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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[A(\Theta)=49\sin 2(\Theta)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you know how to do derivatives?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha some but not with trigonometric functions. I am in precalc

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok...lets try this route; the largest area of any "sin" is a full circle right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then we need to find a circle that has the same "information" is the given equation.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know the area of a sector? when it involves 2 sides and the sin of an angle?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not yet...I think thats in an upcoming chapter.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok... and maybe I saw this wrong to begin with...consider this: the area of a triangle is: (1/2) (base) (height) right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0How do we find the height of a triangle when we know an angle? does sin(t) = y/r?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes...I understand that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0good, 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
 5 years ago
Best ResponseYou've already chosen the best response.0and the biggest sin you can get is 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it help to tell you that I have a rectangle inscribed inside a semicircle with a radius of 7cm.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that is the equation for the area

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0it would..... all the information helps :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah....thatd a been helpful to know lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0which angle is the theta...near the center of up near a corner?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if I recall correctly, the angle needs to be 45 degrees. so 2t = 45 t = 22.5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0so would my dimension them be 7 by 14?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02 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
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get these?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0magic :) first I killed a live chicken; then a spread its blood around the bedposts......

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha seriosuly!...thats how I feel about these problems! You need to do voodoo to get an answer!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we take the equation for a half circle: y = sqrt(7^2x^2) and the Area of the "rectangle" to be maximized. A = xy ; substitute y = sqrt(49  x^2) into this equation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0A = x(sqrt(49x^2)) now find the derivative.... A' = x(x/(sqrt(49x^2))) +1(sqrt(49x^2)) and make that equal to zero.... then solve for x :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its calculus....but it works :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0thanks! have a good night!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0youre welcome :) Gnite
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