※ 引述《Laoda245566 (草莓獸)》之銘言：
: w=x^2+y^2+z^2
: and X^3 -xy+yz+y^3=1
是z^3 - xy + yz + y^3 = 1才對吧
x = [z^3 +yz +y^3 - 1]/y
x = x(y,z)
現在不回去檢查原始考題還真不行
: at(2.-1.1)
是(2, -1, 1)
: find the derative of w with respect x
這題我覺得出得不是很清楚
他沒說到底w = w(x,y,z) 還是w(x,y)
雖然有一個限制條件z^3 - xy + yz + y^3 = 1
按照題意
如果w是三維空間的一個純量場
那@w/@x = 2x = 4
因為w(x,y,z) = x^2+y^2+z^2就是描述w內涵的一個函數
和限制條件無關
既然寫了x^2 + y^2 + z^2就表示有z這個變數
如果在(2, -1, 1)侷部上寫成雙變數函數w(x,y), w(y,z), w(x,z)
例如w = w(x,y), z^3 - xy + yz + y^3 = 1
3z^2(@z/@x) - y + y(@z/@x) = 0
=> (@z/@x)[3z^2 + y] = y
=> @z/@x | = -1/[3 - 1] = -1/2
(2,-1,1)
@w(x,y)/@x = [2x + 2z(@z/@x)]| = 4 + 2(-1/2) = 3......Ans
(2,-1,1)
再看如果w = w(x,z), z^3 - xy + yz + y^3 = 1
-x@y/@x -y + z@y/@x + 3y^2 @y/@x = 0
=> @y/@x[-x + z +3y^2] = y => @y/@x| = -1/(-2+1+3) = -1/2
(2,-1,1)
3z^2 - x@y/@z + y + z@y/@z + 3y^2 @y/@z = 0
=> @y/@z [-x + z + 3y^2] = -3z^2 - y
=> @y/@z | = (-2)/2 = -1
(2,-1,1)
@w(x,z)/@x = [2x + 2y @y/@x]| = 4 - 2(-1/2) = 5
(2,-1,1)
不過我覺得這是題目沒有標清楚
你用@w(x,y,z)/@x = 2x = 4 他應該沒辦法說你錯