Express Area as a function of r

  1. Ok, so I have this picture. A semi-circle at the ends of a rectangle. It tells me that the perimeter is [tex]\frac 1 4[/tex]. So, isn't the total perimeter just the sum of the circle and rectange? [tex]P=2\pi r+2(L+W)[/tex]

    [tex]r=radius[/tex]
    [tex]W=2r[/tex]

    [tex]P=2\pi r+2(L+2r)[/tex]

    And isn't the total area just the sum of the areas? [tex]A=A_{circle}+A_{rectange}\rightarrow A=\pi r^2 + 2rL[/tex]
     
  2. jcsd
  3. Dick

    Dick 25,810
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    Almost. It's only a semi-circle. So the circumerence of that is just pi*r. And the perimeter of partial rectangle is just 2L+r. Why did you double everything?
     
  4. EnumaElish

    EnumaElish 2,483
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    Does it look like this:
    .._____
    (|____|)

    or not? If it does look like the above then total perimeter is not the sum of the circle & rectangle.
     
  5. Yes, that is the correct picture! :-]

    So it's not? eek.
     
  6. EnumaElish

    EnumaElish 2,483
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    If by perimeter you mean "edges exposed to the outside" then it is the two half-circles plus the 2 long edges of the rectangle. The short edges are "internalized."
     
  7. AHH!!! Yes, very true.

    Thanks a lot :-]
     
  8. So even after ignoring the width, I still was unable to solve it. The person I was helping me showed me the solution and the internal parts were included. Blah.
     
  9. Dick

    Dick 25,810
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    Then I guess they tricked you. What was the exact phrasing of the problem?
     
  10. Sorry, I don't have the book, I will post it tomorrow. >:-[
     
  11. HallsofIvy

    HallsofIvy 40,367
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    The wording you gave was clear. You have a rectangle with a semicircle at each end of radius r. That end of the rectangle, then, has length 2r. For the moment call the length of the other sides l. Then the area of the two semi-circles totals [itex]\pi r^2[/itex] and the area of the rectangle is lr. The total area of figure is [itex]\pi r^2+ lr[/itex].

    The perimeter of the figure is the distance around the two semi-circles, [itex]2\pi r[/itex] and the two lengths, 2l: the perimeter is [itex]2\pi r+ 2l= 1/4[/itex]. You can solve that for l as a function of r and replace l by that in the area formula.
     
  12. [tex]\frac 1 4=2\pi r+2l \rightarrow l=\frac{1-8\pi r}{8}[/tex]

    [tex]A=\pi r^2+lr[/tex]

    [tex]A=\pi r^2 +\left(\frac{1-8\pi r}{8}\right)r[/tex]

    [tex]A=\frac r 8[/tex]

    That's still not the answer in the book! Is the book wrong? I will post the actual problem in a few hours, gotta go library.
     
    Last edited: Feb 8, 2008
  13. Wait a minute guys! Is not the area of the rectangle = [itex](2r)l[/itex]
    ?

    This would make

    [tex]A=\pi r^2+2lr[/tex]
     
  14. HallsofIvy

    HallsofIvy 40,367
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    Your right. I have no idea why I wrote rl!
     
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