#$&* course Mth 174 assignment # 8......!!!!!!!!...................................
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Given Solution: ** The key is the antiderivative. For p > 1 the antiderivative is still a negative power and approaches 0 as x -> infinity. For p < 1 it's a positive power and approaches infinity--hence diverges. For p = 1 the antiderivative is ln x, which also approches infinity. ** ** An integral converges if the limit of its Riemann sums can be found and is finite. Roughly this corresponds to the area under the curve (positive for area above and negative for area below the axis) being finite and well-defined (consider for example the area under the graph of the sine function, from 0 to infinity, which could be regarded as finite but it keeps fluctuating as the sine goes positive then negative, so it isn't well-defined). ** ** There is very little difference in the rapidity with which 1/x^1.01 approaches zero and the rapidity of the approach of 1 / x^.99. You would never be able to tell by just looking at the partial sums which converges and which diverges. In fact just looking at partial sums you would expect both to converge, because they diverge so very slowly. If you graph 1 / x^.99, 1 / x and 1 / x^1.01 you won't notice any significant difference between them. The first is a little higher and the last a little lower past x = 1. You will never tell by looking at the graphs that any of these functions either converge or diverge. However if you find the antiderivatives you can tell pretty easily. The antiderivatives are 100 x^.01, ln(x) and -100 x^-.01. On the interval from, say, x = 1 to infinity, the first antiderivative 100 x^.01 will be nice and finite at x = 1, but as x -> infinity is .01 power will also go to infinity. This is because for any N, no matter how great, we can find a solution to 100 x^.01 = N: the solution is x = (N / 100) ^ 100. For this value of x and for every greater value of x, 100 x^.01 will exceed N. The antiderivative function grows without bound as x -> infinity. Thus the integral of 1 / x^.99 diverges. On this same interval, the second antiderivative ln(x) is 0 when x = 0, but as x -> infinity its value also approaches infinity. This is because for any N, no matter how large, ln(x) > N when x > e^N. This shows that there is no upper bound on the natural log, and since the natural log is the antiderivative of 1 / x this integral diverges. On the same interval the third antiderivative -100 x^-.01 behaves very much differently. At x = 1 this function is finite but negative, taking the value -100. As x -> infinity the negative power makes x-.01 = 1 / x^.01 approach 0, since as we solve before the denominator x^.01 approaches infinity. Those if we integrate this function from x = 1 to larger and larger values of x, say from x = 1 to x = b with b -> infinity, we will get F(b) - F(1) = -100 / b^.01 - (-100 / 1) = -100 / b^.01 + 100. The first term approaches 0 and b approaches infinity and in the limit we have only the value 100. This integral therefore converges to 100. We could look at the same functions evaluated on the finite interval from 0 to 1. None of the functions is defined at x = 0 so the lower limit of our integral has to be regarded as a number which approaches 0. The first function gives us no problem, since its antiderivative 100 x^.01 remains finite over the entire interval [0, 1], and we find that the integral will be 100. The other two functions, however, have antiderivatives ln(x) and -100 x^-.01 which are not only undefined at x = 0 but which both approach -infinity as x -> 0. This causes their integrals to diverge. These three functions are the p functions y = 1 / x^p, for p = .99, 1 and 1.01. This example demonstrates how p = 1 is the 'watershed' for convergence of these series either on an infinite interval [ a, infinity) or a finite interval [0, a]. ** More generally: ** If p > 1 then the antiderivative is a negative-power function, which approaches 0 as x -> infinity. So the limit as b -> infinity of the integral from 1 to b of 1 / x^p would be finite. If p < 1 the antiderivative is a positive-power function which approaches infinity as x -> infinity. So the limit as b -> infinity of the integral from 1 to b of 1 / x^p would be divergent. These integrals are the basis for many comparison tests. