Project Work For Additional Mathematic 2009 Part 2 & 3

Part 2 Question A The length of arc (s) of a circle can be found by using the formula where r is the radius. The result is as below:

From the table, we can conclude that Length of arc PQR = Length of arc PAB + Length of arc BCR

Part 2 Question B i Again, we use the same formula to find the length of arc of PQR, PAB, BCD and DER

This time, the table is a big one. You can click on the table to change to the full size view

Again, we can conclude that Length of arc PQR = Length of arc PAB + Length of arc BCD + Length of arc CDR

Part 2 Question B ii Base on the findings in the table in(a) and (b) above, we conclude that: The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number of the inner semicircles. Part 2 Question C

I suggest, Part 2 (c): You need to take at least 2 different values of diameters for the outer semicircle. You need to show your tables for each value of those diameters. This part is only to prove that your generalization stated in Part 2 (b) (ii) is still true. But it is good to make a proof for it…. Below this is just only the equation of formula….make a example off our own by using 2 various if d Diagram above shows a big semicircle with n number of small inner circle. From the diagram, we can see that

The length of arc of the outer semicircle

The sum of the length of arcs of the inner semicircles

Factorise π/2

Substitute We get,

where d is any positive real number We can see that (s out) = ∑ n (s in), n = 2, 3, 4, ...... where, s in = length of arc of inner semicircle s out = length of arc of outer semicircle As a result, we can conclude that The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number of the inner semicircles. This is true for any value of the diameter of the semicircle In other words, for different values of diameters of the outer semicircle, show that the generalisations stated in b (ii) is still true

Part 3 Question A

Part3 Question B

Part3 Question C Linear Law y = -πx^2/4 + 5πx/2 Change it to linear form of Y = mX + C. y/x = -πx/4 + 5π/2 Y = y/x X=x m = -π/4 C = 5π/2 Thus, plot a graph of y/x against x and draw the line of best fit. Find the value of y/x when x = 4.5 m. Then multiply y/x you get with 4.5 to get the actual value of y.

Part 3 Question D We need to get the largest value of y so that the cost of constructing the garden is minimum. Method 1: Differentiation y = -πx^2/4 + 5πx/2 dy/dx = -πx/2 + 5π/2 (d^2)y/dx^2 = -π/2 <--- y has a maximum value. At maximum point, (d^2)y/dx^2 = 0. -πx/2 + 5π/2 = 0 πx/2 = 5π/2 x=5m x = 5 m: maximum value of y = -π(5^2)/4 + 5π(5)/2 = 6.25π m^2 Method 2: Completing the Square y = -πx^2/4 + 5πx/2 = -π/4(x^2 - 10x) = -π/4(x^2 - 10x + 25 - 25) = -π/4[(x - 5)^2 - 25] = -π/4(x - 5)^2 + 25π/4 y is a n shape graph as a = -π/4. Hence, it has a maximum value. When x = 5 m, maximum value of the graph = 6.25π m^2

Part 3 Question E Arithmetic Progression The keywords are: The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. (n = 12) The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m) The diameter of the smallest flower bed is 30 cm. (a = 30 cm = 0.3 m)

The diameter of the flower beds are increased by a constant value successively. (d = ?) S12 = (n/2)[2a + (n - 1)d] 10 = (12/2)[2(0.3) + (12-1)d] = 6(0.6 + 11d) = 3.6 + 66d 66d = 6.4 d = 16/165 Since the first flower bed is 0.3 m, Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are: 131/330 m, 163/330 m, 13/22 m, 227/330 m, 259/330 m, 97/110 m, 323/330 m, 71/66 m, 129/110 m, 419/330 m, 41/30 m