#$&* course MTH272 If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a You would use x0=0, x1=1/2, x2=1 x3=3/2 x4=2. The corresponding values of x^2 sqrt(x^2+1) are 0, 0.5590169943, 1.414213562, 2.704163456, 4.472135954. Using these values: Trap gives you 3.4567. Simpson's rule gives you 3.3922. The exact result, to five significant figures, is int(x sqrt(x^2+1),x,0,2) = 3.3934. This is based on the antiderivative 1/3 * (x^2+1)^(3/2) According to these results Simpson's approximation is within .0012 while trap is within about .064. So the Simpson's approximation is .064 / .0012 = 50 times better, approx.. Theory says that it should be about n^2 = 4^2 = 16 times better. ** To integrate x sqrt( x^2 + 1 ) let u = x^2 so that du = 2 x dx. The integrand becomes 1/2 sqrt(u) du with antiderivative 1/2 ( 2/3 u^(3/2) ) = 1/3 u^(3/2), which with the given limits, calculated to five significant figure, gives you the result 3.3934. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 6.4.21 (7th edition 6.5.21) (was 6.5.19) present value of 6000 + 200 `sqrt(t) at 7% for 4 years by Simpson's rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6000 + 200 `sqrt(t) at 7% and 4 years Modeling after any other value over time C(t)e^(-rt)= present value 6000+200sqrt(t)e^(-.07(4)) (6000+200 sqrt(0)) e^(- 0.07 * 0)=6000*1=6000 (6000+200 sqrt(1/2)) e^(- 0.07 * 1/2)=5930.19*4=23720.76 (6000+200 sqrt(1)) e^(- 0.07 * 1)=5780.84*2=11561.68 (6000+200 sqrt(3/2)) e^(- 0.07 * 3/2)=5622.48*4=22489.92 (6000+200 sqrt(2)) e^(- 0.07 * 2)=5462.04*2=10924.08 (6000+200 sqrt(5/2)) e^(- 0.07 * 5/2)5302.20*4=21208.80 (6000+200 sqrt(3)) e^(- 0.07 * 3)5144.30*2=10288.60 (6000+200 sqrt(7/2)) e^(- 0.07 * 7/2)=4989.09*4=19956.36 (6000+200 sqrt(4)) e^(- 0.07 * 4)=4837.02*1=4837.02 ((b-a)/(4*n))=((4-0)/(3*8))= 4/24=1/6 Total 130987.22*(1/6)=21831.20 Actual 21836.78 Difference between (Actual-Simpson Solution) 5.58 21831.20-6000.00=15831.20 final present value confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a This income stream would be $6000 at the beginning and $6400 at the end of the 4-year period. The income during this period would therefore be between $24,000 and $25,600. The present value of this income will be somewhat less. The present value of [6000 + 200 `sqrt(t)] `dt, for a short time interval t, is [ 6000 + 200 `sqrt(t) ] * e^(-.07 t ) `dt. Evaluating the integral INT( ( 6000 + 200 `sqrt(t) ) * e^(-.07 t ) with respect to t from 0 to 4) we obtain $21,836.98 The Simpson approximation to the integral, using n = 8, is 4/24 * ( (6000 + 200 sqrt(0)) e^(- 0.07 * 0) + 4•((6000 + 200 sqrt(1/2)) e^(- 0.07 * 1/2)) + 2•((6000 + 200 sqrt(1)) e^(- 0.07 * 1)) + 4•((6000 + 200 sqrt(3/2)) e^(- 0.07 * 3/2)) + 2•((6000 + 200 sqrt(2)) e^(- 0.07 * 2))+4•((6000 + 200 sqrt(5/2)) e^(- 0.07 * 5/2)) + 2•((6000 + 200 sqrt(3)) e^(- 0.07 * 3)) + 4•((6000 + 200 sqrt(7/2)) e^(- 0.07 * 7/2)) + (6000 + 200 sqrt(4)) e^(- 0.07 * 4) ) = $21 831. You might go blind trying to read that long expression. Breaking the same approximation down term-by-term: 4/24 * ( (6000 + 200 sqrt(0)) e^(- 0.07 * 0) + 4•((6000 + 200 sqrt(1/2)) e^(- 0.07 * 1/2)) + 2•((6000 + 200 sqrt(1)) e^(- 0.07 * 1)) + 4•((6000 + 200 sqrt(3/2)) e^(- 0.07 * 3/2)) + 2•((6000 + 200 sqrt(2)) e^(- 0.07 * 2))+ 4•((6000 + 200 sqrt(5/2)) e^(- 0.07 * 5/2)) + 2•((6000 + 200 sqrt(3)) e^(- 0.07 * 3)) + 4•((6000 + 200 sqrt(7/2)) e^(- 0.07 * 7/2)) + (6000 + 200 sqrt(4)) e^(- 0.07 * 4) ) STUDENT QUESTION A bit off here, and still unsure with the Simpson rule. In comparison with the midpoint rule, what would I, overall, be doing with the rule? I have it my text, but confusing something here. INSTRUCTOR RESPONSE An 8-interval trapezoidal or midpoint approximation would divide the 4-year interval into 4 equal subintervals, each of length 1. For a midpoint approximation, you would evaluate the function at the midpoint of each subinterval (in this case, at 1/4, 3/4, 5/4 ..., 15/4), the midpoints of the 8 intervals), and multiply each value by the width of the interval (in this case 1/2), to get the approximation to the 'area' for that interval. For a trapezoidal approximation, you would evaluate the function at the endpoints of the intervals (in this case at 0, 1/2, 1, 3/2, 2, 5/2, 3, 7/2 and 4; note that an 8-interval approximation has 9 endpoints). Each interval has two endpoints, and for each you would average these two values, then