大學 微積分問題


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  • 1 month later...

(1) Slope(2,-1,3) = ? for both x = 2 & z =√(2x2 + 3y2 - 2)

F(x,y,z) = √(2x2 + 3y2-2) - z ==> F(2,-1,3) = 3

N = Normal(F) = Graf(F) = (4x/2*3, 6y/2*3, -1) = (4/3, -1 -1) = (4, -3, -3)

z = √(8 + 3y2 - 2) = √(6 + 3y2)  ==>  z' = 6y/(2 * 3) = -1

Set  S = slope = (a, 1, -1)  ==> N ‧S = 4a -3 + 3 = 4a = 0 ==> a = 0

Slope = (0, 1 , -1) at Point (2,-1,3)

 

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