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q explain how we know that the integral from 0 to infinity of e^(-a x) converges for a > 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The convergence of the integral of e^-(ax) from 0 to infinity is plausible because this is a decreasing function. However 1 / x^.99 is also decreasing on the same interval but it doesn't converge. The reason e^(- a x) converges is that it antiderivative is -1/a e^(-a x), and as x -> infinity this expression approaches zero. Thus the integral of this function from 0 to b is -1/a e^(-a b) - (-1/a e^0) = -1/a e^(-ab) + 1. If we allow the upper limit b to approach infinity, e^-(a b) will approach 0 and we'll get the finite result 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Section 7.8 Problem 3 7.8.18 convergence of integral of 1 / sqrt (`theta^2+1) from 1 to infinity YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: On [1, inf] 0<=1/(x^2+1)^(1/2)<1/x Int(1/x dx, 1, inf) is diverges, because p=1 Thus, int(1/(x^2+1)^(1/2) dx, 1, inf) is diverges
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Given Solution: ** If it wasn't for the 1 in the denominator this would be 1 / `sqrt(`theta^2) or just 1 / `theta, which is just 1 / `theta^p for p = 1. As we know this series would diverge (antiderivative of 1 / `theta in ln(`theta), which appropaches infinity as `theta -> infinity). As `theta gets large the + 1 shouldn't matter much, and we expect that 1 /`sqrt (`theta^2 + 1) diverges. If this expression is shown to be greater than something that diverges, then it must diverge. However, 1 / `sqrt ( `theta^2 + 1) < 1 / `theta, and being less than a divergent function that diverges does not prove divergence. We can adjust our comparison slightly: Since 1 / `theta gives a divergent integral, half of this quantity will yield half the integral and will still diverge. i.e., 1 / (2 `theta) diverges. • So if we can show that 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we will have proved the divergence of 1 / `sqrt(`theta^2 + 1). We prove this. Starting with 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we square both sides to get 1 / (4 `theta^2) < 1 / (`theta^2+1). Multiplying by common denominator we get `theta^2 + 1 < 4 `theta^2. Solving for `theta^2 we get 1 < 3 `theta^2 `sqrt(3) / 3 < `theta. This shows that for all values of `theta > `sqrt(3) / 3, or about `theta > .7, the function 1 / `sqrt(`theta^2+1) exceeds the value of the divergent function 1 / (2 `theta). Our function 1 / `sqrt(`theta^2 + 1) is therefore divergent. ** STUDENT ERROR: This integral converges because 1/sq rt(theta^2) approaches 0 rapidly. INSTRUCTOR COMMENT: ** It will indeed converge but your argument essentially says that it converges because it converges. Way too vague. You have to use a comparison test of some kind. ** You have the right idea, but being less than a converging comparison function would prove convergence; being greater than a diverging function would prove divergence. However being greater than a converging function, or being less than a diverging function, does not prove anything at all about the convergence of the original function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):
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Given Solution: On the interval 1 to infinity the integral would converge: 1 / `sqrt(theta^3 + `theta) < 1 / `sqrt(`theta^3) = 1 / `theta^(3/2). This is an instance of 1 / x^p for x =`theta and p = 3/2. The integral of 1 / theta^(3/2) converges by the p test. The given integrand is less than 1 / theta^(3/2), so the original integral also converges, by the comparison test. On the interval from 0 to 1 the result might be surprising, in the context of the previous result: We might expect that the integral diverges, since the integral of 1 / theta^(3/2) diverges on this interval. All we would need to do is show that our original integrand is greater than 1 / theta^(3/2) However this isn't the case, our integrand is less than 1 / theta^(3/2) on the interval 0 < theta < 1. We could patch this up by showing that our original integrand is greater than some fixed multiple of 1 / theta^(3/2) (e.g., maybe 1 / sqrt (theta^3 + theta) is