multiply this average by the width of the interval in order to get the 'area' approximation. Adding all 8 of your approximations you would get your approximation to the 'total area', i.e., the integral. The trapezoidal and midpoint approximations will generally vary slightly, with one tending to overestimate and the other to underestimate the area. Simpson's rule can in fact be stated in terms of the trapezoidal and midpoint approximations, but your text chooses to use a formula (nothing wrong with that choice). In effect what you do when using Simpson's Rule is break the interval into 8 subintervals, and you evaluate the function at each endpoint, just as you do with the trapezoidal approximation. You have 9 endpoints, counting the first point, just as you did with the trapezoidal rule. However you use the 9 values differently: You use the first value 1 time, the second value 4 times, the third value 2 times, the fourth value 4 times, the fifth value 1 time, the sixth value 4 times, the seventh value 2 times, the eighth value 4 times, and the ninth value once. The pattern is simple enough: 1 4 2 4 2 4 ... 2 4 1, where the ... continues as long as necessary to accomodate the value of n. Now if you add up 1, 4, 2, 4, 2, 4, 2, 4 and 1 you get 24, which is 3 times as great as your value of n (which is 8). If you use a larger value of n, you will need to fill in the ... with additional pairs of 2's and 4's, but your total will still be 3 * n. In the n = 8 case you will then have effectively added up 24 numbers, each corresponding to one of the 9 endpoint values. These values represent the values of the function, from left to right, along the original interval. So if you divide this total by 24, you will get a weighted average of the endpoint values. Multiplying this 'weighted-average value' by the width of your original interval will give you a very good estimate of the 'area' corresponding to that interval, i.e., a very good approximation to the integral. You don't need to know this, but in case you are curious the reasons for the pattern 1 4 2 4 ... 2 4 1 are related to the fact that Simpson's rule is based on a quadratic-function approximation to each series of 3 points, rather than the linear approximations used by the trapezoidal and midpoint rules. You do need to know this: The formula for Simpson's Rule is (b - a) / (3 n) ( f(x_0) + 4 f(x_1) + 2 f(x_2) + ... + 2 f(x_(n-2)) + 4 f(x_(n-1)) + f(x_n)). It might be helpful to understand this: ( f(x_0) + 4 f(x_1) + 2 f(x_2) + ... + 2 f(x_(n-2)) + 4 f(x_(n-1)) + f(x_n)) / (3 n) is the weighted average of the 3 n numbers you effectively add when you calculate ( f(x_0) + 4 f(x_1) + 2 f(x_2) + ... + 2 f(x_(n-2)) + 4 f(x_(n-1)) + f(x_n)). This weighted average is a very good approximation to the average value of f(x) on the interval. When you multiply this 'weighted average' by (b - a), which is the width of the interval, you therefore get a very good approximation to the 'area', or integral. Disclaimer: The word 'area' is used loosely in the above. Actual areas are always positive, but integrals can be negative. In the loose sense used here we think of areas below the horizontal axis as negative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): NA ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery distance traveled by pursuer along path y = 1/3 (x^(3/2) - 3 x^(1/2) + 2) over [0, 1]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Come up with distance traveled y = 1/3 (x^(3/2) - 3 x^(1/2) + 2) over [0, 1]. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a This is going to be an arc length integral. The x = 0 point of the curve is (0, 2/3) and the x = 1 point is (1, 0). The straight-line distance between these points is thus sqrt(1^2 + (2/3)^2) = sqrt(13/9) = 1.2 or so. The straight line being the shortest distance between two points, we expect the arc distance to be more than 1.2. However unless the curve gets pretty steep the distance won't be drastically greater than this. ** Integrate to find the arc length of the curve. If a line with slope m lies above an interval `dx on the x axis then the 'run' of the segment above `dx is just `dx, while the rise is m * `dx. The hypotenuse is therefore `sqrt( `dx^2 + ( m `d)^2 ) = `sqrt( 1 + m^2 ) * `sqrt(`dx^2) = `sqrt(1 + m^2) `dx. If a curve y = f(x) lies above the interval `dx then the average slope of the curve is a