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