Equation of an osculating plane to a curve (2 variables) Maybe I am just reading this wrong, but here goes:
Find the equation of the osculating plane to the curve
\begin{cases} x = a\cos(\theta) \\
    y = a\sin(\theta) \\ 
    z = a(1+\sin(\theta)) \end{cases}
For $\theta = \pi/4$.
I completely understand how to do this with a single variable, but that "$a$" throws me off.
 A: That $a$ is just a constant. You can think of it as the number... whatever your favorite number is, as long as you never simplify anything involving it. The osculating plane is the plane spanned by ${\bf T}(\theta)$ (tangent vector) and ${\bf N}(\theta)$ (normal vector). Hence the normal direction to the plane is given by ${\bf B}(\theta)$ (binormal vector), which happens to be: $${\bf B}(\theta) = \frac{{\bf r}'(\theta)\times {\bf r}''(\theta)}{\|{\bf r}'(\theta)\times {\bf r}''(\theta)\|},$$ where ${\bf r}(\theta) = (x(\theta),y(\theta),z(\theta))$ is given in the question. Since we only care about direction, and not magnitude, we can take as the normal to the plane the vector ${\bf r}'(\pi/4)\times{\bf r}''(\pi/4)$, or any convenient multiple of it. Directly: $${\bf r}'(\pi/4) = \left(-\frac{a\sqrt{2}}{2}, \frac{a\sqrt{2}}{2}, \frac{a\sqrt{2}}{2}\right) = \frac{a\sqrt{2}}{2}(-1,1,1), \quad {\bf r}''(\pi/4) = \left(-\frac{a\sqrt{2}}{2}, -\frac{a\sqrt{2}}{2}, -\frac{a\sqrt{2}}{2}\right) = \frac{-a\sqrt{2}}{2}(1,1,1)$$
Hence our normal vector is in the same direction that $$(-1,1,1)\times(1,1,1) = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ -1 & 1 & 1 \\  1 & 1 & 1\end{vmatrix} = (0,2,-2)= 2(0,1,-1)$$ So we can take $(0,1,-1)$ as the normal. The plane is $y-z = c$, where $c$ is a constant that you can discover using that ${\bf r}(\pi/4)$ must satisfy the plane equation.
