# Finding Tangent Plane

• Nov 5th 2009, 04:00 PM
CalcGeek31
Finding Tangent Plane
If a function z = g(x, y) has g(1, 2) = 10, gx(1, 2) = 7 and gy(1, 2) = -5, find the equation of the plane tangent to the surface z = g(x, y) at the point where x = 1 and y = 2.

I understand how to do the question when an equation is given but without it, I am lost. Could someone help me get started through the problem.
• Nov 5th 2009, 05:23 PM
makarooney
Quote:

Originally Posted by CalcGeek31
If a function z = g(x, y) has g(1, 2) = 10, gx(1, 2) = 7 and gy(1, 2) = -5, find the equation of the plane tangent to the surface z = g(x, y) at the point where x = 1 and y = 2.

I understand how to do the question when an equation is given but without it, I am lost. Could someone help me get started through the problem.

The normal vector to the surface is given by $$

All you need for a plane is a normal vector and a point. You have both:
$\overrightarrow{n}=<7, -5, -1>$ and your P is (1,2)

Just plug these into the plane equation and you're done.
• Mar 25th 2013, 04:12 PM
brickford5
Re: Finding Tangent Plane
I know this is an old post, but it is what I am currently studying and I'm curious where the -1 value came from here.

Would anyone mind shedding some light on it for me?
• Mar 25th 2013, 06:17 PM
hollywood
Re: Finding Tangent Plane
Since the equation for the surface is $f(x,y,z) = g(x,y)-z = 0$, you can do a linear estimate:

$f(x,y,z)\approx{f}(x_0,y_0,z_0)+f_x(x_0,y_0,z_0)(x-x_0)+$
$f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)$,

which is the equation of the tangent plane. Of course $f_z$ is just -1 in this case.

- Hollywood