value of y ' for some x in the interval. After a little fancy reasoning we can prove that the arc length is in fact equal to `sqrt( 1 + y'^2) `dx, but proof or not it's easy enough to understand if you understand the thing about m. This leads to the theorem that for a differentiable function y = f(x) the arc length between x = a and x = b is INT ( `sqrt( 1 + y' ^ 2) dx, x from a to b). The derivative of the given function is 1/2 x^.5 - 1/2 x^-.5; the square of this expression is 1/4 ( x - 2 + 1/x) so you integrate `sqrt( 1 + 1/4 ( x - 2 + 1/x) ) from 0 to 1. The integral gives you 1.33327, accurate to 6 significant figures. However you'll probably have to use an approximation technique so your result will probably differ after a few significant figures from this result. STUDENT QUESTION Arc length integral, went an entirely different direction on this problem. I reviewed the solution given, but doesn’t help too much due to my misunderstanding. I may have actually interpreted this wrong, the expression given. Any further elaboration is very much appreciated. Thank you. INSTRUCTOR RESPONSE For the reasons given the arc length on an interval is • integral ( `sqrt( 1 + y' ^ 2) dx, x from a to b). It appears that the integral you calculated is integral( y dx, x from 0 to 1) = integral ( 1/3 (x^(3/2) - 3 x^(1/2) + 2) dx, x from 0 to 1). You have to find y ' by taking the derivative, then form the expression sqrt( 1 + y ' ^ 2 ). It is this expression that gets integrated. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I have not seen any material on this at all could point me to proper place in the book. I don’t remember seeing anything in the CD on this topic ------------------------------------------------ Self-critique Rating:
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Given Solution: `a This is going to be an arc length integral. The x = 0 point of the curve is (0, 2/3) and the x = 1 point is (1, 0). The straight-line distance between these points is thus sqrt(1^2 + (2/3)^2) = sqrt(13/9) = 1.2 or so. The straight line being the shortest distance between two points, we expect the arc distance to be more than 1.2. However unless the curve gets pretty steep the distance won't be drastically greater than this. ** Integrate to find the arc length of the curve. If a line with slope m lies above an interval `dx on the x axis then the 'run' of the segment above `dx is just `dx, while the rise is m * `dx. The hypotenuse is therefore `sqrt( `dx^2 + ( m `d)^2 ) = `sqrt( 1 + m^2 ) * `sqrt(`dx^2) = `sqrt(1 + m^2) `dx. If a curve y = f(x) lies above the interval `dx then the average slope of the curve is a value of y ' for some x in the interval. After a little fancy reasoning we can prove that the arc length is in fact equal to `sqrt( 1 + y'^2) `dx, but proof or not it's easy enough to understand if you understand the thing about m. This leads to the theorem that for a differentiable function y = f(x) the arc length between x = a and x = b is INT ( `sqrt( 1 + y' ^ 2) dx, x from a to b). The derivative of the given function is 1/2 x^.5 - 1/2 x^-.5; the square of this expression is 1/4 ( x - 2 + 1/x) so you integrate `sqrt( 1 + 1/4 ( x - 2 + 1/x) ) from 0 to 1. The integral gives you 1.33327, accurate to 6 significant figures. However you'll probably have to use an approximation technique so your result will probably differ after a few significant figures from this result. STUDENT QUESTION Arc length integral, went an entirely different direction on this problem. I reviewed the solution given, but doesn’t help too much due to my misunderstanding. I may have actually interpreted this wrong, the expression given. Any further elaboration is very much appreciated. Thank you. INSTRUCTOR RESPONSE For the reasons given the arc length on an interval is • integral ( `sqrt( 1 + y' ^ 2) dx, x from a to b). It appears that the integral you calculated is integral( y dx, x from 0 to 1) = integral ( 1/3 (x^(3/2) - 3 x^(1/2) + 2) dx, x from 0 to 1). You have to find y ' by taking the derivative, then form the expression sqrt( 1 + y ' ^ 2 ). It is this expression that gets integrated. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I have not seen any material on this at all could point me to proper place in the book. I don’t remember seeing anything in the CD on this topic ------------------------------------------------ Self-critique Rating: