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I'm doing the following question relating to parametrising surfaces and finding the tangent plane,

Question:

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I parametrised the surface in term of sphereical coordinates, such that $$x=\sin\theta \cos\phi, y = \frac{1}{\sqrt{2}}\sin\theta \sin\phi, z=\cos\theta$$ Using this information I found the normal vector, $\vec{n}$, to be: $$\vec{n}=\left(\frac{\sqrt{2}}{2}\sin^2 \theta \cos\phi\right)\tilde{i}+\left(\sin^2 \theta \sin\phi\right)\tilde{j}+\left(\frac{\sqrt{2}}{2}\sin\theta \cos\theta\right)\tilde{k}$$

From here I used the given point $(\frac{1}{\sqrt{2}},\frac{1}{2},0)$ to find $\theta=\frac{\pi}{2}$ and $\phi=\frac{\pi}{4}$

Which then gave me, $$\vec{n}=(\frac{1}{2},\frac{\sqrt{2}}{2}, 0) $$

Using the relation,$$(x -x_0, y-y_0,z-z_0)\dot{}\vec{n}$$

For part (a), I found the tangent plane to be:$$x + \sqrt{2}y - \sqrt{2}=0$$

But when I try to verify that $c'(\frac{\pi}{4})$ lies in the tangent plane I can never seem to verify it, always getting:$$-\sqrt{2} \neq0$$

Have I done the necessary steps to determine what the problem statement is asking? Could I have made a mistake along the way? Or could there possibly be something wrong with the problem statement?

I can provide more working if needed.

Any help would be greatly appreciated as I have no clue what else I can do. Thanks!

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    $\begingroup$ The tangent vector $c'(t)$ will always be "based at the origin". By "lying in the plane", the question means that the vector $c(t) + c'(t)$ lies in the plane. Since $c(\pi/4)$ definitely lies in the plane, this is equivalent to verifying that $c'(\pi/4)$ is perpendicular to the normal of the plane. $\endgroup$ – Joppy May 18 '17 at 12:57
  • $\begingroup$ Okay, so if it is perpendicular to the normal of the plane does that mean it lies on the surface? $\endgroup$ – Nathan Lowe May 18 '17 at 14:05
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Parametrization :

$x = sin(\theta)cos(\phi)$;

$y = (1/√2)sin(\theta)sin(\phi)$;

$z = cos(\theta)$.

Consider $\vec{r} = (x(\theta,\phi), y(\theta, \phi), z(\theta))$.

1)Tangent vector $\vec{t} $: $\frac{\partial \vec{r}}{\partial \theta}$ and

2) Tangent vector $\vec{s}$: $\frac{\partial \vec{r}}{\partial \phi}$.

$\vec{t} =$

$(cos(\theta)cos(\phi),(1/√2)cos(\theta)sin(\phi), -sin(\theta));$

$\vec{s} =$

$(-sin(\theta)sin(\phi),(1/√2)sin(\theta)cos(\phi),0).$

Note: vectors $t,s$ are independent (not colinear), I.e. they span a plane.

$P: (1/√2,1/2,0)$ corresponds to $\theta = π/2,$ $\phi = π/4$.

Tangent plane at $P: \vec{u} = (1/√2,1/2,0) + a \vec{t} + b \vec{s}$, $a,b \in \mathbb{R}$.

$\vec{t} = (0,0,-1)$; and $\vec{s} = (-1/√2,(1/√2)(1/√2),0)$.

Check:

Normal vector to

$f(x,y,z) = x^2 + 2y^2 + z^2 -1 =0$ is $N = \nabla f(x,y,z)$.

$N = (2x,4y,2z)$ at $P(1/√2,1/2,0)$.

$N = (2/√2,2,0)$.

$N\cdot \vec{t} =0$; and $N\cdot\vec{s} =0$.

Part b) of the problem:

Surface: $x^2 + 2y^2 + z^2 = 1$.

$c(t) = (cos(t), (1/√2)sin(t),0)$ lies on the surface,

$[cos(t)]^2 + 2 (1/2) [sin(t)]^2 = 1$, satisfies eq. of the surface.

$c'(π/4) = $

$((-1/2)√2,(1/√2)(1/2)√2,0) =$

$((-1/2)√2,1/2,0)$.

This is our $\vec{s}$ at $P$, hence $c'(π/4)$ lies in the tangent plane.

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  • $\begingroup$ I don't understand how you got that normal vector? It looks like you just took the grad of S $\endgroup$ – Nathan Lowe May 19 '17 at 9:09

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