Year 12 Spesh Math?

Question

1. a) Find dy/dx of [(x/2)sqrt(16-(x^2))] + [8 sin^(-1) {x/4}]

b) Integrate | [sqrt(16-(x^2))] .dx

Answer 1

Question a)
y = (x/2) (16 - x²)^(1/2) + 8 sin^(-1) (x / 4)
y = ya + yb

dya /dx
(1/2)(16 - x²)^(1/2) + (1/2)(16 - x²)^(-1/2)(-2x)(x/2)
(1/2)(16 - x²)^(1/2) - x (16 - x²)^(-1/2)
(1/2)(16 - x²)^(-1/2) [(16 - x²) - 2x]

dyb / dx
8 / (16 - x²)^(1/2)

dy/dx = dya / dx + dyb / dx
(1/2)(16 - x²)^(-1/2)[16 - x² - 2x + 16(16 - x²)
(1/2)(16 - x²)^(-1/2)[(16 - x²)(17) - 2x ]

Question b
I = ∫ 1 / (4² - x²)^(1/2) dx
I = sin^(-1)(x / 4) + C

Answer 2

a. (x/2) sqrt(16 - x^2) + 8 sin^(-1) [x/4]
= (x/2) (16 - x^2)^(1/2) + 8 sin^(-1) [x/4]

dy/dx = x/2 [(1/2) (16 - x^2)^(-1/2){-2x}] + [(16 - x^2)^(1/2)] [1/2] + 8(-1) {sin^(-2) [x/4]} {(cos x/4) (1/4)}

dy/dx = {(-x^2)/2} {(16 - x^2)^(-1/2)} + {[(16 - x^2)^(1/2)]/2} - 2[{sin^(-2) (x/4)} cos (x/4)]

or {(- x^2) / [2 sqrt (16 - x^2)]} + {[sqrt (16 - x^2)] /2} - {(2 cos x/4)/ [sin^2 (x/4)]}

b. integral of sqrt (16 - x^2) dx

let x= 4 sin a
dx = 4 cos a da, then substitute and integrate