How do you find the equations for the tangent plane to the surface #z=x^2-2xy+y^2# through (1,2,1)?

Question

Samantha Adkison · Accepted Answer

# -2x+2y-z = 1 #

First we rearrange the equation of the surface into the form # f(x,y,z)=0#

# z=x^2-2xy+y^2 #
# :. x^2-2xy+y^2-z = 0 #

And so we define our surface function, #f#, by:

# f(x,y,z) = x^2-2xy+y^2-z #

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

# grad f(x,y,z) = (partial f)/(partial x) hat(i) + (partial f)/(partial y) hat(j) + (partial f)/(partial z) hat(k) #

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

# grad f = ((partial)/(partial x) (x^2-2xy+y^2-z))hat(i) + #
# " " ((partial)/(partial y) (x^2-2xy+y^2-z))hat(j) + #
# " " ((partial)/(partial z) (x^2-2xy+y^2-z))hat(k) #
# " "= (2x-2y)hat(i) + (-2x+2y)hat(j) + (-1)hat(k) #

So for the particular point #(1,2,1)# the normal vector to the surface is given by:

# grad f(1,2,1) = (2-4)hat(i) +(-2+4)hat(j) -hat(k) #
# " " = -2hat(i) +2hat(j) -hat(k) #

So the tangent plane to the surface # z=x^2-2xy+y^2 # has this normal vector and it also passes though the point #(1,2,1)#. It will therefore have a vector equation of the form:

# vec r * vec n = vec a * vec n #

Where #vec r=((x),(y),(z))#; #vec n=( (-2), (2), (-1) )#, is the normal vector and #a# is any point in the plane

Hence, the tangent plane equation is:

# ((x),(y),(z)) * ( (-2), (2),(-1) ) = ((1),(2),(1)) * ( (-2), (2),(-1) ) #
# :. (x)(-2) + (y)(2) + (z)(-1) = (1)(-2) + (2)(2) + (1)(-1) #
# :. -2x+2y-z = -2+4-1 #
# :. -2x+2y-z = 1 #

We can confirm this graphically: Here is the surface with the normal vector:

Samantha Adkison · Answer

As an alternative approach, we can parameterise the surface in terms of #alpha# and #beta# as:

#mathbf r(alpha, beta) = langle alpha, beta, alpha^2 - 2 alpha beta + beta^2 rangle#

Noting that to first order :

#d mathbf r_alpha = (partial mathbf r)/(partial alpha) d alpha = langle 1,0,2 (alpha - beta) rangle d alpha#
#d mathbf r_beta = (partial mathbf r)/(partial beta) d beta = langle 0,1,- 2 (alpha - beta) rangle d beta #
....we have this:

The shaded area is a vector, #d mathbf A #, whose direction is the normal to the surface at that point, and given by the vector cross product :

#d mathbf A = dA \ color(red)( mathbf e_n ) =d mathbf r_alpha xx d mathbf r_beta#
# = det ((mathbf e_x, mathbf e_y, mathbf e_z),(1,0,2 (alpha - beta)),(0,1,- 2 (alpha - beta))) dalpha dbeta#
#= langle - 2(alpha - beta), 2 (alpha - beta), 1 rangle \ dalpha \ dbeta#
We are not really bothered by the magnitide of #d mathbf A#, its the direction (of any normal vector #mathbf n#, ie we don't even need to normalise to #mathbf e_n#) that matters, so we can forget about # \ dalpha \ dbeta# and evaluate this as one of many normal vectors:

#mathbf n = langle - 2(alpha - beta), 2 (alpha - beta), 1 rangle_{(alpha = 1, beta = 2)} = langle 2, -2, 1 rangle#
Now we know from the scalar dot product that, for a plane surface:

#(mathbf r - mathbf r_o) cdot mathbf n = 0#
#implies mathbf r cdot mathbf n = mathbf r_o cdot mathbf n#
#implies langle x, y, z rangle cdot langle 2, -2, 1 rangle = langle x_o, y_o, z_o rangle cdot langle 2, -2, 1 rangle#
So we plug in #mathbf r_0 = langle 1, 2, 1 rangle# and we get:

# 2x - 2 y + z = -1#

Elijah Abram · Answer

undefined To find the equation for the tangent plane to the surface z=x^2-2xy+y^2 through the point (1,2,1), we can use the following steps:

1. Calculate the partial derivatives of the surface equation with respect to x and y.
   ∂z/∂x = 2x - 2y
   ∂z/∂y = -2x + 2y

2. Evaluate the partial derivatives at the given point (1,2,1).
   ∂z/∂x = 2(1) - 2(2) = -2
   ∂z/∂y = -2(1) + 2(2) = 2

3. The equation for the tangent plane can be written as:
   z - z₀ = (∂z/∂x)(x - x₀) + (∂z/∂y)(y - y₀)

Plugging in the values:
   z - 1 = -2(x - 1) + 2(y - 2)

4. Simplify the equation:
   z - 1 = -2x + 2 + 2y - 4
   z = -2x + 2y - 1

Therefore, the equation for the tangent plane to the surface z=x^2-2xy+y^2 through the point (1,2,1) is z = -2x + 2y - 1.