# Thread: Equation for Tangent Plane

1. ## Equation for Tangent Plane

2. Find an equation for the plane tangent to the torus (donut)

X(s,t) = (5+2*cost)*coss, (5+2*cost)*sins, 2*sint

at the point (5-sqrt(3))/sqrt(2), (5-sqrt(3))/sqrt(2), 1

I'm trying to get a head start on the lecture Monday. I'm still trying to figure out how to get ya'll some of the tuition that I've paid, your doing a great job

As always I thank you for taking your time and helping

2. Am I going in the right direction?

Because Z = 2*sint and Z = 1 then t = pi/6 or 5pi/6

now X and Y both equal (5-sqrt(3))/sqrt(2) so going back to X(s,t) I can see that sins and coss need to equal each other. This happens at pi/4 or 5pi/4

So s and t are pi/4 and 5pi/6

Am I headed in the right direction???????????????????

3. I was thinking about this question and I have never seen how one can draw tangent plane to a differenciable surface defined parametrically (like here).

My suggestion is to try to "eliminate the parameter" and express $f(x,y,z)=C$. Meaning remove $s,t$ so you end with some function of $x,y,z$. Henceforth, you can use the property of the gradient.

4. Yes, you're going in the right direction. Your values for s and t are correct.
The equation of the plane tangent to X at the point P(a,b,c) is given by:

$\vec S = \vec p + \lambda \left. {\frac{{\partial \vec X}}{{\partial s}}} \right|_{\vec p} + \mu \left. {\frac{{\partial \vec X}}{{\partial t}}} \right|_{\vec p}$

Here, lambda and mu are scalars, the partial derivatives which have to be evaluated in p are the directional vectors.
This works as long as both directional vectors are non-zero and not parallel, because then they span the tangent plane.

5. I still dont understand the steps we need to take to find the equation for the tangent plane...?!

6. Now that we now s, t here is what we need to do.

We need the derivative for X(s,t) with respect to s and with respect to t

Once we do that, evauate at the points s and t

Ts and Tt (below) are already evauated at s and t (please excuse any bad notation)

7. Okay, so apparently I'm loosing my mind & cant remember what to do after taking the cross product of T(s) and T(t) with (x,y,z).

I came out with:
-sqrt(6)*(5-sqrt(3))/2 X + sqrt(6)*(sqrt(3)-5)/2 Y + (5-sqrt(3)) Z

If you can remember , what did you do next? (I know I cant end up with a scalar because the problem asks for an equation.)

8. @ Ranger SVO: your directional vectors are correct, now fill in the formula for the tangent plane I gave.

@ AbbyNormal: where are you stuck, what's the problem? Do you understand the equation of the tangen plane?

9. No, I dont understand the equation of the tangent plane.
Ranger told me to get it in terms of (x,y,z) through the cross product, but I dont understand what I need to do with that / why it needs to be in cartesian coordinates.

10. You don't have to switch to cartesian coordinates, if you use the equation I gave you can use the parametric equation.
The tangent plane is spanned by two (linearly independent) tangent vectors, the directional vectors.
These can be found as the partial derivatives of the surface with respect to the parameters s and t.
Evaluating these partial derivatives at the given (s,t) yields the directional vectors, now just plug it in the equation.