course Mth 272 7/28 11:45 am 020.
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 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: To find the present value using the Simpson’s Rule, we must first identify that our domain exists from [0,4]. We can now set up our present value equation, which we find id present value = Int((6000+2000sqrt(t)) e^-.07t)dt. Since we are solving an integral, we can now use Simpson’s Rule to solve. Because n = 8, we end up with the new intervals of [0,.5][.5,1][1,1.5][1.5,2][2,2.5][2.5,3][3,3.5][3.5,4]. When we use the Simpson’s Rule formula, we find that the integral = 4/(3*8)(f(0) + 4f(.5) + 2f(1)+ 4f(1.5) + 2f(2) + 4f(2.5) +2f(3) +4f(3.5) + f(4)). When we evaluate this, we find that our integral = 21964.60. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer is off from the given answer. I think this is because I rounded throughout my work, which made it not as accurate as the given answer. ------------------------------------------------ Self-critique Rating: ********************************************* 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: To find the distance, we can first integrate the fnction and find the area. Since this function is easily integrated using the power rule, our integral = 1/3(2/5x^(5/2) – 2x^(3/2) +2x) + C. Since we are finding the area over [0,1], we can use the Fundamental Theorem of Calculus to find the area. We find that the area under the arc is 2/15. Now we can use our knowledge about triangles to solve the rest of the problem. Since we know that the area of a triangle is equal to ½ base * height, we can estimate the distance traveled by using this formula. I am stuck after this point, I looked up the distance of an arc and found the formula dealing with radians, but have no idea how to use it to find the distance. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. "