greater than .000000001 * 1 / theta^(3/2)) on this interval), but it just isn't so. Whatever constant multiple we choose, there is some neighborhood of x = 0 where the original function is less than our comparison function (you are invited to prove this). It turns out that as we approach zero, it is theta^3 that becomes insignificant, not theta. So the integrand must be compared 1 / theta^(1/2), or at least with a constant multiple of this expression. The integral 1 / theta^(1/2) on 0 <= theta <= 1 is convergent (its antiderivative is 2 theta^(1/2), which can be evaluated at 0 and 1; the value of the integral is just 2). Conveniently, 1 / (theta^3 + theta) is less than 1 / theta^(1/2), so the convergence of the latter ensures the convergence of the former.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):
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Given Solution: ****** ****** FOR HORIZONTAL STRIPS The solution for the x of the equation x^2 + y^2 = 10 is x = +- sqrt(10 - y^2). In the first quadrant we have x > = 0 so the first-quadrant solution is x = +sqrt(10 - y^2). A vertical strip at position y extends from the y axis to the point on the curve at which x = sqrt(10 - y^2), so the ‘altitude’ of the strip is sqrt(10 - y^2). If the width of the strip is `dy, then the strip has area `dA = sqrt(10 - y^2) `dy. The curve extends along the y axis from y = 0 to the x = 0 point y^2= 10, or for first-quadrant y values, to y = sqrt(10). If the y axis from y = 0 to y = sqrt(10) is partitioned into subintervals the the total area of the subintervals is approximated by A = sum(`dA) = sum ( sqrt(10 - y^2) `dy), and as interval width approaches zero we obtain the area A = integral ( sqrt(10 - y^2) dy, y, 0, sqrt(10)). The integral is performed by letting y = sqrt(10) sin(theta) so that dy = sqrt(10) cos(theta) * dTheta; 10 - y^2 will then equal 10 - 10 sin^2(theta) = 10 ( 1 - sin^2(theta)) = 10 sin^2(theta) and sqrt(10 - y^2) becomes sqrt(10) cos(theta); y = 0 becomes sin(theta) = 0 so that theta = 0; y = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 0, giving us theta = pi / 2. The integral is therefore transformed to Int(sqrt(10) cos(theta) * sqrt(10) cos(theta), theta, 0, pi/2) = Int(10 cos^2(theta) dTheta, theta, 0, pi/2). The integral of cos^2(theta) from 0 to pi/2 is pi/4, so that the integral of our function is 10 * pi/4 = 5/2 pi. Note that this is ¼ the area of the circle x^2 + y^2 = 10. FOR VERTICAL STRIPS The solution for the y of the equation x^2 + y^2 = 10 is y = +- sqrt(10 - x^2). In the first quadrant we have y > = 0 so the first-quadrant solution is y = +sqrt(10 - x^2). A vertical strip at position x extends from the x axis to the point on the curve at which y = sqrt(10 - x^2), so the ‘altitude’ of the strip is sqrt(10 - x^2). If the width of the strip is `dx, then the strip has area `dA = sqrt(10 - x^2) `dx. The curve extends along the x axis from x = 0 to the y = 0 point x^2= 10, or for first-quadrant x values, to x = sqrt(10). If the x axis from x = 0 to x = sqrt(10) is partitioned into subintervals the the total area of the subintervals is approximated by A = sum(`dA) = sum ( sqrt(10 - x^2) `dx), and as interval width approaches zero we obtain the area A = integral ( sqrt(10 - x^2) dx, x, 0, sqrt(10)). The integral is performed by letting x = sqrt(10) sin(theta) so that dx = sqrt(10) cos(theta) * dTheta; 10 - x^2 will then equal 10 - 10 sin^2(theta) = 10 ( 1 - sin^2(theta)) = 10 sin^2(theta) and sqrt(10 - x^2) becomes sqrt(10) cos(theta); x = 0 becomes sin(theta) = 0 so that theta = 0; x = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 0, giving us theta = pi / 2. The integral is therefore transformed to Int(sqrt(10) cos(theta) * sqrt(10) cos(theta), theta, 0, pi/2) = Int(10 cos^2(theta) dTheta, theta, 0, pi/2). The integral of cos^2(theta) from 0 to pi/2 is pi/4, so that the integral of our function is 10 * pi/4 = 5/2 pi. Note that this is ¼ the area of the circle x^2 + y^2 = 10. DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: Query problem 8.1.12. Half disk radius 7 m thickness 10 m, `dy slice parallel to rectangular base. