Multi variable calculus, equation tangent to plane question

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!

• 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 $3$. – John Omielan Mar 9 at 8:06

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.
• Yes, the tangent plane at the graph of $f$ at $(1,1,3)$ is $z = 2x + 2y - 1$ (do not use $f(x,y,z)$). – Robert Z Mar 9 at 8:36