Normal Vector of a Surface

Question

\begin{align} \mbox{How to find the normal vector of on}\quad & z = \left(25 - x^{2} - y^{2}\right)^{1/2}\quad \mbox{at the point}\quad \left(4,-3, 0\right)\ \mbox{?}. \\[3mm] \mbox{The normal vector to the plane would be}\quad & n = \left({\partial z \over \partial x},{\partial z \over \partial y}, -1\right) \end{align} But taking both partial derivatives and evaluating at the point gives division by zero ( the point is on the $xy$ plane circle trace ).

Answer 1

Would it be correct to find the tangent line of x^2+ y^2 = 25 at (4, -3) and it'd would be the plane in 3 space of the surface.

Answer 2

Why not just look at it as half of the surface of a sphere? http://www.wolframalpha.com/input/?i=z+%3D+(25-x%5E2-y%5E2)%5E1%2F2

Namely, since it's defined as z = (...)^1/2, you're only looking at the upper-z-half of the surface of a sphere of radius 5 (in this case). Picture what's going on - if we take the x and y coordinates of 4 and -3, and a slightly small value in z. If we do that, you can see that the normal should have a z-coordinate of 0 right? In this case we can't use the canned formula you just gave for the normal vector of a surface in the form $z = f(x,y)$.

One way to approach this is to look at it as already being solved in "z". We know the z-coordinate of the normal vector is going to be zero, what are the other coordinates? If we let z = 0 (which we get from the point (4,-3,0) in the equation for the surface, we get: $25 = x^2 + y^2$. The derivative is $\frac{dy}{dx} = -x/y $ which we want to know at the point (4,-3). But the derivative is tangent to the circle, so we actually want the negative reciprocal to make sure we get the normal to the surface instead of the tangent. Normal = $y/x$. At (4,-3) it's: -3/4.

How do we interpret this? For every positive step in x, y changes by -4/3. So we need the equation of the line that goes through the point (4,-3) with slope -3/4. The formula we need is $y-y_1 = m (x - x_1)$. We get the line $y+3 = -\frac{3}{4}(x-4)$. Plot them with the circle. Yep, this goes through the centre, so it should be the normal vector.

Now you're at the edge of the surface, and I'm not actually sure if the normal vector is really defined at the edge of a surface. But it exists as a limit for sure, even if it doesn't exist at that point. http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D25,+y%2B3+%3D+-3%2F4(x-4)

Picture embedding that in the 3D image from above. The normal vector at the point (4,-3,0) is: n = (1,-3/4,0).

Answer 3

First convert the surface spherical coordinates: Let $$x=5\cos(\theta)\cos(\phi),y=5\cos(\theta)\sin(\phi),z=5\sin(\theta)$$ and suppose $$r(\theta,\phi)=(x,y,z)$$ Verify that $||r||^2=x^2+y^2+z^2=25$, with the given parameterizations. Now determine $\theta,\phi$ such that $r=(4,-3,0)$. From the third coordinate, it is easy to see $\theta=0$. You can also find $\phi=\arctan(\frac{-3}{4})$.

Now we compute the normal vector by taking the cross product: $$\frac{\partial r}{\partial \theta} \times\frac{\partial r}{\partial \phi}$$

Upon computation, we see this product is $$\left(-25(\cos(\theta))^2 \cos(\phi),-25(\cos(\theta))^2 \sin(\phi),-25\cos(\theta) \sin(\theta)\right)$$

Plug in the computed $\theta=0$ and $\phi=\arctan(\frac{-3}{4})$ to get the normal vector to be $(-20,15,0)$. Normalize this vector to get the unit normal vector: $\left(\frac{-4}{5},\frac{3}{5},0 \right)$