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: H^2 + (w/2)^2 = 49 W = 2sqrt(49-h^2) V= SUM(10*2sqrt(49-h^2) dh) V= int(10*2sqrt(49-h^2) dh 0, 7) = 20int(sqrt(49-h^2) dh 0, 7) = (490 arcsin(h/7) 0, 7) = 490arcsin1/2=796.690 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT SOLUTION here we used the pythagorean Theorem which is y^2 + (7/2)^2 = 7^2 y = square root of 49 - (7/2) Then we use the area of a semicircle which is w * delta h = 49 - (7/2) * delta h cm^2 then we get 49 arcsin 1 = 49 / 2 pi = 65.23 cm^2 INSTRUCTOR RESPONSE You have to do the Riemann sum, get the integral then perform the integration. A slice of the region parallel to the y axis is a rectangle. If the y coordinate is y = c_i then the slice extends to an x coordinate such that x^2 + (c_i)^2 = 7^2, or x = sqrt(49 - (c_i)^2). So if the y axis, from y = -7 to y = 7, is partitioned into subintervals y_0 = -7, y1, y2, ..., y_i, ..., y_n = 7, with sample point c_i in subinterval number i, the corresponding 'slice' of the solid has thickness `dy, width sqrt(49 - (c_i)^2) and length 10 so its volume is 10 * sqrt(49 - (c_i)^2) * `dy and the Riemann sum is sum(10 * sqrt(49 - (c_i)^2) * `dy, i from 1 to n). The limit of this sum, as x approaches infinity, is then integral ( 10 * sqrt(49 - y^2) dy, y from 0 to 7) = 10 integral (sqrt(49 - y^2) dy, y from 0 to 7). Using the trigonometric substitution y = 7 sin(theta), analogous to the substitution used in the preceding solution, we find that the integral is 49 pi / 2 and the result is 10 times this integral so the volume is 10 * 49 pi / 2 = 490 pi / 2 = 245 pi / 2. A decimal approximation to this result is between 700 and 800. Compare this with the volume of a 'box' containing the figure: The 'box' would be 14 units wide and 7 units high, and 10 units long, so would have volume 14 * 7 * 10 = 980. The solid clearly fills well over half the box, so a volume between 700 and 800 appears very reasonable. DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: query problem 8.2.14 (previously 8.2.11) arc length x^(3/2) from 0 to 2
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Arc length = int(sqrt(1+(dy/dx)^2) 0, 2) = int(sqrt(1+9x/4 0, 2) = int( sqrt(1+9x/4) 0, 2) = 3.526 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: On an interval of length `dx, containing x coordinate c_i, the ‘slope triangle’ at the top of the approximating trapezoid has slope approximately equal to f ‘ (c_i). The hypotenuse of this triangle corresponds to the arc length. A triangle with ‘run’ `dx and slope m has ‘rise’ equal to m * `dx. So its hypotenuse is sqrt((`dx)^2 + (m `dx)^2) = sqrt( (1 + m^2) * (`dx)^2) = sqrt( 1 + m^2) * `dx. For small `dx, the hypotenuse is very close to the curve so its length is very near the arc length of the curve on the given interval. Since the slope here is f ‘(c_i), we substitute f ‘ (c_i) for m and find that the contribution to arc length is `dL_i = sqrt(1 + f ‘ ^2 (c_i) ) * `dx So that the Riemann sum is Sum(`dL_i) = sum ( sqrt(1 + f ‘ ^2 (c_i) ) * `dx ), where the sum runs from i = 1 to i = n, with n = (b - a) / `dx = (2 - 0) / `dx. In other words, n is the number of subintervals into which the interval of integration is broken. This sum approaches the integral of sqrt(1 + (f ' (x)) ^2, over the interval of integration. In general, then, the arc length is arc length = integral(sqrt(1 + (f ' (x)) ^2) dx, x from a to b). In this case f(x) = x^(3/2) so f ' (x) = 3/2 x^(1/2), and (f ' (x))^2 = (3/2 x^(1/2))^2 = 9/4 x. Thus sqrt( 1 + (f ‘ (x))^2) = sqrt(1 + 9/4 x) and we find the integral of this function from x = 0 to x = 2. The integral is found by letting u = 1 + 9/4 x, so that u ‘ = 9/4 and dx = 4/9 du, so that our integral becomes Integral ( sqrt( 1 + 9/4 x) dx, x from 0 to 2) = Integral ( sqrt(u) * 4/9 du, x from 0 to 2) = 4/9 Integral ( sqrt(u) du, x from 0 to 2) Our antiderivative is 4/9 * 2/3 u^(3/2), which is the same as 8/27 (1 + 9/4 x) ^(3/2). Between x = 0 and x = 2, the change in this antiderivative is 8/27 ( 1 + 9/4 * 2) ^(3/2) - 8/27 ( 1 + 9/4 * 0) ^(3/2) = 8/27 ( ( 11/2 )^(3/2) - 1) = 3.526, approximately. Thus the arc length is integral(sqrt(1 + (f ' (x)) ^2) dx, x from a to b = Integral ( sqrt( 1 + 9/4 x) dx, x from 0 to 2) = 3.526. