Question

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

Answer 1

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

Explanation:

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:

and here is the surface with the tangent plane and the normal vector:

Answer 2

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`