3u Mathematics Marathon V 1.0 (1 Viewer)

[richz]

Answer to Question 38:

Spoiler

Posed originally by xrtzx

Part (a):

limits of integration at x = 0, x = 1, middle ordinate is: x = 0.5

ie. Area = pi*DInt[(ArcCos(x))^2 dx] = pi*[(1/3)((ArcCos1)^2 + 4(ArcCos0.5)^2 + (ArcCos0)^2)]
= (pi/3)[(0)^2 + 4(pi/3)^2 + (pi/2)^2] = (pi^3/3)(4/9 + 1/4) = (pi^3/3)(25/36)
= (25pi^3)/108 units^3

Part (b):

assuming the balloon is spherical (which was not mentioned in question); let its radius = r

ie. Volume of balloon = (4/3)pi*r^3 ; dV/dr = 4pi.r^2 = surface area of balloon.
ie. d(SA)/dr = 8pi.r
we know dV/dt = 15 when r = 5; but, dV/dt = dV/dr * dr/dt
---> 15 = 4pi.(5)^2 * dr/dt ---> dr/dt = 3/(20pi)

and, d(SA)/dt = d(SA)/dr * dr/dt = 8pi.(5) * 3/(20pi) = 6
therefore, surface area of balloon is increasing at 6 cm^2/s

Next question... will be Question 37.

xrtzx's solutions to both parts of the question were, unfortunately, incorrect... xrtzx, do not let the function f(x) = mx + b .
the formula y = mx + b is a function that describes the relationship between successive output values of Newton's method on f(x). ie. 'y' is 'x(2)' and is output value, while 'x' denotes 'x(1)' and is input value.

Here's the question again for ppl:

Question 37:

Given that when applying the iterative method of Newton's to find a zero of a function f(x), the output values of successive iterations follow a linear pattern, irrespective of the starting input value, given by:
y = mx + b ; where 'y' is the output value, and 'x' is the input value. and 'm' & 'b' are any reals.

Find:

(a) ONE such function f(x) where m = 1 and b is non-zero.

(b) ONE such function f(x) where b = 0 and m is non-zero.

[HINT: f(x) need not be a linear function.]

wlm, for part a) i think u shud halve ur answer again, simpsons rule is h=(a-b)/strips(1/3) etc etc so h=1/2

 

[who_loves_maths]

^ oops my bad, nicely spotted xrtzx, thanks for that. it'll be rectified.