Give an equation of the tangent plane to the paraboloid $z = x^2 +y^2$ at a point $(x_0, y_0, x^2_0 + y^2_0)$

I think I am on the right track but one thing confuses me. First I find the partial derivatives with respect to $x $ and $y$.

$Fx(a,b) = 2x$

$Fy(a,b) = 2y$

I plug in the values...

$Fx(x_0, y_0) = 2x_0$

$Fy(x_0, y_0) = 2y_0$

Now I am confused. I do not know what to do with $z$. This is a 3rd order function, am I supposed to find the partial derivative of $z$ as well?

I know I am supposed to plug in these values...

$z = 2x_0(x-x_0) + 2y_0(y-y_0)$

But I do not know what to do with $z$. I've never worked with 3rd order functions. Can someone clear this up for me?


You are almost there. Your equation will be $z-z_0=2x_0(x-x_0)+2y_0(y-y_0)$, $z_0=f(x_0,y_0)$.

  • $\begingroup$ so would $z_0$ be equal to $x^2_0 + y^2_0$? so $z - (x^2_0 + y^2_0) = ...$ $\endgroup$
    – buydadip
    Sep 14 '17 at 15:58
  • $\begingroup$ This is correct $\endgroup$
    – Vasya
    Sep 14 '17 at 16:00
  • $\begingroup$ Just to clarify a little bit: your tangent point should satisfy the equation of the tangent plane. The general equation of the plane containing $(x_0,y_0,z_0)$ will be $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$, that's where $z-z_0$ is coming from, $\endgroup$
    – Vasya
    Sep 14 '17 at 16:08

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