Question

Consider the paraboloid surface z = x 2 + y 2 .

Answer 1

a. We don't really need calculus to do this

$z = f(x,y) = x^2 + y^2$

We're interested in f near (1,1,f(1,1))

$f(x,y)=f(1+(x-1), 1+(y-1)) = (1+(x-1))^2 + (1+(y-1))^2$

$f(x,y)=2 + 2(x-1) + 2(y-1) + (x-1)^2 + (y-1)^2$

The tangent plane at (1,1,2) is the best linear approximation to f at (1,1).  We just drop the squared terms:

$z =2 + 2(x-1) + 2(y-1)$

$z=2x + 2y - 2$

Answer: 2x + 2y - z = 2

b.

We repeat the above in general, near f(r,s)

$f(x,y)=f(r + (x-r), s+(x-s))$

$f(x,y) = (r + (x-r))^2 + (s + (x-s))^2$

$f(x,y)= r^2 + s^2 + 2r(x-r) + 2s(y-s) + (x-r)^2 + (y-s)^2$

Again the tangent plane at (r,s,f(r,s)) is gotten by dropping the squared terms,

$z = r^2 + s^2 + 2r(x-r) + 2s(y-s)$

This has to contain

$(x,y,z)=(t, 2-2t, -1)=(0,2,-1) + t(1,-2,0)$

$-1 = r^2 + s^2 + 2r(0-r) + 2s(2-s)$

$r^2 + s^2 - 4s - 1 = 0$

The line is perpendicular to the normal of the plane, so a zero dot product.

$2r x + 2sy - z = r^2 + s^2$

$(2r, 2s, -1) \cdot (1, -2, 0) = 0$

$2r - 4s=0$

$r = 2s$

$(2s)^2 + s^2 - 4s -1=0$

$5s^2 - 4s -1 = 0$

$(5s + 1)(s - 1) = 0$

$s=1 \textrm{ or } s=-1/5$

$r=2, s=1 \textrm{ or } r=-2/5, s=-1/5$

Two tangent planes contain the line.  

$2r x + 2sy - z = r^2 + s^2$

$4x + 2y - z = 5 \textrm{ and } (-4/5)x - (2/5)y - z = (-2/5)^2+(-1/5)^2$

Answer:  4x + 2y - z = 5 and  4x + 2y + 5z = -1

Let's check two points on the line are in our planes, (0,2,-1), (1,0,-1)

Looks good.