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: query problem 8.2.41 (31 4th edition, 21 3d edition) volume of region bounded by y = e^x, x axis, lines x=0 and x=1, cross-sections perpendicular to the x axis are squares YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: L = e^x V = int((e^x)^2 dx, 0,1) = int(e^(2x) dx, 0,1) = e^(2x)/2 0, 1) = 1/2 (e^2 - e^0) = 3.1945 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: x runs from 0 to 1. At a given value of x the 'altitude' of the y vs. x graph is e^x, and a slice of the solid figure is a square with sides e^x so its area is (e^x)^2 = e^(2x). If we partition the interval from x = 0 to x = 1 into subintervals defined by x_0 = 0, x_1, x_2, ..., x_i, ..., x_n, with interval number i containing sample point c_i, then the slice corresponding to interval number i has the following characteristics: the thickness of the 'slice' is `dx the cross-sectional area of the 'slice' at the sample point x = c_i is e^(2x) = e^(2 * c_i) so the volume of the 'slice' is e^(2 * c_i) * `dx. The Riemann sum is therefore sum(e^(2 * c_i * `dx) and its limit is integral(e^(2 x) dx, x from 0 to 1). Our antiderivative is easily found to be 1/2 e^(2 x) and the change in our antiderivative is 1/2 e^(2 * 1) - 1/2 e^(2 * 0) = 1/2 e^2 - 1/2 = 1/2 (e^2 - 1). The approximate value of this result about 3.19. A 'box' containing this region would have dimensions 1 by e by e, with volume e^2 = 7.3 or so. The figure fills perhaps half this box, so our results are reasonably consistent. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: **** 8.1.28 (3d edition 8.1.29) volume of dam base 1400 m long, 160 m wide, 150 m high, narrows to 10 m wide at top. **** What is the volume of the dam? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10/160 = L/150+L L = 10, L+150 = 160 Wi/160 = (160-h)/160 Wi = 160-h V = int(1400(160-h) dh, 0, 150) = 1400(160h - h^2/2, 0, 150) = 1400(160*150 - 150^2/2) = 1.125e7 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The width at height y is a linear function equal to 160 when y = 0 and to 10 when y = 150. This linear function is 160 - y. So a strip of width `dyi at altitude yi is (160 - yi) `dyi, and the volume corresponding to this strip is 1400 (160 - yi) `dyi. This leads to the integral of 1400 (160 - y) with respect to y, for y = 0 to y = 150. Antiderivative is 1400 (160 y - y^2/2); evaluating at limits we get 1400 (160 * 150 - 150^2 / 2) - 0 = 1400 ( 24000 - 22500/2) = 1400 (24000 - 11250) = 1400 ( 22750) = 30 million, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 1400 (24000 - 11250) = 1400 ( 22750), here is a calculation mistake. So the answer should be 11 million rather than 30 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: problem 7.8.24 convergence of integral from 1 to infinity of (2x^2+1)/(4x^4+4x^2-2) and [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: On [1, inf] 0<=(2x^2+1)/(4x^4+4x^2-2<= 1/x^2 Int(1/x^2 1, inf) is converges because of the p here is 2 which is greater than 1. Therefore, int((2x^2+1)/(4x^4+4x^2-2 1, inf) is converges On [1, inf] 0<=[ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4)<=1/x^(1/2) Int(1/x^(1/2) dx, 1, inf) is diverges because of p here is (1/2) which is less than 1 Therefore [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) is diverges
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Given Solution: ** The first thing we see is that (2x^2+1)/(4x^4+4x^2-2) acts for large x like the ratio of the leading terms in the numerator and denominator: 2 x^2 / ( 4 x^4) = 1 / (2 x^2), which is quite convergent so we expect that the integral will converge. All we have to do is spell out the details to be sure. There's a little sticking point here, because of the -2 in the denominator and +1 in the numerator we can't say that the given expression is always less than 1 / (2 x^2). So let's solve to see where (2x^2 + 1) / ( 4 x^4 + 4 x^2 - 2 ) < 1 / ( 2 x^2). This expression is equivalent to 2 x^2 ( 2 x^2 + 1) < 4 x^4 + 4 x^2 - 2, or to 4 x^4 + 2 x^2 < 4 x^4 + 4 x^2 - 2, which is in turn equivalent to 0 < 2 x^2 - 2, equiv to 0 < x^2 - 1 which occurs for x > 1 or for x < -1. So for x > 1 we have (2x^2+1)/(4x^4+4x^2-2) < 1 / (2 x^2). Noting that 1 / ( 2 x^2) < 1 / x^2, and that in [1, infinity) x^2 is a convergent p series, we see that (2x^2+1)/(4x^4+4x^2-2) is convergent on [1, infinity). Now [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) acts a whole lot like ( 1 / (2x^2) )^(1/4) = 