Modelling with differentiation Flashcards
1
Q
Figure 3 shows a flowerbed.
Its shape is a quarter of a circle of radius x metres with two equal rectangles attached to it along its radii.
Each rectangle has length equal to x metres and
width equal to y metres.
( Figure shows a quarter of a circle with a radius of x, rectangles have a length of x and a width of y )
Given that the area of the flowerbed is 4 m^2,
show that y = 16 - pi x^2 / 8x
A
- Area of flowerbed = 4 m^2
- Kr^2 + Cxy = 4 ( Where k is the section of the circle and C is the amount of rectangles )
- 1 / 4 pi x^2 + 2xy = 4
- y = ( 4 - 1 / 4 pi x^2 ) / 2x
= 16 - pi x^2 / 8x
2
Q
Hence show that the perimeter P metres of the flowerbed is given by the equation.
P = 8 / x + 2x
( y = 16 - pi x^2 / 8x )
A
- P = 2x + Cy + K pi x^2 ( K = 1 / 2 ( pi d = the perimeter of the whole circle, pi r = the perimeter of a semi - circle, so perimeter of quarter of the circle is 1 / 2 pi r )
( C = 4 , the number of y values ) - P = pi x / 2 + 2x + 4 ( 16 - pi x^2 / 8x )
- P = pi x / 2 + 2x + 8 / x - pi x / 2
- P = 8 / x + 2x