# Thread: Tangent plane, parametric equation of the normal

1. ## Tangent plane, parametric equation of the normal

I must calculate 2 tangent vectors (not parallel between themselves) to the surface $\displaystyle z$ in $\displaystyle (1,2,4)$. Then find the tangent plane's to the surface $\displaystyle z$ in $\displaystyle (1,2,4)$ and I must find the parametric equation of the normal to the plane in the point $\displaystyle (1,2,4)$.
$\displaystyle z=x^2y^2$.
My attempt : I've done everything but I know I'm wrong and I misunderstand something.
A vector tangent to the surface $\displaystyle z$ in $\displaystyle (1,2,4)$, $\displaystyle \bold {u_1}=(1,2,4)+(8,4,-1)=(9,6,3)$.
I multiplied $\displaystyle (8,4,-1)$ by $\displaystyle \cos \frac{\pi}{4}=\frac{\sqrt 2}{2}(8,4,-1)$ in order to find $\displaystyle \bold {u_2}=(1,2,4)+(4\sqrt 2 , 2 \sqrt 2 , -\frac{\sqrt 2}{2})$. Here I realize that these 2 vectors must span the tangent plane to $\displaystyle z$ in $\displaystyle (1,2,4)$ so I could do the cross product to get the normal to the plane, but I did it another way.
$\displaystyle (\bold x - \bold {x_0}) \cdot \nabla F(\bold {x_0})=0 \Leftrightarrow 8x+4y-z-15=0$. Which is the asked equation.

Here comes the big problem. According to my textbook (Calculus by Leithold), the implicit equation of the normal curve of the tangent plane in (1,2,4) can be found by the formula $\displaystyle \frac{x-x_0}{F_x(x_0,y_0,z_0)}=\frac{y-y_0}{F_y(x_0,y_0,z_0)}=\frac{z-z_0}{F_z(x_0,y_0,z_0)}$.
So I get $\displaystyle r(t)=(t,\frac{t+3}{2},\frac{t+31}{8})+(1,2,4)$.
Obviously this curve doesn't pass by $\displaystyle (1,2,4)$. What happens?!
I put $\displaystyle x=t$ and wrote $\displaystyle y$ and $\displaystyle z$ in function of $\displaystyle t$ in order to get $\displaystyle r(t)$.

Now that I think I could take the gradient of $\displaystyle F(x,y,z)=x^2y^2-z$ and evaluate it in $\displaystyle (1,2,4)$ as I did to get $\displaystyle \bold {u_1}$.
I reach $\displaystyle (8,4,-1)t+(1,2,4)$ and this time the curve pass by $\displaystyle (1,2,4)$. But why it doesn't work using Leithold's method?

Almost edit (I didn't post yet) I see my error! Actually I must not add the $\displaystyle (1,2,4)$ vector to $\displaystyle r(t)=(t,\frac{t+3}{2},\frac{t+31}{8})$. It works great!
Well, if it is not too much asked, are my answers correct? (For $\displaystyle \bold {u_1}$ and $\displaystyle \bold {u_2}$ I'm curious, especially for $\displaystyle \bold {u_2}$.)

2. You got the wrong equation of the tangent plane in $\displaystyle \left( {1,2,4} \right)$ to $\displaystyle z = {x^2}{y^2}$.

Should be

$\displaystyle z - {z_0} = {\left. {\frac{{\partial z}}{{\partial x}}} \right|_{\left( {{x_0},{y_0}} \right)}}\left( {x - {x_0}} \right) + {\left. {\frac{{\partial z}}{{\partial y}}} \right|_{\left( {{x_0},{y_0}} \right)}}\left( {y - {y_0}} \right).$

$\displaystyle \frac{{\partial z}}{{\partial x}} = 2x{y^2} \Rightarrow {\left. {\frac{{\partial z}}{{\partial x}}} \right|_{\left( {1,2} \right)}} = 8.$

$\displaystyle \frac{{\partial z}}{{\partial y}} = 2{x^2}y \Rightarrow {\left. {\frac{{\partial z}}{{\partial y}}} \right|_{\left( {1,2} \right)}} = 4.$

$\displaystyle z - 4 = {\text{8}}\left( {x - 1} \right) + 4\left( {y - 2} \right) \Leftrightarrow 8x + 4y - z - 12 = 0.$