1 / 2^(1/4) * 1 / x^(1/2) . We are thus looking at comparison with a p series with p = .5, which diverges on [1, infinity). We therefore suspect that our function is divergent. The problem is that [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) < ( 1 / (2x^2) )^(1/4) = 1 / 2^(1/4) * 1 / x^(1/2), and being < something that diverges doesn't imply divergence. However the function is so close to the divergent function that we know in our hearts that it doesn't matter. Our hearts don't prove a thing, but they can sometimes lead us in the right direction. We need to show that [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) is greater than something that diverges. Fortunately this is easy to do. All we need is something just a little smaller than 1 / 2^(1/4) * 1 / x^(1/2). If we change the 1 / 2^(1/4) to 1/2 we get 1 / (2 x^(1/2)), which is still divergent but a bit smaller than the original divergent expression. Now we hypothesize that [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) > 1 /(2 x^(1/2) ). Solving this inequality for x we first take the 4th powe of both sides to obtain [ (2x^2+1)/(4x^4+4x^2-2) ] > 1 /(16 x^2 ) which we rearrange to get (2x^2+1) * (16 x^2) > (4x^4+4x^2-2) which we expand to get 32 x^4 + 16 x^2 > 4 x^4 + 4x^2 - 2 or 28 x^2 + 12 x + 2 > 0. If we evaluate the discriminant of the quadratic function on the left-hand side we find that it is negative so that it can never be zero. Since for x = 0 the quadratic is equal to 2, it must always be positive. Thus the inequality is satisfied for all x. It follows that [ (2x^2+1)/(4x^4+4x^2-2) ] ^ (1/4) > 1 /(2 x^(1/2) ) for all x and, since the right-hand side is a multiple of a divergent p-series, the original series diverges. ** 17:28:47 Note: The questions below are 'followup questions'. You need not insert answers to these questions, but you may do so. If you do, please precede your answer with **** and add #$&* in the line immediately after your insertion. You will be able to recognize similar 'followup questions' on many questions in subsequent assignments.
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does the first integral converge? If so what is an upper bound for the integral?
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17:28:47 17:29:01
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Does the second integral converge? If so what is an upper bound for the integral?
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17:29:01 17:29:07
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Explain how you obtained your results.
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17:29:07 I graphed x^2/(x^4 + x^2) which approaches 0 ** 1/x approaches zero too but its integral doesn't converge. Approaching zero doesn't show anything. And you can't tell about convergence from a graph on your calculator. ** 17:30:09
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Is it true that (2x^2+1)/(4x^4+4x^2-2) < 1 / (2 x^2) for 1 < x?
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17:30:09 17:32:06
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The given inequality is true. How the you use this inequality to place an upper bound on the first integral?
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17:32:06 ** the integral from 1 to infinity of 1 / (2x^2) is convergent. Neither the integral nor the value of the function approaches infinity. ** ** The integral from 1 to infinity of 1/`sqrt(x) approaches infinity and 1 / x is less than 1/ `sqrt(x). However, the integral from 1 to infinity of 1/x still aproaches infinity. Being less than something that approaches infinity doesn't imply finiteness. Roughly speaking, being less in magnitude than something finite does imply finiteness, and being greater in magnitude than something infinite implies inifiteness; be sure you make these ideas precise, though, in the sense of the text and the class notes. ** 17:36:10
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How do use the same inequality to show that the second integral is convergent?
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17:36:10 17:36:43
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Why would you expect this inequality to occur to someone trying to solve the problem?
