 # 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.
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Solve tfor the partial derivative of f with respect to x
${f}_{x}\left(x,y\right)=-4x+1$
${f}_{y}\left(x,y\right)=6y$
At the point (2,1) evaluate
${f}_{x}=\left(2,-1\right)=-7$
At the point (2,-1) evaluate
${f}_{y}\left(2,-1\right)=-6$
Frame the equation of the tangent plane
$z+3={f}_{x}\left(2,-1\right)\left[x-2\right]+{f}_{y}\left(2,-1\right)\left[y+1\right]$
Plug in respective values
$z+3=-7x+14-6y-6$
Simplify
$-7x-6y+5=z$

###### Not exactly what you’re looking for? Jeffrey Jordon

You can use the partial derivatives to find a normal vector to the surface. At the point (2,-1,-3) you have ${f}_{x}\left(2,-1\right)=-7$ and ${f}_{y}\left(2,-1\right)=-6$ so that one normal vector is <-7,-6,-1>.

The equation of the tangent plane is given by $-7\left(x-2\right)-6\left(y+1\right)-\left(z+3\right)=0$

which simplifies to $z=-7x-6y+5$