a. We don't really need calculus to do this
[tex]z = f(x,y) = x^2 + y^2[/tex]
We're interested in f near (1,1,f(1,1))
[tex]f(x,y)=f(1+(x-1), 1+(y-1)) = (1+(x-1))^2 + (1+(y-1))^2[/tex]
[tex]f(x,y)=2 + 2(x-1) + 2(y-1) + (x-1)^2 + (y-1)^2[/tex]
The tangent plane at (1,1,2) is the best linear approximation to f at (1,1). We just drop the squared terms:
[tex]z =2 + 2(x-1) + 2(y-1)[/tex]
[tex]z=2x + 2y - 2[/tex]
Answer: 2x + 2y - z = 2
b.
We repeat the above in general, near f(r,s)
[tex]f(x,y)=f(r + (x-r), s+(x-s))[/tex]
[tex]f(x,y) = (r + (x-r))^2 + (s + (x-s))^2[/tex]
[tex]f(x,y)= r^2 + s^2 + 2r(x-r) + 2s(y-s) + (x-r)^2 + (y-s)^2[/tex]
Again the tangent plane at (r,s,f(r,s)) is gotten by dropping the squared terms,
[tex]z = r^2 + s^2 + 2r(x-r) + 2s(y-s)[/tex]
This has to contain
[tex](x,y,z)=(t, 2-2t, -1)=(0,2,-1) + t(1,-2,0)[/tex]
[tex]-1 = r^2 + s^2 + 2r(0-r) + 2s(2-s)[/tex]
[tex]r^2 + s^2 - 4s - 1 = 0[/tex]
The line is perpendicular to the normal of the plane, so a zero dot product.
[tex]2r x + 2sy - z = r^2 + s^2[/tex]
[tex](2r, 2s, -1) \cdot (1, -2, 0) = 0[/tex]
[tex]2r - 4s=0[/tex]
[tex]r = 2s[/tex]
[tex](2s)^2 + s^2 - 4s -1=0[/tex]
[tex]5s^2 - 4s -1 = 0[/tex]
[tex](5s + 1)(s - 1) = 0[/tex]
[tex]s=1 \textrm{ or } s=-1/5[/tex]
[tex]r=2, s=1 \textrm{ or } r=-2/5, s=-1/5[/tex]
Two tangent planes contain the line.
[tex]2r x + 2sy - z = r^2 + s^2[/tex]
[tex]4x + 2y - z = 5 \textrm{ and } (-4/5)x - (2/5)y - z = (-2/5)^2+(-1/5)^2[/tex]
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.