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17:36:43 ** dividing a polynomial with degree 2 by a polynomial with degree 4 should give you something a lot like a polynomial with degree 2 in the denominator ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: problem 8.2.23 was 8.1.12 volume of region bounded by y = e^x, x axis, lines x=0 and x=1, cross-sections perpendicular to the x axis are squares YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: L = e^x V = int((e^x)^2 dx, 0,1) = int(e^(2x) dx, 0,1) = e^(2x)/2 0, 1) = 1/2 (e^2 - e^0) = 3.1945 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** At coordinate x = .3, for example, the y value is e^.3, so the cross-section is a square with dimensions e^.3 * e^.3 = e^.6. At general coordinate x the y value is e^x so the cross-section is a square dimensions e^x * e^x = e^(2x). So you integrate e^(2x) between the limits 0 and 1. The antiderivative is .5 e^(2x) so the integral is .5 e^(2 * 1) - .5 e^(2 * 0) = .5 (e^2 - 1) = 3.2, approx. ** INCORRECT STUDENT ANSWER integral from 0 to 1 (e^2x dx) = x^2 * e^2x from 0 to 1 = 7.3891 - 0 = 7.3891 INSTRUCTOR RESPONSE ** you have the right integral but your result is incorrect ** ** Your integration is faulty. x^2 e^(2x) is not an antiderivative of e^(2x). The derivative of x^2 e^(2x) = 2x e^(2x) + 2 x^2 e^(2x), not e^(2x). ** Note: The questions below are 'followup questions'. You need not insert answers to these questions, but you may do so. If you do, please precede your answer with **** and add #$&* in the line immediately after your insertion. You will be able to recognize similar 'followup questions' on many questions in subsequent assignments.
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what is the volume of the region?
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17:39:44 17:39:48
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What integral did you evaluate to get the volume?
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17:39:48 17:39:50
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What is the cross-sectional area of a slice perpendicular to the x axis at coordinate x?
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17:39:50 17:41:16
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What is the approximate volume of a thin slice of width `dx at coordinate x?
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17:41:16 e^2xdx ** The thin slice has thickness `dx matching the increment on the x axis. It is located near coordinate x so its cross-sectional area is close to e^(2x), as seen above. So its volume is volume = area * thickness = e^(2x) * `dx. ** 17:41:51
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How the you obtain the integral from the expression for the volume of the thin slice?
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17:41:51 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: problem 8.2.11 was 8.1.20 are length x^(3/2) from 0 to 2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Arc length = int(sqrt(1+(dy/dx)^2) 0, 2) = int(sqrt(1+9x/4 0, 2) = int( sqrt(1+9x/4) 0, 2) = 3.526 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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What integral do you evaluate to obtain the arc length?
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17:43:21 integral from 0 to 2 (1 +(9/4)x)^1/2 dx ** The formula is integral( `sqrt( 1 + (f'(x))^2 ) dx). If f(x) = x^(3/2) then f'(x) = 3/2 * x^(1/2) so you get integral( `sqrt( 1 + (3/2 * x^(1/2)))^2 ) dx, x from 0 to 2). You should understand the reason for the formula. Think of the top of a trapezoid running from x to x + `dx. The width of the trapezoid is `dx. Its slope is f'(x). The little triangle at the top therefore has run `dx and slope f'(x), so its rise is rise=slope*run = f'(x) * `dx. Its hypotenuse, which approximates the arc distance, is thus hypotenuse = `sqrt(rise^2 + run^2) = `sqrt( (f'(x) `dx)^2 + `dx^2) = `sqrt( 1 + f'(x)^2) `dx. That leads directly to the formula. ** 17:43:30
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What is the approximate arc length of a section corresponding to an increment `dx near coordinate x on the x axis?
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17:43:30 student answer: integral from x to (x+dx) (1 + (9/4)x)^1/2 dx instructor response: ** this is exact. We need the approximate length, which is a product of the length `dx of the interval with a factor that gives the approximate length of the arc. This is a geometric situation. ** 17:43:47
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What is the slope of the graph near the graph point with x coordinate x?
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17:43:47 17:44:01
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How is this slope related to the approximate arc length of the section?
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17:44:01 student answer: it is the derivative of f(x) instructor response: ** this is what the slope is but it doesn't explain the relationship between slope and arc length. This explanation involves a picture of a triangle with 'run' `dx and slope m. You need to find the length of the hypotenuse, which requires that you find the 'rise' then use the Pythagorean Theorem. ** 17:44:19
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