汪汪 asked in 科學數學 · 1 decade ago

微積分的Parametric curves的問題2?

一、Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.

1. x= 2t^2+1 , y=(1/3)t^3-t ; t =3

二、Find dy/dx and d2y/dx2.For which values of t is the curve concave upward?

2. x=t^3-12t , y=t^2-1

三、Find the points on the curve where the tangent is horizontal or vertical.If you have a graphing device,graph the curve to check your work.

3. x=2t^3+3t^2-12t , y=2t^3+3t^2+1

Update:

沒人回答?

Update 2:

我算出

1. 2x-3y=20

對嗎?

Update 3:

第2題 2/3 *[ (t^2- 4) - 2t^2] /[3(t^2-4)^3]

看不懂...........

Update 4:

懂了

Update 5:

凹向上是什麼意思?圖形向上?

2 Answers

Rating
  • 1 decade ago
    Favorite Answer

    Q1:

    m=dy/dx= y'(t)/x'(t)=(t^2 -1)/(4t)

    t=3, (x, y, m)=(19, 6, 2/3) => (y-6)= 2/3 *(x-6) => 2x-3y= 20

    Q2:

    dy/dx=y'(t)/x'(t)=(2t)/(3t^2 -12)= 2/3 * t/(t^2 - 4) 設為 m(t)

    d^2 y/dx^2 = dm/dx = m'(t)/ x'(t)= 2/3 *[ (t^2- 4) - 2t^2] /[3(t^2-4)^3]

    = -2/9 (t^2 + 4) /(t^2- 4)^3

    凹向上, d^2 y/dx^2 >0 => (t^2 - 4) <0 => -2 < t< 2

    Q3:

    dy/dx= y'(t)/x'(t)= (6t^2 + 6t)/(6t^2 + 6t-12)

    = t(t+1)/[(t+2)(t-1)]

    水平切點 t=0, (x, y)=(0, 1)

    t=-1, (x, y)=(13, 2)

    鉛直切點 t=-2, (x, y)=(20, -3)

    t=1, (x, y)=(-7, 6)

    2009-05-05 22:31:00 補充:

    凹向上: concave upward, 斜率函數遞增, 二階導函數>0

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