Find an equation of the tangent plane to the given surface at the specified point. z=3y^2-2x^2+x,\ (2,-1,-3)

avissidep 2021-05-27 Answered
Find an equation of the tangent plane to the given surface at the specified point. z=3y22x2+x, (2,1,3)
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Aamina Herring
Answered 2021-05-28 Author has 85 answers

Solve tfor the partial derivative of f with respect to x
fx(x,y)=4x+1
fy(x,y)=6y
At the point (2,1) evaluate
fx=(2,1)=7
At the point (2,-1) evaluate
fy(2,1)=6
Frame the equation of the tangent plane
z+3=fx(2,1)[x2]+fy(2,1)[y+1]
Plug in respective values
z+3=7x+146y6
Simplify
7x6y+5=z

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Jeffrey Jordon
Answered 2021-10-14 Author has 2262 answers

You can use the partial derivatives to find a normal vector to the surface. At the point (2,-1,-3) you have fx(2,1)=7 and fy(2,1)=6 so that one normal vector is <-7,-6,-1>. 

The equation of the tangent plane is given by 7(x2)6(y+1)(z+3)=0

which simplifies to z=7x6y+5

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