Determine the cartesian equation of the tangent plane to M in p. Given the torus and given the point p $\in$ M corresponding to the parameters $s=\frac{\pi }{4}$ and $t=\frac{\pi }{3}$.
Determine the cartesian equation of the tangent plane to M in p.
$\begin{cases} x=\left(3+\sqrt{2}cos\left(s\right)\right)cos\left(t\right) \\  y=\left(3+\sqrt{2}cos\left(s\right)\right)sin\left(t\right) \\  z=\sqrt{2}sin\left(s\right) \end{cases}$
Could someone give me a hint or help me? I'm not sure if I firstly should go from the given parametric equation to a cartesian equation.
 A: First find the Cartesian equation for the torus $$F(x,y,z)=0$$ Second, find the point as $(x_1,y_1,z_1)$. Third Differentiate $F$ to find $$\nabla F=[\partial F/\partial x,\partial F/\partial y,\partial F/\partial z]$$ Third substitute $x=x_1,y=y_1,z=z_1$ in $\nabla F$ to find $v$. The tangent plane is $$v \bullet [x-x_1,y-y_1,z-z_1]=0$$
A: Since you have the parametric equation of the surface, which is
$ \mathbf{P}(t, s) = ( x(t,s), y(t,s), z(t,s) ) $
Then find $\dfrac{\partial \mathbf{P}}{\partial t} $ and $\dfrac{\partial \mathbf{P}}{\partial s } $ as follows
$ \dfrac{\partial \mathbf{P}}{\partial t} =( - (3 + \sqrt{2} \cos(s) ) \sin(t) , (3 + \sqrt{2} \cos(s)) \cos(t) , 0 )$
$ \dfrac{\partial \mathbf{P}}{\partial s } =( - \sqrt{2} \sin(s) \cos(t) , - \sqrt{2} \sin(s) \sin(t) , \sqrt{2} \cos(s) ) $
Evaluate $\mathbf{P},\dfrac{\partial \mathbf{P}}{\partial t},\dfrac{\partial \mathbf{P}}{\partial s }$  at $s = \dfrac{\pi}{4} $ and $ t = \dfrac{\pi}{3} $, you will get
$ \mathbf{P} = (2, 2 \sqrt{3}, 1 ) $
$ \dfrac{\partial \mathbf{P}}{\partial t} = (- 2 \sqrt{3}, 2 , 0 ) $
$ \dfrac{\partial \mathbf{P}}{\partial s } = ( - \dfrac{1}{2} , -\dfrac{\sqrt{3}}{2} , 1) $
Next, find $\mathbf{N}$ (the normal vector to the plane)
$ \mathbf{N} = \dfrac{\partial \mathbf{P}}{\partial t} \times \dfrac{\partial \mathbf{P}}{\partial s } = (-2 \sqrt{3}, 2, 0) \times (- \dfrac{1}{2}, - \dfrac{\sqrt{3}}{2}, 1 ) = (2 , 2 \sqrt{3} , 4)$
Now the equation of the tangent plane is
$  \mathbf{N} \cdot ( \mathbf{r} - \mathbf{P} ) = 0 $
where $ \mathbf{r} = (x, y, z) $
Hence, the equation of the plane is
$ (2, 2\sqrt{3}, 4) \cdot (x - 2 , y - 2 \sqrt{3}, z - 1) = 0 $
which reduces to
$ 2 x + 2 \sqrt{3} y + 4 z - 20 = 0 $
and further to,
$ x + \sqrt{3} y + 2 z - 10 = 0 $
And this is the cartesian equation of the tangent plane at $ s = \dfrac{\pi}{4}, t = \dfrac{\pi}{3} $
