A surface is defined by the following equation: z(x,y)= 2x/(1+y^2) - exp(3 ln(x+y))?

[匿名]

2016-05-16 7:41 am

A) Find the equation of the tangent plane to the surface at the point p(3,-1)

B) Find the total differential of function z(x,y) at that point?

回答 (1)

Lopez

2016-05-16 2:09 pm

✔ 最佳答案

(A)
z(x,y)
= 2x/(1+y^2) - exp( 3 * ln(x+y) )
= 2x/(1+y^2) - (x+y)e^3

z( 3 , - 1 ) = 3 - 2*e^3

z = 2x/(1+y^2) - (x+y)e^3
2x/(1+y^2) - (x+y)e^3 - z = 0

Let f(x,y,z) = 2x/(1+y^2) - (x+y)e^3 - z
∂f / ∂x = 2/(1+y^2) - e^3
∂f / ∂y = - 4xy/(1+y^2) - e^3
∂f / ∂z = - 1
∇ f = [ 2/(1+y^2) - e^3 ] i + [ - 4xy/(1+y^2) - e^3 ] j - k

∇ f , at ( 3 , - 1 , 3 - 2e^3 )
= ( 1 - e^3 , - 3 - e^3 , - 1 )

tangent plane :
( 1 - e^3 )( x - 3 ) + ( - 3 - e^3 )( y + 1 ) - 1( z - 3 + 2e^3 ) = 0
( 1 - e^3 )x - ( 3 + e^3 )y - z = 3 ..... Ans

(B)
at ( 3 , - 1 , 3 - 2e^3 ) :

dz
= ( ∂f / ∂x )dx + ( ∂f / ∂y )dy
= ( 1 - e^3 )dx + ( - 3 - e^3 )dy
= ( 1 - e^3 )dx - ( 3 + e^3 )dy ..... Ans

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