Callahan 6.2:  11, 12, 14, 15, 16, 19, 20, 27, 28

[Graphics:Images/CalcWk7_gr_1.gif]

11)  Calculate the left endpoint Riemann sums for the function:

[Graphics:Images/CalcWk7_gr_2.gif]

using 10, 100, and 1000 equally spaced subintervals.  The Riemann sums are all negative.  Why?

[Graphics:Images/CalcWk7_gr_3.gif]
[Graphics:Images/CalcWk7_gr_4.gif]
[Graphics:Images/CalcWk7_gr_5.gif]
[Graphics:Images/CalcWk7_gr_6.gif]

    The Riemann sums are all negative because the function is negative in this region.

[Graphics:Images/CalcWk7_gr_7.gif]

12)

    a)  Calculate midpoint Riemann sums for the function:

[Graphics:Images/CalcWk7_gr_8.gif]

    using 10, 100, and 1000 equally spaced subintervals.  The Riemann sums are all zero.  Why?

[Graphics:Images/CalcWk7_gr_9.gif]

    The Riemann sums are all zero because the function has an equal amount of negative and positive area in this interval.  The errors in the sum cancel symmetrically.

from visual.graph import*


def f(x):
    y = x**3
    return y

xinitial = -2
xfinal = 2

graph = gdisplay()

exponent = 3

for j in range(1,exponent+1) :
    accumulation = 0

    numberofsteps = 10**j
    deltax = (xfinal - xinitial)/float(numberofsteps)
    x = xinitial + (deltax/2)

    pcurve = gcurve(color = (1, 0, 0))
    pcurve.plot(pos = (x, f(x)))

    for k in range(1,numberofsteps+1) :
        deltaS = f(x) * deltax
        accumulation = accumulation + deltaS
        x = x + deltax
        #print k, x, f(x), deltaS, accumulation
        pcurve.plot(pos = (x, f(x)))

    print "Number of steps:", numberofsteps, "  Area:", accumulation
    

    b)  Repeat part (a) using left endpoint Riemann sums.  Are the results still zero?  Can you explain the difference, if any, between these two results?

[Graphics:Images/CalcWk7_gr_10.gif]

    They are no longer zero.  The difference can be explained because the errors in the sum no longer cancel symmetrically.

[Graphics:Images/CalcWk7_gr_11.gif]

14)  The volume of a sphere with radius [Graphics:Images/CalcWk7_gr_12.gif] is exactly [Graphics:Images/CalcWk7_gr_13.gif].

    a)  Using the formula, determine the volume of the sphere whose radius is 3.  Give the numerical value to four decimal places accuracy.

[Graphics:Images/CalcWk7_gr_14.gif]

    One way to get a sphere of radius 3 is to rotate the graph of the semicircle:

[Graphics:Images/CalcWk7_gr_15.gif]

    around the x-axis.  Every cross section perpendicular to the x-axis is a circle.  At the point x, the radius of the circle is r(x), and its area is:

[Graphics:Images/CalcWk7_gr_16.gif]

    You can thus get estimates for the volume of the sphere by constructing Riemann sums for [Graphics:Images/CalcWk7_gr_17.gif] on the interval [Graphics:Images/CalcWk7_gr_18.gif].

    b)  Calculate a sequence of estimates for the volume of the sphere that use more and more slices, until the value of the estimate stabilizes out to four decimal places.  Does this value agree with the value given by the formula in part (a)?

[Graphics:Images/CalcWk7_gr_19.gif]
[Graphics:Images/CalcWk7_gr_20.gif]

15)

    a)  Rotate the graph of [Graphics:Images/CalcWk7_gr_21.gif], with [Graphics:Images/CalcWk7_gr_22.gif] around the x-axis.  What do you get?  Describe it precisely, and find its volume using an appropriate geometric formula.

    This object will be a cone.  The line of slope [Graphics:Images/CalcWk7_gr_23.gif]passes through the origin and is swept around the x-axis.

[Graphics:Images/CalcWk7_gr_24.gif]
[Graphics:Images/CalcWk7_gr_25.gif]

    b)  Calculate a sequence of estimates for the volume of the same object by constructing Riemann sums for the area function [Graphics:Images/CalcWk7_gr_26.gif].  Continue until your estimates stabilize out to four decimal places.  What value do you get?

