corresponds to the interval between the vertical lines near the left and right sides of the picture. By the methods of this section we can find the Fourier series for this function, which has coefficients a_1 = 1 and a_5 = .3. All other coefficients are 1. Thus the Fourier series for this function is f(x) = 1 cos(1x) + .3 cos(5x), or just f(x) = cos(x) + .3 cos(5x). Here's a graph of y = cos(x): Here's a graph of y = .3 cos(5x) Here's a graph of both functions together. If the y coordinates of each point of this graph are added they will give us the y coordinates of the original graph, so we say that the sum of these two graphs is the original graph. The amplitude of the cos(x) term is greater than that of the cos(5x) term, so we say that the cos(x) term has more energy. The energy is in fact determined by amplitude. The first term has amplitude 1, so its energy relative energy is 1. The first term has amplitude .3, so its relative energy is .3. If we had a general Fourier series with a_k and b_k terms, then they would be the coefficients of a cosine and a sine function. • Think of the cosine and the sine contributions as being represented by the legs of a right triangle. The hypotenuse represents what we get when we combine them. The hypotenuse would be sqrt(a_k^2 + b_k^2). • The actual argument is a little more specific than this, but this at least makes it plausible that the energy term is sqrt(a_k^2 + b_k^2). The energy for the k = 0 term is simply sqrt(A_0^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 174 017. `query 17Cal 2
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Given Solution: Substitute y = cos(omega * t) into the equation. The goal is to find the value of omega that satisfies the equation: We first calculate y ‘’ y = cos(omega*t) so y' = -omega*sin(omega*t) and y"""" = -omega^2*cos(omega*t) Now substituting in y"""" + 9y = 0 we obtain -omega^2*cos(omega*t) + 9cos(omega*t) = 0 , which can be easily solved for omega: -omega^2*cos(omega*t) = -9cos(omega*t) omega^2 = 9 omega = +3, -3 Both solutions check in the original equation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: Query problem 11.1.18 (4th edition 11.1.14 3d edition 11.1.13) (formerly 10.6.1) P = 1 / (1 + e^-t) satisfies dP/dt = P(1-P) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = 1/(1+e^-t) = e^t/(e^t + 1) dP/dt = [(e^t)^2 + e^t - (e^t)^2]/(e^t+1)^2 = e^2/(e^t+1)^2 1 - P = 1 - e^t/(e^t+1) = 1/(e^t+1) P (1 - P) = e^t/(e^t+1) * 1/(e^t+1) = e^2/(e^t+1)^2 = dP/dt confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: RESPONSE --> P= 1/(1+e^-t) ; multiplying numerator and denominator by e^t we have P = e^t/(e^t+1) , which is a form that makes the algebra a little easier. dP/dt = [ (e^t)’ ( e^t + 1) - (e^t + 1) ‘ e^t ] / (e^t + 1)^2, which simplifies to dP/dt = [ e^t ( e^t + 1) - e^t * e^t ] / (e^t+1)^2 = e^t / (e^t + 1)^2. Substituting: P(1-P) = e^t/(e^t+1) * [1 - e^t/(e^t+1)] . We simplify this in order to see if it is the same as our expression for dP/dt: e^t/(e^t+1) * [1 - e^t/(e^t+1)] = e^t / (e^t + 1) - (e^t)^2 / (e^t + 1)^2. Putting the first term into a form with common denominator e^t ( e^t + 1) / [ (e^t + 1) ( e^t + 1) ] - (e^t)^2 / (e^t + 1 ) ^ 2 = (e^(2 t) + e^t) / (e^t + 1)^2 - e^(2 t) / (e^t + 1) ^ 2 = (e^(2 t) + e^t - e^(2 t) ) / (e^t + 1)^2 = e^t / (e^t + 1)^2. This agrees with our expression for dP/dt and we see that dP/dt = P ( 1 - P). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: Query problem (omitted from 5th edition but should be worked and self-critiqued) (4th edition 11.1.16 3d edition 11.1.15 formerly 10.1.1) match one of the equations y '' = y, y ' = -y, y ' = 1 / y, y '' = -y, x^2 y '' - 2 y = 0 with each of the solutions (cos x, cos(-x), x^2, e^x+e^-x, `sqrt(2x) ) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = cos x, f`(x) = -sin x, f``(x) = -cos x, in which y`` = -y f(x) = cos(-x), f`(x) = sin(-x), f``(x) = -cos(-x), in which y`` = -y f(x) = x^2, f`(x) = 2x, f``(x) = 2, in which x^2y`` - 2y = 0 e^x+e^-x = f(x), f`(x) = e^x - e^-x, f``(x) = e^x+e^-x, in which y`` = y f(x) = `sqrt(2x), f`(x) = 1/2sqrt(2x) *2 = 1/`sqrt(2x), in which y` = 1/y confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: . y '' = y would give us (x^2) '' = x^2. Since (x^2)'' = 2 we would have the equation 2 = x^2. This is not true, so y = x^2 is not a solution to this equation. B) y' = -y and its solution can be (II) y = cos(-x) C) y' = 1/y and its solution can be (V) y = sqrt(2x) D) y'' = -y and its solution can be (II) y = cos(-x) E) x^2y'' - 2y = 0 and its solution can be (IV) y = e^x + e^-x
