course Mth 174 ??w????O?^?????assignment #008
......!!!!!!!!...................................
15:14:26 query explain the convergence or divergence of series (no summary needed)
......!!!!!!!!...................................
RESPONSE --> Both of these terms look at functions as they tend towards infinity, checking to see if there is a limit to the function. In the case of convergence, if the limit exists, then the value function is said to converge on the limit. In the case of divergence, the limit does not exist and the value of function increases (or decreases) continually. ** 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]. **
.................................................
......!!!!!!!!...................................
15:32:02 explain how we know that the integral from 1 to infinity of 1 / x^p converges if p > 1, diverges for p < 1, diverges for p = 1
......!!!!!!!!...................................
RESPONSE --> In the case of p > 1, the value of the function quickly moves to a very small number (1 being divided by a large number). Since it quickly moves to a finite number, it then converges. In the case of p < 1, the value of the function grows without bound, tending towards infinity itself as the upper limit of the integral goes towards infinity. Since there is no finite value, the function is said to diverge. In the case of p = 1, the value of the function slowly moves towards a finite value. Since it does not quickly move towards a finite number, the function is said to diverge.
.................................................
......!!!!!!!!...................................
15:45:40 explain how we know that the integral from 0 to 1 of 1 / x^p diverges if p > 1, converges for p < 1, diverges for p = 1
......!!!!!!!!...................................
RESPONSE --> In the case of p > 1, the value of function is 1 divided by a small number. This returns a larger number which shrinks over the interval and does not tend towards a fixed value, therefore diverging. In the case of p < 1, the function grows slowly towards a fixed value, thereby converging. In the case of p = 1, the value of the function shrinks over the interval, but does not tend towards a fixed value, therefore diverging.
.................................................
......!!!!!!!!...................................
15:49:15 explain how we know that the integral from 0 to infinity of e^(-a x) converges for a > 0.
......!!!!!!!!...................................
RESPONSE --> Because for successively higher values of x, the value of the function tends towards 0 and moves quickly towards it. 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. **
.................................................
......!!!!!!!!...................................
15:58:36 query problem 7.8.18 integral of 1 / (`theta^2+1) from 1 to infinity
......!!!!!!!!...................................
RESPONSE --> The function as listed in the book is: 1 / 'sqrt(`theta^2+1) This can be compared to the function 1 / 'sqrt('theta^2) = 1 / 'theta. This function can be considered as a member of the family 1 / x^p, which we know how that acts for different values of p. Since the value of p in this case is 1, then all we have to do is consider the interval of integration, which in this case is 1 to infinity. In this case it diverges. Since the square root of any number plus one is greater than the square root of that number, then our comparison function is less than the original in value. Therefore since the comparison function diverges, the original function diverges as well.
.................................................
......!!!!!!!!...................................
15:58:41 does the integral converge or diverge, and why?
......!!!!!!!!...................................
RESPONSE --> See prior answer.
.................................................
......!!!!!!!!...................................
15:58:51 If you have not already stated it, with what convergent or divergent integral did you compare the given integral?
......!!!!!!!!...................................
RESPONSE --> 1 / 'theta
.................................................
......!!!!!!!!...................................
16:10:00 query problem 7.8.19 (3d edition #20) convergence of integral from 0 to 1 of 1 / `sqrt(`theta^3 + `theta)
......!!!!!!!!...................................
RESPONSE --> Here we can compare the original function with 1 / 'sqrt( 'theta^3 ), which is the same as 1 / 'theta^( -3/2 ). Since the value of p in this case is less than one and the interval is from 0 to 1, the comparison function converges.
.................................................
......!!!!!!!!...................................
16:10:05 does the integral converge or diverge, and why?
......!!!!!!!!...................................
RESPONSE --> See prior.
.................................................
......!!!!!!!!...................................
16:10:16 If you have not already stated it, with what convergent or divergent integral did you compare the given integral?
......!!!!!!!!...................................
RESPONSE --> See prior.
.................................................
......!!!!!!!!...................................
16:46:43 Query problem 8.1.5. Riemann sum and integral inside x^2 + y^2 = 10 within 1st quadrant, using horizontal strip of width `dy.
......!!!!!!!!...................................
RESPONSE --> Riemann sum: 'sqrt( 10 - y^2 ) 'dy Integral: int( 'sqrt( 10 - y^2 ) 'dy, y, 0, 'pi ).
.................................................
......!!!!!!!!...................................
16:46:52 Give the Riemann sum and the definite integral it approaches.
