Tangent plane and Parametrized Surface

Question

"Given a sphere of radius $2$ centered at the origin, find the equation for the plane that is tangent to it at the point $(1,1,\sqrt[]{2})$ by considering the sphere as a surface parametrzed by $\Phi(\theta, \phi) = (2\cos(\theta)\sin(\phi), 2\sin(\theta)\sin(\phi), 2\cos(\phi))$"

I'm not really sure what I'm supposed to do here. The question is asking for a plane tangent to a point, but why do we care about the radius and parametrization of the sphere? Couldn't I find a tangent plane to the point without that information?

Answer 1

Regarding to what we know about the Curvilinear coordinates try to show that the Gradient vector can be described as follows in spherical coordinates:

$$\nabla\Phi=\left(\frac{\partial\Phi}{\partial r},\frac{1}r\frac{\partial\Phi}{\partial \theta},\frac{1}{r\sin\theta}\frac{\partial\Phi}{\partial \phi}\right).$$

Answer 2

As pointed out by Muphrid his/her answer, we are looking for the equation of the plane which is tangent to the sphere of radius $2$ centered at $(0, 0, 0)$.

This thing being said:

The first thing we need "to do here" is figure out which parameter values $\theta, \phi$ map to the point $(1, 1, \sqrt 2)$ under $\Phi(\theta, \phi) = (2(\cos \theta)(\sin \phi), 2(\sin \theta)(\sin \phi), 2(\cos \phi))$; but that's easy if we assume that $\phi, \theta$ correspond to standard spherical coordinates (with which they are incidentally at least consistent), so that $0 \le \phi \le \pi$ and $0 \le \theta \le 2 \pi$, and observe that we must have $2 \cos \phi = \sqrt 2$, or $\cos \phi = \sqrt 2 / 2$; thus $\phi = \pi / 4$ and $\sin\phi = \sqrt 2 / 2$ as well; knowing this, it follows from $2(\cos \theta)(\sin \phi) = 1$ and $ 2(\sin \theta)(\sin \phi) = 1$ that $\sin \theta = \cos \theta = \sqrt 2 / 2$ and so in fact we also have $\theta \ = \pi / 4$. Having $\phi$ and $\theta$ at our disposal, we compute the surface normal at $\Phi(\pi / 4, \pi / 4)$ as follows: the function $\Phi(\theta, \phi) = (2(\cos \theta)(\sin \phi), 2(\sin \theta)(\sin \phi), 2\cos \phi)$, when considered in each variable $\theta, \phi$ separately, describes two families of curves in its spherical image surface, viz. $\Phi(\theta_0, \phi)$ and $\Phi(\theta, \phi_0)$, where $\theta_0$ and $\phi_0$ are held constant along the curves while $\phi, \theta$, respectively, are allowed to vary individually. We may compute the tangent vectors to these curves by taking $\Phi_\phi(\theta_0, \phi)$ and $\Phi_\theta(\phi, \theta_0)$, where the subscript denotes the corresponding partial derivative, e.g. $\Phi_\phi = \partial \Phi / \partial \phi$ etc. Thus we obtain:

$\Phi_\theta(\theta, \phi_0) = (-2(\sin \theta)(\sin \phi_0), 2(\cos \theta)(\sin \phi_0), 0) \tag{1}$

and

$\Phi_\phi(\theta_0, \phi) = (2(\cos \theta_0)(\cos \phi), 2(\sin \theta_0)(\cos \phi), -2\sin \phi); \tag{2}$

evaluating these at the computed parameter values $\phi, \theta = \pi / 4$ yields

$\Phi_\theta(\pi / 4, \pi / 4) = (-1, 1, 0) \tag{3}$

and

$\Phi_\phi(\pi / 4, \pi / 4) = (1, 1, -\sqrt 2). \tag{4}$

(3) and (4) are tangent vectors to the surface $\Phi(\theta, \phi)$ at the point $\Phi(\pi / 4, \pi / 4)$; we obtain a normal vector $\mathbf N$ there by taking their vector cross product:

$\mathbf N = \Phi_\phi(\pi / 4, \pi / 4) \times \Phi_\theta(\pi / 4, \pi / 4) = (\sqrt 2, \sqrt 2, 2). \tag{5}$

Having obtained the surface normal $\mathbf N$ at the point $(1, 1, \sqrt 2)$, we may write the equation of the tangent plane as

$((x, y, z) - (1, 1, \sqrt 2)) \cdot \mathbf N = 0, \tag{6}$

which with a little algebraic maneuvering may be transformed to the standard format

$\sqrt 2 x + \sqrt 2 y + 2 z - 4 \sqrt 2 = 0. \tag{7}$

Some might choose to normalize $\mathbf N$ before forming the equation for the tangent plane. Indeed, we have $\Vert \mathbf N \Vert = 2 \sqrt 2$, so if we set

$\mathbf n = (1 / \Vert \mathbf N \Vert)\mathbf N = (1 / 2, 1 / 2, \sqrt 2 / 2) \tag{8}$

the equation of the tangent plane becomes

$(1 / 2)x + (1 / 2)y + (\sqrt 2 / 2)z - 2 = 0. \tag{9}$

Having ground through the above derivation, I cannot help but note that by using the geometrical fact that the normal vector to a sphere is the same as its radius vector at any given point we can obtain the requisite normal vector without any real work at all: the normal at $(1, 1, \sqrt 2)$ is exactly that: $(1, 1, \sqrt 2)$. The equation for the tangent plane at $(1, 1, \sqrt 2)$ then follow immediately. Of course, it appears this question is an exercise, note I did not say homework exercise, in calculation the normal given a parametric patch. As such, the computational path taken above cannot really be avoided when answering.

In closing, I would like to add that I don't yet see how showing the gradient vector attains the given form in spherical coordinates, as espoused by B.S. in his answer, really makes things any easier, especially since (i.) calculating the gradient of $R^2 = x^2 +y^2 + z^2$ is so simple; and (ii.) the transformation to spherical coordinates is somewhat compuation intensive in and of itself. If someone cares to leave remarks with further explanation, I would appreciate it.

Answer 3

Lots of planes run through any given point. You're being asked for a plane tangent to a surface that happens to contain a particular point. That's why you care about the sphere's radius and parameterization. There are many spheres that contain the given point also; most of them have different tangent planes.

The partial derivatives of the parameterization function--that is, $\partial \Phi/\partial \theta$ and $\partial \Phi/\partial \phi$--form a basis for the plane tangent to the surface at a given point. The cross product of these coordinate basis vectors yields a normal vector to the surface, which you should be able to use to find an equation for the plane.