? 發問於 科學及數學數學 · 1 十年前

amaths

1. A light on the ground is 10m from a wall.A man 2m tall walks at 8km/h from the light towards the wall. what is the rate of change of the height of the man's shadow on the wall when he is 4m from the light? Ans= -10km/h

2. if v = 2t - 4, find the distance that the particle moves

a) from t=0 to t=2

Ans=4

b) from t=0 to t=5

Ans=13

3.y = tanx(secx)^2

find dy/dx in terms of secx.

hence find 積(0 to 派/3) (secx)^4 dx. Ans=2根3

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2 個解答

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  • 1 十年前
    最愛解答

    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

    資料來源: My Maths Knowledge
  • 1 十年前

    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

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