Sunday, January 12, 2014

2-3 隱函數的微分

2-3 隱函數的微分

隱函數的微分


(甲) 隱函數的微分 






2x^2+xy+y^2- 4 = 0 對其做x的微分, 可得 
4x + y + xy' + 2yy' = 0 
(x+2y)y' = - (4x+y)
y' = -(4x+y) / (x+2y) 
dy/dx| (x,y) = (-1,2) = -(-4+2)/ -1+4 = 2/3  
所以其切線為 y- 2 = 2/3(x-(-1)) 



  





dy/dx 為對式中x做一次微分, 求得 
xy'+y+2yy'-2x = 0 
(x+2y)y' = 2x -y 
y' = (2x - y) / (x+2y)
x的2次微分,求得 
(1+2y')y' + (x+2y)y" = 2 - y' 
(x+2y)y"  = 2 - (2+2y)y' 
 y" = 2 [ 1- (1+y)y' ] / (x+2y)  再將 y' 代入

法2 : 
將 y' 再對x微分一次, 使用 (f/g)' 也可以求得. 








原式對x做一次微分, 可得: 

3y^2y' + 3y + 3xy' + 3x^2 = 0 
3(x+y^2)y'  = -3(y+x^2) 
(x+y^2)y' = - (y+x^2)
y ' =  - (y+x^2) / (x+y^2)


  



原式對x做一次微分, 可得:  


6x - 2(y+xy') - 2yy' = 0 
6x - 2y - 2xy' - 2yy' = 0 
6x - 2y - 2(x+y)y' = 0 
6 - 0 - 2(1+0)y' = 0 
-2y' = -6  
y' = 3 











(1)  2x - 2y - 2xy' - 6yy' + 2 - y' = 0 
(2) 2 + 2 - 2y' +6y' + 2 - y' = 0 , 3y' = -6 , y' = -2  







先求對x一次的微分. 
2/3x^-1/3 + 2/3y^-1/3y' = 0 
y' =  - 2/3x^-1/3 /  2/3y^-1/3  =  x^-1/3 / y^-1/3 = (y/x)^1/3 , m = 1 at x , y  = (2^1/2)/4 . 
y = mx  + b , (2^1/2)/ 4 = (2^1/2)/ 4 + b , b = 0 , y = x  . 










先求x的微分, 

2ax+b(y+xy') + 2yy' + d + ey' = 0 
2ax + d + by + (bx+2y+e)y' = 0 
(bx+2y+e)y' = - (2ax + d + by) 
y' = - (2ax + d + by) / (bx+2y+e) 

Again ? 








To find the tangent line , we must find its slope at (-1,2) . the slope of curve is m . 
4x + y+ xy' + 2yy' = 0 
-(4x+y) = (x+2y)y' 
y' = - (4x+y) /(x+2y)  
m = - (-4+2) / (-1+4) = 2 / 3 , So the tangent line of it is y = 2/3 x + b 
2 = -2/3 + b , b = 8/3 . 
so y = 2/3x + 8/3 is the tangent line . 




   




由題意, x- 4y + 11 = 0 為其切線. 
 假設其切點為(a,b)
-4y = - (x+11) , 4y = (x+11) , y = 1/4x + 11/4 . 所以 m = 1/4  at (a,b) .
接著求x的微分, 2x + 8yy' + 2 = 0 , 8yy' = -2x -2  , y' = -(x+1)/(4y) .
-(a+1)/4b = 1/4 , 4b = -4(a+1), b = -(a+1) . 將其帶入曲線式中
a^2 + 4(a+1)^2 + 2a - 19 = 0
a^2 + 4 (a^2+ 2a + 1) + 2a - 19 = 0
5a^2 + 10 a  - 15 = 0
a^2 + 2a -3 = 0
(a+3)(a-1) =  0

a= - 3 or 1

a = -3 , b = 2 => 2 = -3/4 + 11/4
a = 1 ,  b = -2 =>  -2  = 1/4 + 11/4 不合
所以  (-3,2) 為所求.





先求對x一次的微分. 
2x - 8yy' - 16y'  = 0 
-8(y+2)y' = -2x 
y'  = x/4(y+2)
令 y = mx +b ,  m為其切線 at (1,-2)'s slope . 
- 2 = m + b 
- (2+b) = m , 

  







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