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

Question

Find an equation of the tangent plane to the given surface at the specified point.

z = 3 y^{2} - 2 x^{2} + x, (2, - 1, - 3)

Answer 1

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

Answer 2

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

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

which simplifies to $z = - 7 x - 6 y + 5$

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

Expert Answer

Not exactly what you’re looking for?

Not exactly what you’re looking for?

You might be interested in

New questions