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Homework Help: Trouble creating a conjecture, relating to graphs and complex functions

  1. May 26, 2009 #1
    Need to investgate the ratio of the areas formed when y=xn is graphed between two arbirary parameters x=a and x=b such that a<b

    1. Given the funtion y=x2, consider the region formed by this function from x=0 to x=1 and the x-axis. Label this area B. Label the region form y=0 to y=1 and the y-axis area A.

    Find the ratio of area A : area B

    now this is where i had trouble

    Calculate the ratio of the ares for other functions of the type y=xn, n [tex]\in[/tex] Z+ between x=0 an x=1. (not to complex but) Make a conjecture and test your conjecture for other subsets of the real numbers.

    2.Does your conjecture hold only for areas between x=0 and x=1? Examine for x=0 and x=2, x=1 and x=2, etc.

    3. Is your conjecture true for the general case y=xn from x=a to x=b such that a<b and for the regions defned below? If so prove it; if not explain why not.

    Area A: y=xn, y=an, y=bn and the y-axis
    Area B: y=xn, x=a, x=b and the x-axis

    4. Are there general formulae for the ratios of the volumes of revolution generated by the regions A and B when they are each rotated about
    a) the x-axis?
    b) the y-axis?

    State and prove your conjecture

    I can do the very first part of this rather easily but I'm having trouble making the conjcture because although it doesn't have to be correct, I do have to prove it later on. Questions 3 and 4 are the ony real problems so ny help with this would be of a major assistance to me. Much Thanks
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2


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    You calculated some ratios in part (1), for n=2 and for other values of n. What were the results? Maybe we can help you spot the pattern.
  4. May 28, 2009 #3
    ok this is the work i have done so far

    y=x2 [tex]\int[/tex] x2 dx = x3/3 betwen x=1 and x=0
    gives dy/dx = 1/3

    1-1/3 = 2/3
    so the ratio between area A and B is 1/3 : 2/3

    then again with this time with y=x3 [tex]\int[/tex] x3 dx = x4/4 between x=1 and x=0
    gives dy/dx = 1/4

    1-1/4 = 3/4
    so the ratio is 1/4 : 3/4

    then for when y=x4

    the ratio becomes 1/5 : 4/5

    so everytime the exponent increases so does the ratio they end up being simple ratios like
    1:2, 1:3, 1:4 etc.

    then as i moved on i got to:
    [tex]\int[/tex] xn dx = xn+1/n between x=1 and x=0
    [tex]\int[/tex] y1/n dy = y1/n+1/(1/n+1) between y=1 and y=0

    this is most of the work i have done and some small scratchwork as well, im still working on it but any help will be greatly appreciated
  5. May 28, 2009 #4


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    Okay, but why have you not found those values and their ratio?

  6. May 28, 2009 #5
    alright so an update on things that i have done and the ratio to go with it

    it should say
    [tex]\int[/tex] xn dx = xn+1/n+1 between x=0 and x=1



    [tex]\int[/tex] y1/n dy = y1/n+1/(1/n +1) between y=0 and y=1


    which gives a ratio of



  7. May 28, 2009 #6
    and i was working on doing it for between bn, an and b and a.

    [tex]\int[/tex] xn dx = xn+1/n+1 between x=b and x=a


    [tex]\int[/tex] y1/n dy = y1/n + 1(1/n + 1) between y= bn and y= an

    for y, i got

    nbn+1/(n+1) - nan+1/(n+1)

    but for x i dont know where to go after

    bn+1/n - an+1/n

    any advice available on what step to take next?
  8. Jun 1, 2009 #7
    me again, started doing Q4, but am not entirely sure where to go next, can anyone point me in the right direction?

    so for x, volume about a solid is V= [tex]\int[/tex] [tex]\pi[/tex] y2 dy between x=b and x=a

    which gives -

    V = [tex]\int[/tex] [tex]\pi[/tex] x2 dx = [tex]\pi[/tex] (x3/3) between x= b and x= a

    V = [tex]\pi[/tex](b3/3) - (a3/3)

    then for y

    V = [tex]\int[/tex] [tex]\pi[/tex] [tex]\sqrt{y}[/tex] dy = [tex]\pi[/tex] (y1.5) between y=bn and y=an

    V = [tex]\pi[/tex](b1.5n) - (a1.5n)

    (pi isn't to a power, its just how the refrencing tool put it)

    thats what i have so far but im not 100% sure on what to do next, any tips are welcome =)
  9. Jun 1, 2009 #8


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    I'm still not clear on what areas you are talking about. Given y= xn, between x= a and x= b, you want the areas below the curve and above the curve, both between x= a, x= b and, of course, y= an, y= bn? If that is true, after calculating the area under the curve by
    [tex]\int_a^b x^n dx= \frac{b^{n+1}- a^{n+1}}{n+1}[/tex]
    the area above the cuve is just that subtracted from the area of the entire rectangle with vertices at (a, 0), (a, bn), (b, bn), (b, 0). That is, of course, "height x width"= (b- a)bn so the "area above the curve" is
    [tex](b- a)b^n- \frac{b^{n+1}- a^{n+1}}{n+1}[/tex].