......!!!!!!!!...................................
RESPONSE --> See prior answer.
.................................................
......!!!!!!!!...................................
16:47:42 Give the exact value of your integral.
......!!!!!!!!...................................
RESPONSE --> Exact value: 5 pi / 2
.................................................
......!!!!!!!!...................................
16:52:56 Query problem 8.1.12. Half disk radius 7 m thickness 10 m, `dy slice parallel to rectangular base.
......!!!!!!!!...................................
RESPONSE --> Riemann: 10 ( 2 'sqrt(49 - y^2) ) 'dy Integral: int(10 ( 2 'sqrt(49 - y^2) ) 'dy, y, 0, 7)
.................................................
......!!!!!!!!...................................
16:53:01 Give the Riemann sum and the definite integral it approaches.
......!!!!!!!!...................................
RESPONSE --> See above.
.................................................
......!!!!!!!!...................................
16:53:53 Give the exact value of your integral.
......!!!!!!!!...................................
RESPONSE --> 490 'pi / 2 m^3
.................................................
......!!!!!!!!...................................
16:57:35 query problem 8.2.11 arc length x^(3/2) from 0 to 2
......!!!!!!!!...................................
RESPONSE --> I found the integral to be: int( 'sqrt( 1 + (-3/2 'sqrt(x) )^2 ) 'dx, x, 0, 2 ) The approximated value: 3.526
.................................................
......!!!!!!!!...................................
16:57:42 what is the arc length?
......!!!!!!!!...................................
RESPONSE --> See approximated value.
.................................................
......!!!!!!!!...................................
16:57:49 What integral do you evaluate obtain the arc length?
......!!!!!!!!...................................
RESPONSE --> See above.
.................................................
......!!!!!!!!...................................
17:00:49 What is the approximate arc length of a section corresponding to an increment `dx near coordinate x on the x axis?
......!!!!!!!!...................................
RESPONSE --> Wouldn't this just be: 'sqrt( 1 + (-3/2 'sqrt(x) )^2 ) 'dx
.................................................
......!!!!!!!!...................................
17:04:23 What is the slope of the graph near the graph point with x coordinate x?
......!!!!!!!!...................................
RESPONSE --> For a general approximation in a small area, I think all that we need to do is divide the rise by the run: f' (x) 'dx / 'dx = f' (x)
.................................................
......!!!!!!!!...................................
17:07:04 How is this slope related to the approximate arc length of the section?
......!!!!!!!!...................................
RESPONSE --> If I got it right, then the approximate arc length of each tiny section is simply the slope of that section multiplied by the length of the section.
.................................................
......!!!!!!!!...................................
17:13:22 query problem 8.2.31 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
......!!!!!!!!...................................
RESPONSE --> I found the volume integral to be: int( e^(2x), x, 0, 1 ) This evaluated to exactly: (1/2) e^2 - 1/2 Approximation: 3.1945 In the answer section in the back of the book, the answer to this problem is listed as: V = (e^2 - 1/2) = 3.195 Did they just leave out the 1/2 on the e^2? When I punched their numbers into the calculator, it didn't match their approximate integral.
.................................................
......!!!!!!!!...................................
17:13:42 what is the volume of the region?
......!!!!!!!!...................................
RESPONSE --> V = 3.195
.................................................
......!!!!!!!!...................................
17:13:48 What integral did you evaluate to get the volume?
......!!!!!!!!...................................
RESPONSE --> See above.
.................................................
......!!!!!!!!...................................
17:14:23 What is the cross-sectional area of a slice perpendicular to the x axis at coordinate x?
......!!!!!!!!...................................
RESPONSE --> The cross sectional area would be: (e^x)^2 or just e^(2x)
.................................................
......!!!!!!!!...................................
17:14:48 What is the approximate volume of a thin slice of width `dx at coordinate x?
......!!!!!!!!...................................
RESPONSE --> Simply enough, area multiplied by depth ('dx): e^(2x) 'dx
.................................................
......!!!!!!!!...................................
17:18:39 How the you obtain the integral from the expression for the volume of the thin slice?
......!!!!!!!!...................................
RESPONSE --> By taking the Riemann sum of the overall area and converting it to an integral with the limits of 0 and 1 since 0 and 1 are the bounds on the left and right. Riemann: e^(2x) 'dx Integral: int(e^(2x) 'dx, x, 0, 1)
.................................................
......!!!!!!!!...................................
17:19:15 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> Interesting stuff. Downright cool at times. So that's how they figured out those formulas for volume!
.................................................
"