1. ## Taylor Expansion x^2+y^2+z^2=9

Hi!

I have a calculus exam tomorrow, and one of the questions in the 'Series Expansion for Functions of Two Variables' is confusing me a little...

"A Sphere is defined by the equation x^2 +y^2+z^2=9 Determine the equation of the tangent plane on its surface when x=1, and y=2"

I assumed I had to jiggle the equation, so i have z=-x-y+3, and start my partial differentiations from there, but my numbers end up silly, and the answer is way wrong when I do that.

Any help?

2. Welcome to the forum!

Consider the function

$f(x,y,z)=x^2+y^2+z^2-9$

which has partial derivatives

$f_x(x,y,z)=2x,\ f_y(x,y,z)=2y,\ f_z(x,y,z)=2z$

So if we consider its first Taylor polynomial about the point $(1,2,2)$ :

$P(x,y,z)=f(1,2,2)+f_x(1,2,2)(x-1)+f_y(1,2,2)(y-2)+f_z(1,2,2)(z-2)$

$= 2(x-1)+4(y-2)+4(z-2)$

Now notice that your sphere is the solution space of $f(x,y,z)=0$; and the tangent plane at the point $(1,2,2)$ will be the solution space of $P(x,y,z)=0$, i.e.

$2(x-1)+4(y-2)+4(z-2)=0$
or
$x+2y+2z=9.$

3. Thanks for the help!

Just a couple of questions, what made you choose the point z=2? As the question only included (x,y), as in (1,2), not (1,2,2).

Also, at the end, where did you get the 9 from? you went from
2(x-1)+4(y-2)+4(z-2)=0
to

sorry if I'm being blind, its just I have to say this in front of a panel tomorrow! best to know what I'm talking about!

4. You are right - there are two points $z$ corresponding to those values of $x,y$ (namely $z=2$ and $z=-2$). In that sense, the problem is unclear.

As for how I went from $2(x-1)+4(y-2)+4(z-2)=0$ to $x+2y+2z=9$... I just rearranged the equation. Divide everything by two, and put the constants on the right hand side.