1617-257/TUT-R-7
On 10/28/16, we discussed the following problem:
Suppose [math]\displaystyle{ F : \mathbb{R}^2 \to \mathbb{R}^2 }[/math] is [math]\displaystyle{ C^1 }[/math] with [math]\displaystyle{ F(t^4, e^{t^2}) = (4,5) }[/math] for all [math]\displaystyle{ t \in \mathbb{R} }[/math]. Prove that [math]\displaystyle{ DF(0,1) }[/math] is not invertible.
Let [math]\displaystyle{ \gamma(t) := (t^4, e^{t^2}) }[/math]. At the end of the tutorial, a couple of students pointed out that it's also true that [math]\displaystyle{ DF(\gamma(t)) }[/math] is not invertible for any [math]\displaystyle{ t \in \mathbb{R} }[/math]. It's easy to check that [math]\displaystyle{ \gamma'(t) \neq 0 }[/math] when [math]\displaystyle{ t \neq 0 }[/math] and that [math]\displaystyle{ \gamma'(t) }[/math] is in the kernel of [math]\displaystyle{ DF(\gamma(t)) }[/math].
It takes some more care to prove what the problem is asking because one can't immediately deduce that [math]\displaystyle{ DF(\gamma(t)) }[/math] has non-trivial kernel with [math]\displaystyle{ \gamma'(0) = 0 }[/math]. A key observation here is that since [math]\displaystyle{ F }[/math] is [math]\displaystyle{ C^1 }[/math], invertibility of [math]\displaystyle{ DF }[/math] at [math]\displaystyle{ \gamma(0) }[/math] implies invertibility of [math]\displaystyle{ DF }[/math] at a neighborhood of [math]\displaystyle{ \gamma(0) }[/math].