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RESPONSE --> Only solution IV corresponds to y"""" = y y = e^x + e^(-x) y' = e^x - e^(-x) y"""" = e^x + e^(-x) Thus, y = y"""" Solution I, y = cos x, y' = -sin x, y""""= -cos x, not = y Solution II, y = cos(-x), y' = sin(-x), y"""" = - cos(-x), not = y Solution III, y = x^2, y' = 2x, y"""" = 2, not = y Solution V, y = sqrt (2x), y' = 1/sqrt(2x), y"""" = -1/sqrt[(2x)^3], not = y which solution(s) correspond to the equation y' = -y and how can you tell
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RESPONSE --> None of the solutions correspond to y' = -y Solution I: y = cos x, y' = -sin x, not = -y Solution II: y = cos(-x), y' = sin x, not = -y Solution III: y = x^2, y' = 2x, not = -y Solution IV: y = e^x + e^(-x), y' = e^x - e^(-x), not = -y Solution V: y = sqrt(2x), y' = 1/sqrt(2x), not = -y which solution(s) correspond to the equation y' = 1/y and how can you tell
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RESPONSE --> Solution V corresponds to y' = 1/y y = sqrt(2x), y' = 1/sqrt(2x), y' = 1/y None of the other solutions correspond (see previous responses). which solution(s) correspond to the equation y''=-y and how can you tell
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RESPONSE --> Solutions I and II correspond to y"""" = -y Solution I: y = cos x, y' = -sin x, y"""" = -cos x, y"""" = -y Solution II: y = cos(-x), y' = sin(-x), y"""" = -cos(-x), y"""" = -y which solution(s) correspond to the equation x^2 y'' - 2y = 0 and how can you tell
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RESPONSE --> Solution III corresponds to x^2*y"""" - 2y = 0 Solution III: y = x^2, y' = 2x, y"""" = 2 Substituting: x^2*(2) - 2(x^2) = 2x^2 - 2x^2 = 0 None of the other solutions correspond. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: Query problem 11.2.8 (4th edition 11.2.5 3d edition 11.2.4) The slope field is shown. Plot solutions through (0, 0) and (1, 4). Describe your graphs. Include in your description any asymptotes or inflection points, and also intervals where the graph is concave up or concave down. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the solution through (0,0) is concave upward to about the point (3,5) and then became concave downward, and approach y = 10 as an asymptotes. the solution through (1,4) is concave downward from interval (1,inf), it is concave upward in interval (-1, 1) and approach y = 10 as an asymptotes confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: RESPONSE --> Solution through (0,0): concave up from (0,0) to about (3,3) then concave down for t>3, point of inflection at (3,3), horizontal asymptote at P=10. Graph appears to be P=0 for all t<0. Since givens include P>=0, no graph for negative P. Solution through (1,4): concave up from (-1,0) to about (2,3) , then concave down for t>2, point of inflection at (2,3), horizontal asymptote at P=10. Graph appears to be P=0 for at t<-1. Since givens include P>=0, no graph for negative P.
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14:13:09 Query problem 11.2.10 (was 10.2.6) slope field
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14:22:17 describe the slope field corresponding to y' = x e^-x
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RESPONSE --> Slope field corresponding to y' = x e^-x: Slope field III: slope increasingly negative for x<0, slope increasing positive (at decreasing rate) to about x=1, then decreasing positive, slope appears to have a limit of 0 for large x.
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14:25:41 describe the slope field corresponding to y' = sin x
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RESPONSE --> Slope field for y' = sin x: Slope field I: Slope ranges from 0 to 1 to 0 to -1 to 0 repeatedly, with slope = 0 at (0,0), period = 2pi
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14:28:17 describe the slope field corresponding to y' = cos x
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RESPONSE --> Slope field corresponding to y' = cos x: slope field II. Same as slope field for y' = sin x, except slope = 1 at x = 0 (cyled shifted pi/2 to left).
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14:32:27 describe the slope field corresponding to y' = x^2 e^-x
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RESPONSE --> Slope field corresponding to y' = x^2 e^-x: Slope field IV: Slope increasingly positive as x decreases from 0, slope 0 at x=0, slope increasingly positive (at decreasing rate) from x=0 to x=2, then decreasingly positive, appears to approach 0 for large x.
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14:36:20 describe the slope field corresponding to y' = e^-(x^2)
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RESPONSE --> Slope field corresponding to y' = e^-(x^2): Slope field V: slope maximum at x= 0, decreasing positive as |x| icreases, slope appears to approach 0 as |x| gets large (slope is very small with |x|>2).
