Asked by Alexis
so its this picture of a rectangle inscribed in a circle. the circle has a radius of 2. the vertexes of the circle are present in every quadrant, but you cant tell exactly whgat point they represent. point Q (x,y) is the vertex of the rectangle in Quadrant I and is on the circle. the equation of the circle is : x^2 + y^2 = 4
a.) express the area of the rectangle as a function of x.
b.) give the domain of the area
c.) epxress the perimeter as a function of x.
d.) give the domain of Q
PLEASE HELP IM SO LOST. and friday afternoon is my scheduled math homework day. (dont ask im psycho) please this is bother ing me like crazyy
a.) express the area of the rectangle as a function of x.
b.) give the domain of the area
c.) epxress the perimeter as a function of x.
d.) give the domain of Q
PLEASE HELP IM SO LOST. and friday afternoon is my scheduled math homework day. (dont ask im psycho) please this is bother ing me like crazyy
Answers
Answered by
Damon
a)
x^2 + y ^2 = 4
area = 4 |x y| (so it works in each quadrant)
y = sqrt(4-x^2)
area = 4 x sqrt(4-x^2)
b)
the area can be 0 (at x = 0 and x = 2)and it has a maximum at x = y = sqrt 2
to show that take derivative and set to zero
0 = -4 x^2/sqrt(4-x^2) +4 sqrt(4-x^2)
0 = -4x^2 + 16 - 4 x^2
8 x^2 = 16
x = sqrt 2
so domain of area is 0 to x = sqrt 2 where the area is 8
c)
L = 2 x + 2 y
= 2 x + 2 sqrt(4-x^2)
= 2 ( x + sqrt (4-x^2))
d)
Q goes from x = 0 to x = 2 in Quadrant 1
x^2 + y ^2 = 4
area = 4 |x y| (so it works in each quadrant)
y = sqrt(4-x^2)
area = 4 x sqrt(4-x^2)
b)
the area can be 0 (at x = 0 and x = 2)and it has a maximum at x = y = sqrt 2
to show that take derivative and set to zero
0 = -4 x^2/sqrt(4-x^2) +4 sqrt(4-x^2)
0 = -4x^2 + 16 - 4 x^2
8 x^2 = 16
x = sqrt 2
so domain of area is 0 to x = sqrt 2 where the area is 8
c)
L = 2 x + 2 y
= 2 x + 2 sqrt(4-x^2)
= 2 ( x + sqrt (4-x^2))
d)
Q goes from x = 0 to x = 2 in Quadrant 1
Answered by
Jordan
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