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anonymous

  • 5 years ago

A tough one. Find the largest possible area for A(theta)=49sin(2theta)

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  1. anonymous
    • 5 years ago
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    \[A(\Theta)=49\sin 2(\Theta)\]

  2. amistre64
    • 5 years ago
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    you know how to do derivatives?

  3. anonymous
    • 5 years ago
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    haha some but not with trigonometric functions. I am in precalc

  4. amistre64
    • 5 years ago
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    ok...lets try this route; the largest area of any "sin" is a full circle right?

  5. anonymous
    • 5 years ago
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    yes, 2pi

  6. amistre64
    • 5 years ago
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    then we need to find a circle that has the same "information" is the given equation.

  7. amistre64
    • 5 years ago
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    Do you know the area of a sector? when it involves 2 sides and the sin of an angle?

  8. anonymous
    • 5 years ago
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    not yet...I think thats in an upcoming chapter.

  9. amistre64
    • 5 years ago
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    ok... and maybe I saw this wrong to begin with...consider this: the area of a triangle is: (1/2) (base) (height) right?

  10. anonymous
    • 5 years ago
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    yes

  11. amistre64
    • 5 years ago
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    How do we find the height of a triangle when we know an angle? does sin(t) = y/r?

  12. anonymous
    • 5 years ago
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    yes

  13. amistre64
    • 5 years ago
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    like this

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  14. anonymous
    • 5 years ago
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    yes...I understand that

  15. amistre64
    • 5 years ago
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    good, then we need to determine the value of "t" that will give us the largest area for the triangle....or thats how I understand this problem

  16. anonymous
    • 5 years ago
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    and the biggest sin you can get is 1

  17. anonymous
    • 5 years ago
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    would it help to tell you that I have a rectangle inscribed inside a semicircle with a radius of 7cm.

  18. anonymous
    • 5 years ago
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    and that is the equation for the area

  19. amistre64
    • 5 years ago
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    i keep seeing this 2 different ways...for example: the area of any given sector of a circle is define as (theta)(r^2)/2

  20. amistre64
    • 5 years ago
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    it would..... all the information helps :)

  21. anonymous
    • 5 years ago
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    sorry about that

  22. amistre64
    • 5 years ago
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    like this?

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  23. amistre64
    • 5 years ago
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    like this?

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  24. anonymous
    • 5 years ago
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    exactly!

  25. amistre64
    • 5 years ago
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    yeah....thatd a been helpful to know lol

  26. anonymous
    • 5 years ago
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    haha...wow...sorry

  27. amistre64
    • 5 years ago
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    which angle is the theta...near the center of up near a corner?

  28. anonymous
    • 5 years ago
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    near the center

  29. amistre64
    • 5 years ago
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    if I recall correctly, the angle needs to be 45 degrees. so 2t = 45 t = 22.5

  30. amistre64
    • 5 years ago
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    the largest area you can get is formed by a square.... and you need 2 squares side by side to make this the max area under the half circle

  31. anonymous
    • 5 years ago
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    so would my dimension them be 7 by 14?

  32. amistre64
    • 5 years ago
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    2 sides equal 7sin(45); and the other 2 sides are 28sin(45) 4.95 is one side... 9.90 is the other side...... it gives you an area about equal to 49.005

  33. anonymous
    • 5 years ago
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    how did you get these?

  34. amistre64
    • 5 years ago
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    magic :) first I killed a live chicken; then a spread its blood around the bedposts......

  35. anonymous
    • 5 years ago
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    haha seriosuly!...thats how I feel about these problems! You need to do voodoo to get an answer!!

  36. amistre64
    • 5 years ago
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    we take the equation for a half circle: y = sqrt(7^2-x^2) and the Area of the "rectangle" to be maximized. A = xy ; substitute y = sqrt(49 - x^2) into this equation

  37. amistre64
    • 5 years ago
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    A = x(sqrt(49-x^2)) now find the derivative.... A' = x(-x/(sqrt(49-x^2))) +1(sqrt(49-x^2)) and make that equal to zero.... then solve for x :)

  38. anonymous
    • 5 years ago
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    thanks!

  39. amistre64
    • 5 years ago
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    its calculus....but it works :)

  40. amistre64
    • 5 years ago
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    the other way is just to notice that the largest area that can be produced in a quarter of a circle is a square....a square has a 45 degree angle where we need it down there

  41. anonymous
    • 5 years ago
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    gotcha!!

  42. anonymous
    • 5 years ago
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    thanks! have a good night!

  43. amistre64
    • 5 years ago
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    youre welcome :) Gnite

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