[Graphics:Images/CalcWk7_gr_27.gif]
[Graphics:Images/CalcWk7_gr_28.gif]
[Graphics:Images/CalcWk7_gr_29.gif]

16)  Determine the numerical value of each of the following:

    a)

[Graphics:Images/CalcWk7_gr_30.gif]

    b)

[Graphics:Images/CalcWk7_gr_31.gif]

    c)

[Graphics:Images/CalcWk7_gr_32.gif]
[Graphics:Images/CalcWk7_gr_33.gif]

19)  Express the following sums using summation notation:

    a)

[Graphics:Images/CalcWk7_gr_34.gif]

    b)

[Graphics:Images/CalcWk7_gr_35.gif]

    c)

[Graphics:Images/CalcWk7_gr_36.gif]

    d)

[Graphics:Images/CalcWk7_gr_37.gif]
[Graphics:Images/CalcWk7_gr_38.gif]

20)  Express each of the following as a sums written out term by term.

    a)

[Graphics:Images/CalcWk7_gr_39.gif]

    b)

[Graphics:Images/CalcWk7_gr_40.gif]

    c)

[Graphics:Images/CalcWk7_gr_41.gif]
[Graphics:Images/CalcWk7_gr_42.gif]

27)

    a)  What is the length of the hyperbola [Graphics:Images/CalcWk7_gr_43.gif] over the interval [Graphics:Images/CalcWk7_gr_44.gif]?  Obtain an estimate that is accurate to four decimal places.

    Length function:

[Graphics:Images/CalcWk7_gr_45.gif]

    So to find the length we need [Graphics:Images/CalcWk7_gr_46.gif]:

[Graphics:Images/CalcWk7_gr_47.gif]
[Graphics:Images/CalcWk7_gr_48.gif]

from visual.graph import*


def f(x):
    y = sqrt(1+(-1/(x**2))**2)
    return y

xinitial = 1
xfinal = 4

graph = gdisplay()

exponent = 4

for j in range(1,exponent+1) :
    accumulation = 0

    numberofsteps = 10**j
    deltax = (xfinal - xinitial)/float(numberofsteps)
    x = xinitial + (deltax/2)

    pcurve = gcurve(color = (1, 0, 0))
    pcurve.plot(pos = (x, f(x)))

    for k in range(1,numberofsteps+1) :
        deltaS = f(x) * deltax
        accumulation = accumulation + deltaS
        x = x + deltax
        #print k, x, f(x), deltaS, accumulation
        pcurve.plot(pos = (x, f(x)))

    print "Number of steps:", numberofsteps, "  Length:", accumulation

    b)  What is the area under the hyperbola over the same interval?  Obtain an estimate that is accurate to four decimal places.

[Graphics:Images/CalcWk7_gr_49.gif]
[Graphics:Images/CalcWk7_gr_50.gif]

28)  The graph of [Graphics:Images/CalcWk7_gr_51.gif] is a semicircle whose radius is 2.  The circumference of the whole circle is [Graphics:Images/CalcWk7_gr_52.gif], so the length of the part of the circle in the first quadrant is exactly [Graphics:Images/CalcWk7_gr_53.gif].

[Graphics:Images/CalcWk7_gr_54.gif]

    So to find the length we need [Graphics:Images/CalcWk7_gr_55.gif]:

[Graphics:Images/CalcWk7_gr_56.gif]
[Graphics:Images/CalcWk7_gr_57.gif]
[Graphics:Images/CalcWk7_gr_58.gif]
[Graphics:Images/CalcWk7_gr_59.gif]
[Graphics:Images/CalcWk7_gr_60.gif]

    But we don't want imaginary numbers, so:

[Graphics:Images/CalcWk7_gr_61.gif]
[Graphics:Images/CalcWk7_gr_62.gif]
[Graphics:Images/CalcWk7_gr_63.gif]
[Graphics:Images/CalcWk7_gr_64.gif]

    a)  Using left endpoint Riemann sums, estimate the length of the graph [Graphics:Images/CalcWk7_gr_65.gif] over the interval [Graphics:Images/CalcWk7_gr_66.gif] in the first quadrant.  How many subintervals did you need in order to get an estimate that has the value 3.14159...?

    b)  There is a technical problem that makes it impossible to use right endpoint Riemann sums.  What is the problem?


Converted by Mathematica      March 3, 2004