    It might help to remember that
    [tex]b^n- a^n= (b- a)\left(b^{n-1}+ ab^{n-2}+ a^2b^{n-3}+ \cdot\cdot\cdot+ a^{n-2}b+ a^{n-1}\right)[/tex]
  10. Jun 1, 2009 #9
    @HallsofIvy: Here are the questions: http://www.eca.com.ve/hsclassweb/kurtsupplee/Higher%20Level/PROJECTS/Ratio%20areas-volumes.pdf" [Broken]
    I'm also stuck on #4...
    But for the other ones, you forgot to state the limitations of the conjecture. (ALSO!!! the ratio is of A:B so that means the conjecture is supposed to be n:1 and not 1:n; you just have the form B:A but all of it is correct (referring to CaptainK's work).) Remember that you have to test ALL of the subsets of real numbers so that means that you should state your conjecture in such a way to incorporate the most subsets as possible.

    Here is mine:
    Following the form y=x^n, the ratio of area A:area B is n:1 where A=the integral of the ABSOLUTE VALUE of the function (y^(1/n))dy from a^n to b^n and B=the integral of the ABSOLUTE VALUE of the function (x^(n))dx from a to b. The conjecture only works with n=Positive Numbers and n=0. (The absolute values essentially limit all testing to the first quadrant since the function y=x^n is essentially just reflected over the y-axis for n=Even Numbers and reflected over both the x-axis and y-axis for n=Odd Numbers, so you won't need to do extensive tests on Negative a's and b's since they'll still give the same ratios (but you need to prove this!).) (The conjecture does not work for n=Negative Numbers since you cannot define A and B when using them.)
    And HallsofIvy, it isn't asking for above the curve to form a rectangle, it wants A=area between y=x^n , y=a^n , y=b^n , and the y-axis ; and B=area between y=x^n , x=a , x=b , and the x-axis. Just to clarify. :)
    So for the area questions (1,2,3) I have:
    Setup A=integral of (y^(1/n))dy from a^n to b^n
    Result A=( (n(b^n)((b^n)^(1/n))) - (n(a^n)((a^n)^(1/n))) ) / (n+1)
    Setup B=integral of (x^n)dx from a to b
    Result B=( (b^(n+1)) - (a^(n+1)) ) / (n+1)
    Dividing Result A by Result B yields n therefore making the ratio n:1.
    And for the volume question (4) I have: (V=pi(r^2) where the function replaces r)
    Setup A=pi(integral of ((b^n)^2)dx from 0 to b) - pi(integral of ((x^n)^2)dx from a to b) - pi(integral of ((a^n)^2)dx from 0 to a)
    Result A=(b^(2n+1))(pi - (pi/(2n+1))) + (a^(2n+1))( (pi/(2n+1)) - pi ) / (2n+1)
    Setup B=pi(integral of ((x^n)^2)dx from a to b)
    Result B=(pi( (b^(2n+1)) - (a^(2n+1)) )) / (2n+1)
    Dividing Result A by Result B yields 2n therefore making the ratio 2n:1.
    Setup A=pi(integral of ((y^(1/n))^2)dy from a^n to b^n)
    Result A=((pi)n((b^(n+2)) - (a^(n+2))) / (n+2)
    Setup B=pi(integral of (b^2)dy from 0 to b^n) - pi(integral of ((y^(1/n))^2)dy from a^n to b^n) - pi(integral of (a^2)dy from 0 to a^n)
    Result B=(((pi)n((a^(n+2)) - (b^(n+2))) / (n+2)) - pi((a^(n+2)) - (b^(n+2)))
    Dividing Result A by Result B yields (1/2)n therefore making the ratio (n/2):1.
    Now that the analytical derivation is out of the way, you just need to prove it with concrete examples and be sure to test the odd stuff like irrational, rational, negative, etc. (Negative won't work as I said above but you need to prove that again for volumes!)
    PS: Thanks in advance and this is due Wednesday June 3, 2009 for me.
    Also, what program do you guys use to make the mathematical notation?
    (If you can't read the computer notation, copy and paste it into http://www.wolframalpha.com/" [Broken]))

    Okay, so I ended up figuring it all out (or at least I think it works (thx CaptainK and HallsofIvy for making me think analytically rather than numerically)), so I'd greatly appreciate if someone could check my work before I have to type it all up and get screengrabs of my calculators, etc. :)
    Last edited by a moderator: May 4, 2017
  11. Jun 2, 2009 #10
    So, this assignment is due tomorrow for me, so can anyone check my work in the above post?

    TIA. :)
  12. Jun 5, 2009 #11
    y=x^n, the ratio of area A:area B is n:1, i tested this conjecture but it only works if the limits of x start at 0 therefore the conjecture does not apply in question 3. how do i answer question 3?
  13. Jun 5, 2009 #12
    you show an example of a n=negative number to show that the conjecture does not hold as a general formula and that this is a limitation of your conjecture.

    this also applies to question 4, since it is the same idea, starting off with the same areas, just rotated either around the x-axis or y-axis, which just yields a different ratio.
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