I have parametrized the surface of a 3d-cardioid. The parametrization is:

$$x(t,s)=(1+\cos(t))\cos(t) $$ $$y(t,s)=(1+\cos(t))\sin(t)\cos(s)$$ $$z(t,s)=(1+\cos(t))\sin(t)\sin(s)$$

Now I want to calculate the equation of the tangent plane in $(s,t)=(\frac{1}{2}(1+\sqrt{2}),\frac{1}{2}(1+\sqrt{2}))$.

I think a correct method would be calculating the vectors $$\frac{\partial}{\partial s}(x(s,t)i+y(s,t)j+z(s,t)k)$$ $$\frac{\partial}{\partial t}(x(s,t)i+y(s,t)j+z(s,t)k)$$

Now the cross product of these two vectors will be the normal vector of the tangent plane to the surface. Finally plugging the values of $(\frac{1}{2}(1+\sqrt{2}),\frac{1}{2}(1+\sqrt{2}))$ into the parametric equations I will have the tangent point.

Is this method correct? Is there another method to calculate the tangent plane? Calculating the cross vector in this case is very tedious.

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    $\begingroup$ Yes this is the method for calculating the tangent plane, I'm certain this is essentially the way to do it $\endgroup$ – Triatticus Jan 4 '18 at 21:48
  • $\begingroup$ If you were somehow able to re-express the surface into the form $F(x,y,z)=C$, then you can take the normal vector as $\nabla F$. $\endgroup$ – Dylan Jan 4 '18 at 22:03
  • $\begingroup$ Based on your other question it looks like you’re generating this surface by rotating a plane cardioid. It will be a lot less tedious to find the tangent plane to a point in the $x$-$y$ plane (i.e., $s=0$) and then rotate it into place. $\endgroup$ – amd Jan 5 '18 at 0:57
  • $\begingroup$ @amd As you pointed, I have calculated the tangent plane at the $x-y$ plane, and it is $-0.4x-1.58y=-2.16$, to do so I calculated it finding a point of the curve where $t=\frac{1}{2}(1+\sqrt{2})$ which gives the point $(0.47,1.25,0)$. And using the cross product of the vectors $(0,0,1)$, $(\frac{dx}{dt},\frac{dy}{dt},0)$ I get the normal vector. How do I rotate this plane? $\endgroup$ – John Keeper Jan 5 '18 at 1:23
  • $\begingroup$ Rotate its normal about the axis of rotation that you’re using to generate the surface. $\endgroup$ – amd Jan 5 '18 at 1:26

You could try converting the parametric equations into an implicit equation for the surface and then computing that function’s gradient, but since you’re working with a surface of revolution, it might be easier to compute the normal for the original plane curve and then rotate it into place.

This surface is generated by rotating the cardioid $x(t)=(1+\cos t)\cos t$, $y(t)=(1+\cos t)\sin t$, $z=0$ about the $x$-axis. This rotation is represented by the parameter $s$, and this curve corresponds to $s=0$. A normal to this curve is $$\left( {\partial y\over\partial t}, -{\partial x\over\partial t}, 0 \right) = \left( \cos t+\cos 2t, \sin t+\sin 2t, 0 \right).$$ When the curve is rotated, it drags its normals along with it, so to find the normal at any point on the surface, we need only rotate this normal through an angle of $s$: $$\begin{bmatrix}1&0&0 \\ 0&\cos s&-\sin s\\0&\sin s&\cos s\end{bmatrix} \begin{bmatrix} \cos t+\cos 2t \\ \sin t+\sin 2t \\ 0 \end{bmatrix} = \begin{bmatrix} \cos t+\cos 2t \\ (\sin t+\sin 2t)\cos s \\ (\sin t+\sin 2t)\sin s \end{bmatrix}.$$ The equation of the tangent plane can then be found in the usual ways. It is approximately $$0.391195x+0.568984y+1.49488z=1.83799$$

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You have $$ x^2 + y^2 + z^2 = (1+\cos t)^2 $$ $$ \frac{x^2}{x^2 + y^2 + z^2} = \cos^2 t $$

Therefore $$ \sqrt{x^2 + y^2 + z^2} = 1 + \frac{x}{\sqrt{x^2+y^2+z^2}} $$ $$ x^2+y^2+z^2 - \sqrt{x^2+y^2+z^2} - x= 0 $$

Let $$ F(x,y,z) = x^2+y^2+z^2 - \sqrt{x^2+y^2+z^2} - x $$

Then $$ \frac{\partial F}{\partial x} = 2x - \frac{x}{\sqrt{x^2+y^2+z^2}}-1 $$ $$ \frac{\partial F}{\partial y} = 2y - \frac{y}{\sqrt{x^2+y^2+z^2}} $$ $$ \frac{\partial F}{\partial z} = 2z - \frac{z}{\sqrt{x^2+y^2+z^2}} $$

You can now compute $\nabla F$ for the corresponding $(x,y,z)$


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