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Homework Help: Domain and Range of a function

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data
    If the maximum allowable width, w, of the wall is 100 cm, what are the domain and rand of the function? Remember to include units for both the domain and range.

    2. Relevant equations
    w(d)=100, so
    100=2d, there fore

    3. The attempt at a solution
    Domain: d (less than or equal to) 50cm
    Range: all real numbers

    Sorry, I'm a bit rusty on finding D and R. Thanks for the help, here is the original question via PDF, I got part (a). I am just unsure about my D and R ( I am a perfectionist)

    ALSO: For part (a) I got : A(d)=((Pi)(d^2))/2, correct?


    Attached Files:

  2. jcsd
  3. Oct 28, 2012 #2
    For part a I got [itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}=d^{2}(1+\frac{\pi}{4})[/itex]

    I hope I'm understanding the problem correctly, [itex]d[/itex] is also the diameter of the semi-circle on either side, so [itex]r=\frac{d}{2}[/itex]. I think you are neglecting to add the area of the square inbetween those two semi-circles in your function for part a.

    Edit: squaring mistake :/
    Last edited: Oct 28, 2012
  4. Oct 28, 2012 #3
    Area of cross section is (Area of Square=s^2)+(2 Semi-circles, which is equivalent to one circle, so Area=(Pi)(r^2))

    Area of Square + Area of Circle = Area of Cross section

    Since ,d, in diagram is also the sides of the square I have accounted for the area of a square.

    So area of a circle in terms of Area of Circle(d)= ((Pi)((d/2)^2))
    there fore,

    Area of Cross section (d)= (d^2)+((Pi)((d/2)^2))
    simplified: A(d)=((Pi)(d^2))/2
  5. Oct 28, 2012 #4
    I'm not sure I follow your simplification to: [itex]A(d)=\frac{\pi d^{2}}{2}[/itex]

    My Reasoning:
    Area of square: [itex]d^{2}[/itex]
    Area of circle: [itex]\pi r^{2}[/itex]
    Radius of circle: [itex]r=\frac{d}{2}[/itex]
    [itex]A(d)=d^{2}+\pi r^{2}[/itex]
    [itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}[/itex]
  6. Oct 28, 2012 #5
    Which step are you doing from
    [itex]A(d)=d^{2}+\frac{\pi d^{2}}{4}[/itex]
  7. Oct 29, 2012 #6
    I'm just factoring the [itex]d^{2}[/itex] out of [itex]d^{2}+\frac{\pi d^{2}}{4}[/itex]. Multiply [itex]d^{2}(1+\frac{\pi}{4})[/itex] out using the distributive property and you'll get back to [itex]d^{2}+\frac{\pi d^{2}}{4}[/itex].
  8. Oct 29, 2012 #7
    Wow, I'm embarrassed and you are great.
    Thanks a lot.
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