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14:40:55 describe the slope field corresponding to y' = e^-x
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RESPONSE --> Slope field corresponding to y' = e^-x: Slope field VI: Slope increasing positive as x decreases, slope 1 at x=0, slope approaches 0 for large positive x (slope very small for x>about 2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 174 Calculus IIAsst # 16
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Given Solution: We first shift the interval (- pi , pi) to the interval (0, 1). To shift the interval (-pi, pi) to (-1/2, 1/2) you would replace x by 2 pi x. To then shift the interval to (0, 1) you would in addition substitute x-1/2 for x. The function sin(k x) would become sin(2 pi k ( x - 1/2) ) = sin(2 pi kx - k pi). For even k this would be just sin(2 pi k x); for odd k this would be -sin(2 pi k x). (recall that replacing x with x - b shifts the graph b units in the x direction) The integral of the function itself over the interval is 1, so you would have a0 = 1/2. The integral of x * sin(2 pi k x) from 0 to 1 is easily found by integration by parts to be 1 / (2 pi k) (details of the integration: let u = x, dv = sin(2 pi k x) dx; v = -1 / (2 pi k) cos(2 pi k x), so the integral of v du is a multiple of sin(2 pi k x) and hence yields 0 at both limits; the u v term is -x cos(2 pi k x) / (2 pi k), which yields 0 at the left limit and -1 / (2 pi k) at the right limit). It follows that the first five terms of the series would be 1/2, -1/(2 pi), -1/(4 pi), -1 / (6 pi) and -1 / (8 pi). This yields the Fourier polynomial 1/2 - (1/pi)sin(2pix) - 1/(2pi)sin(4pix) - 1/(3pi)sin(6pix) - 1/(4pi)sin(8pix). The graph below depicts this function on the approximate interval (-1, 2) in red, as well as the function y = x (in blue). On the interval (0, 1) the two functions agree very well. You can see that the Fourier series repeats with period 1. . **** What substitution did you use to compensate for the fact that the period of the function is not 2 `pi?
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17:19:41 student solution: f has period b = 1, so I let x = bt/2'pi = t/2'pi &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: **** Query problem 10.5.25 (4th edition 10.5.24 was 9.5.24) integral of cos^2(mx) from -`pi to `pi is `pi **** which formula from the table did you use to establish your result and what substitution did you use?
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21:32:53 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: int(cos^2(mx)) = 1/2 cos x sin x/m + 1/2 x + C int(cos^2(mx) dx, -`pi, `pi) = cos`pi sin`pi/2m + `pi/2 - cos(-`pi)sin(-`pi)/2m +`pi/2 = `pi 0/2m - 0/2m = 0 thus, int( cos^2(mx) dx, -`pi, `pi) = `pi confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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21:32:53 ......!!!!!!!! STUDENT QUESTION on Problem #18: """"Justify the formula b(sub k) = (1/pi)integral f(x)sin(kx) dx (-pi to pi)..."""" I was able to complete this problem using the methodology on p.511-512. But I don't totally understand their justification for reducing the right side to one term (section above first boxed formula on p. 512. sin(kx) cos(mx) is the product of an even and an odd function, so is itself an odd function. Its integral over any region symmetric with respect to the origin must be 0. Note that cos(kx) goes through k complete cycles between x = -pi and x = pi, while cos(mx) goes through m complete cycles. Assuming k > m (which can be assumed without loss of generality since the values of k and m can be interchanged without changing the integrand), cos(kx) and cos(mx) will go 'in and out of phase' k - m times between x = -pi and x = pi, each time on an interval of length 2 pi / (k - m). This is easy to see by graphing two cosine functions for different integer values of k and m. Partition the interval [-2 pi, 2 pi] into subintervals each having length 2 pi / (k - m). On each such subinterval the product cos(kx) * cos(m x) is antisymmetric with respect to the midpoint of the interval and hence contributes 0 to the integral. Since the interval [-pi, pi] is partitioned into such subintervals, the integral from -pi to pi is 0. This can be easily but somewhat messily 'nailed down' with algebraic-trigonometric manipulations, which however don't add much to this geometric explanation. The function depicted below is a superposition of sine and cosine functions. The function is defined on the interval -pi < x < pi, which corresponds to the interval between the vertical lines near the left and right sides of the picture. By the methods of this section we can find the Fourier series for this function, which has coefficients a_1 = 1 and a_5 = .3. All other coefficients are 1. Thus the Fourier series for this function is f(x) = 1 cos(1x) + .3 cos(5x), or just f(x) = cos(x) + .3 cos(5x). Here's a graph of y = cos(x): Here's a graph of y = .3 cos(5x) Here's a graph of both functions together. If the y coordinates of each point of this graph are added they will give us the y coordinates of the original graph, so we say that the sum of these two graphs is the original graph. The amplitude of the cos(x) term is greater than that of the cos(5x) term, so we say that the cos(x) term has more energy. The energy is in fact determined by amplitude. The first term has amplitude 1, so its energy relative energy is 1. The first term has amplitude .3, so its relative energy is .3. If we had a general Fourier series with a_k and b_k terms, then they would be the coefficients of a cosine and a sine function. • Think of the cosine and the sine contributions as being represented by the legs of a right triangle. The hypotenuse represents what we get when we combine them. The hypotenuse would be sqrt(a_k^2 + b_k^2). • The actual argument is a little more specific than this, but this at least makes it plausible that the energy term is sqrt(a_k^2 + b_k^2). The energy for the k = 0 term is simply sqrt(A_0^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!