anonymous
  • anonymous
A tough one. Find the largest possible area for A(theta)=49sin(2theta)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[A(\Theta)=49\sin 2(\Theta)\]
amistre64
  • amistre64
you know how to do derivatives?
anonymous
  • anonymous
haha some but not with trigonometric functions. I am in precalc

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amistre64
  • amistre64
ok...lets try this route; the largest area of any "sin" is a full circle right?
anonymous
  • anonymous
yes, 2pi
amistre64
  • amistre64
then we need to find a circle that has the same "information" is the given equation.
amistre64
  • amistre64
Do you know the area of a sector? when it involves 2 sides and the sin of an angle?
anonymous
  • anonymous
not yet...I think thats in an upcoming chapter.
amistre64
  • amistre64
ok... and maybe I saw this wrong to begin with...consider this: the area of a triangle is: (1/2) (base) (height) right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
How do we find the height of a triangle when we know an angle? does sin(t) = y/r?
anonymous
  • anonymous
yes
amistre64
  • amistre64
like this
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anonymous
  • anonymous
yes...I understand that
amistre64
  • amistre64
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
anonymous
  • anonymous
and the biggest sin you can get is 1
anonymous
  • anonymous
would it help to tell you that I have a rectangle inscribed inside a semicircle with a radius of 7cm.
anonymous
  • anonymous
and that is the equation for the area
amistre64
  • amistre64
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
amistre64
  • amistre64
it would..... all the information helps :)
anonymous
  • anonymous
sorry about that
amistre64
  • amistre64
like this?
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amistre64
  • amistre64
like this?
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anonymous
  • anonymous
exactly!
amistre64
  • amistre64
yeah....thatd a been helpful to know lol
anonymous
  • anonymous
haha...wow...sorry
amistre64
  • amistre64
which angle is the theta...near the center of up near a corner?
anonymous
  • anonymous
near the center
amistre64
  • amistre64
if I recall correctly, the angle needs to be 45 degrees. so 2t = 45 t = 22.5
amistre64
  • amistre64
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
anonymous
  • anonymous
so would my dimension them be 7 by 14?
amistre64
  • amistre64
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
anonymous
  • anonymous
how did you get these?
amistre64
  • amistre64
magic :) first I killed a live chicken; then a spread its blood around the bedposts......
anonymous
  • anonymous
haha seriosuly!...thats how I feel about these problems! You need to do voodoo to get an answer!!
amistre64
  • amistre64
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
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
thanks!
amistre64
  • amistre64
its calculus....but it works :)
amistre64
  • amistre64
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
anonymous
  • anonymous
gotcha!!
anonymous
  • anonymous
thanks! have a good night!
amistre64
  • amistre64
youre welcome :) Gnite
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