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So i think i may have it right but now sure...please check and help me to see if i got it right thanks!

Question: $f(x,y) = 1 + x^2 + y^2$, find vector $v$ tangent to plane of graph at $f(1,1,3)$.

Answer: $f(x,y) = 1 + x^2 + y^2$ and $g(x,y,z) = 1 + x^2 + y^2 – z$

$g(x,y,z) = f(x,y) – z$

$0 = f(x,y) – z$

$z = f(x,y) $

$z = 1 + x^2 + y^2 $

$z = 1 + 1 + 1$ (sub in 2 points i know already $f(1,1,3)$ for $x$ and $y$).

$z = 3 $

$g(x,y,z) = 1 + x^2 + y^2 – 3z$

$g(x,y,z) = 1 + x^2 + y^2 –3z$

$\text{grad} g(x,y,z) = 1 + x^2 + y^2 – 3z$

$\text{grad} g(x,y,z) = (2x, 2y, –3) $

$(x,y,z)\cdot \text{grad} g(1,1,3) = (1,1,3))\cdot \text{grad} g(1,1,3) $

$(x,y,z)\cdot (2,2,-3) = (1,1,3))\cdot (2,2,-3) $

$2x+ 2y -3z = 2 + 2 – 9 $

$2x+ 2y -3z = -5 $

$-3z = -5 - 2x -2y $

$3z = 2x + 2y +5 $

Is this right? Thanks!

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  • $\begingroup$ You originally start with "g(x,y,z) = f(x,y) – z" which would simplify to $g(x,y,z) = 1 + x^2 + y^2 - z$, but then when you determine $z$ would be $3$ at this point, you add a factor of $3$ in front of $z$ to get "g(x,y,z) = 1 + x2 + y2 – 3z". This is why your $z$ coordinate for the gradient is off by a factor of $3$. $\endgroup$ – John Omielan Mar 9 at 8:06
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No, your gradient is not correct (and, in your statement, $f(1,1,3)$ is misleading since $f$ is a function of TWO variables).

Let $f(x,y) = 1 + x^2 + y^2$ then the gradient of $$g(x,y,z):=f(x,y)-z=1 + x^2 + y^2 – z$$ at $(1,1,f(1,1))=(1,1,3)$, is $$\left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},\frac{\partial g}{\partial z}\right)_{(1,1,3)}= (2x,2y,-1)_{(1,1,3)}=(2,2,-1).$$ Now choose a vector ${\bf v}$ such that the scalar product ${\bf v}\cdot (2,2,-1)$ is zero.

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  • $\begingroup$ Thanks, ok got it the answer was z = 2x + 2y - 1 the tangent to f(x,y,z) at (1,1,3) $\endgroup$ – Shaun Weinberg Mar 9 at 8:32
  • $\begingroup$ Yes, the tangent plane at the graph of $f$ at $(1,1,3)$ is $z = 2x + 2y - 1$ (do not use $f(x,y,z)$). $\endgroup$ – Robert Z Mar 9 at 8:36
  • $\begingroup$ @ShaunWeinberg BTW, since you are new here, please take a few moments for a quick tour math.stackexchange.com/tour $\endgroup$ – Robert Z Mar 9 at 8:38

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