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

• Oct 28th 2009, 09:58 AM
Chris9620
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?
• Oct 28th 2009, 12:33 PM
Bruno J.
Welcome to the forum!

Consider the function

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

which has partial derivatives

\$\displaystyle 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 \$\displaystyle (1,2,2)\$ :

\$\displaystyle 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) \$

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

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

\$\displaystyle 2(x-1)+4(y-2)+4(z-2)=0\$
or
\$\displaystyle x+2y+2z=9.\$
• Oct 28th 2009, 01:30 PM
Chris9620
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
http://www.mathhelpforum.com/math-he...e3053087-1.gif

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!
• Oct 28th 2009, 05:26 PM
Bruno J.
You are right - there are two points \$\displaystyle z\$ corresponding to those values of \$\displaystyle x,y\$ (namely \$\displaystyle z=2\$ and \$\displaystyle z=-2\$). In that sense, the problem is unclear.

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