amaths

1) Suppose the distance between the man and the wall be ym and the height of the wall is hm.

By similar triangles,

( 10 - y ) / y = 2 / h

20 = 10h - hy

When the man is 4m from the light, y = 10 - 4 = 6, so

20 = 10h - 6h

h = 5

Differentiate both sides with respect to t,

( d / dt ) 20 = ( d / dt ) 10h - ( d / dt ) hy

10( dh / dt ) - h ( dy / dt ) - y ( dh / dt ) = 0

( 10 - 6 )( dh / dt ) = ( 5 )( - 8 )

dh / dt = -10

So the rate of change is -10km/h.

2a) ds / dt = v

s = ∫v dt

= ∫( 2t - 4 ) dt

= 2t^2 / 2 - 4t + C

t^2 - 4t + C

When t = 0, s = 0 and so C = 0, then

s = t^2 - 4t

s = ( 2 )^2 - 4 ( 2 )

= - 4

So the distance moved is 4.

b) When t = 2, v = 0 and the particle is going to change its moving direction.

When t = 5,

s = ( 5 )^2 - 4 ( 5 )

= 5

But it's only the displacement travelled but not distance travelled, so the sum of displacement from t = 0 to 2 and t = 2 to 5 ( s1 ) will be the distance travelled, i.e.

- 4 + s1 = 5

s1 = 9

So the distance travelled: 4 + 9 = 13

3) y = ( tan x )( sec^2 x )

dy / dx = ( tan x )( d / dx )( sec^2 x ) + ( sec^2 x )( d / dx )( tan x )

= 2 tanx secx( tanx secx) + sec^2 x ( sec ^2 x )

= 2 tan^2 x sec^2 x + sec^4 x

= 2 ( sec^2 x - 1 )( sec^2 x ) + sec^4 x

= 3 sec^4 x - 2 sec^2 x

∫( 3 sec^4 x - 2 sec^2 x ) dx = tan x ( sec^2 x )

∫3 sec^4 x dx = tan x ( sec^2 x ) + 2 tanx

∫sec^4 x dx = tan x sec^2 x / 3 + 2 tan x / 3

So define the integral from 0 to pi / 3,

∫sec^4 x dx = tan(pi/3)sec^2(pi/3)/3+2tan(pi/3)/3-tan(0)sec^2(0)/3-2tan(0)/3

= 4sqr3/3+2sqr3/3

= 6sqr3/3

= 2sqr3

when he is x km from the light
height of the man's shadow = h
h/(10/1000) = (2/1000)/x
h = 1/(50000x)
dh/dt = (1/50000)(-1/x^2) dx/dt
when x = 4/1000 = 0.004
dh/dt = (1/50000)(-1/0.004^2) (8) = -10 km/hr
--------------------------------------------------------------------------------
2a)
ds/dt = 2t - 4
- int (2t - 4) dt, where t from 0 to 2
= -(-4)
= 4
b)
- int (2t - 4) dt, where t from 0 to 2 + int (2t - 4) dt, where t from 2 to 5
= 4 + 9
= 13
-------------------------------------------------------------
y = tanx (secx)^2
dy/dx
= (secx)^4 + 2(tanx)^2 (secx)^2
= (secx)^4 + 2 [(sec)^2 - 1] (secx)^2
= 3(secx)^4 - 2 (secx)^2
y = int [ 3(secx)^4 - 2 (secx)^2 ] dx
= 3 int (secx)^4 dx - 2 int (secx)^2 dx
= 3 int (secx)^4 dx - 2 tanx
int (secx)^4 dx = [ tanx(secx)^2 + 2 tanx ]/3
int (secx)^4 dx, where x from 0 to pi/3
= ( 4sqrt(3) + 2sqrt(3) )/3
= 2sqrt(3)

2007-11-08 21:33:24 補充：
oh, 慢左, 做得好辛苦, 記得投我一票, 唔該