Differential Geometry review questions. Need help I have a final coming up in Differential Geometry and we got a review worksheet and I am having serious trouble with two problems. I'm still chugging along at them but I need help understanding. I know we learned the 2nd problem (#5) while I was sick last week, so I'm deep in my book trying to comprehend what i missed. Any help would be appreciated!





I have been working on 5 and I have come up with the following, which makes it seem way easier than it should be so I'm probably wrong

plug in t=0 which leads to both being (0,0). I end up using the equation cosx=F/sqrt(EG) and plugging in the values for the first fundamental form i get pi/2, which i remember reading if F=0 then all curves are orthogonal at their intersections. Am i right about any of this?

I'm stuck on 2, but i Haven't put much effort into it yet compared to 5.
The counter example I have in mind is a curve you built in two steps. Consider a smooth curve $\alpha$ defined on $[0,5]$ that satisfies $$\alpha(t)=\begin{cases} (t,0,0) &amp; \text{for }t\in[0,1] \\(2,1+t,0) &amp; \text{for }t\in[2,3] \\(2,3,1+t) &amp; \text{for }t\in[4,5] \end{cases}$$
and has $\alpha([0,3])\subset\lbrace z=0\rbrace$ and $\alpha([2,5])\subset\lbrace x=2\rbrace$. Smooth curves satisfying these constraints can be constructed, one could even give explicit formulas using integrals of smooth bump functions for the parts of the curve that I have not described.

Such a regular curve (in your sense) will satisfy $(\alpha'\times\alpha'')\cdot\alpha'''\equiv0$ yet is manifestly not planar. The reason that quantity is constant equal to $0$ is that for every point $x\in[0,5]$, there is an open neighborhood $V$ of $x$ with $\alpha(V)\subset\mathcal{P}$ where $\mathcal{P}$ is one of the planes $\lbrace z=0\rbrace$ and $\lbrace x=2\rbrace$. The problem should disappear if we impose that at every point $x$ of the domain of $\alpha$, the first and second derivative $\alpha'(x)$ and $\alpha''(x)$ be linearly independent.
The counter example I have in mind is a curve you built in two steps. Consider a smooth curve $\alpha$ defined on $[0,5]$ that satisfies $$\alpha(t)=\begin{cases} (t,0,0) &amp; \text{for }t\in[0,1] \\(2,1+t,0) &amp; \text{for }t\in[2,3] \\(2,3,1+t) &amp; \text{for }t\in[4,5] \end{cases}$$
and has $\alpha([0,3])\subset\lbrace z=0\rbrace$ and $\alpha([2,5])\subset\lbrace x=2\rbrace$. Smooth curves satisfying these constraints can be constructed, one could even give explicit formulas using integrals of smooth bump functions for the parts of the curve that I have not described.

Such a regular curve (in your sense) will satisfy $(\alpha'\times\alpha'')\cdot\alpha'''\equiv0$ yet is manifestly not planar. The reason that quantity is constant equal to $0$ is that for every point $x\in[0,5]$, there is an open neighborhood $V$ of $x$ with $\alpha(V)\subset\mathcal{P}$ where $\mathcal{P}$ is one of the planes $\lbrace z=0\rbrace$ and $\lbrace x=2\rbrace$. The problem should disappear if we impose that at every point $x$ of the domain of $\alpha$, the first and second derivative $\alpha'(x)$ and $\alpha''(x)$ be linearly independent.