Suppose I've a function $f(x,y,z)$ and a helix. There are parameterised points on the helix (using the parameter $t$) such that-
$\:\:\:\:\:\:\:\:\:$ $x=cos(t)$ , $y=sin(t)$ , $z=t$
Now, the partial derivatives of $f$ at the given points are given by-
$\:\:\:\:\:\:\:\:\:$ $f_x=cos(t)$ , $f_y=sin(t)$ , $f_z=t^2$
So, at what points on the helix can $f$ take an extreme value?
My approach:
If the helix is represented by the curve $c(t)$, then $f(x,y,z)$ must be lying on $c(t)$. As a result, $f(cos(t),sin(t),t)$ must satisfy the helix $c(t)$, where each input in $f$ is a function of $t$. In that case, $f'(t)=\nabla f c'(t)$, thus fetching me $t^2$ as the output of $f'(t)$. For an extrema, $f'(t)=0$ is valid, so setting $t^2$ equal to zero, I get the required points as $(1,0,0)$.
My doubt:
In reality, $f(x,y,z)$ doesn't have an extremum. But, how is this justifiable, when I do get a value of $t(=0)$ corresponding to an extreme value of $f$? Is it somewhat, an inflection point at $(1,0,0)$? Can someone help me out in this with a lucid explanation? Thanks.
Your situation involves a function $(x,y,z)\mapsto f(x,y,z)$ and a helix $t\mapsto {\bf c}(t):=(\cos t,\sin t,t)$ in ${\mathbb R}^3$. You are told to study the composed function $$g(t):=f\bigl({\bf c}(t)\bigr)\ .$$ You can view $f(x,y,z)$ as temperature at the point $(x,y,z)$ and $t\mapsto{\bf c}(t)$ as orbit of a space craft. The function $t\mapsto g(t)$ is then the temperature measured on the space craft.
By the chain rule and information given about $f$ one has $$g'(t)=\nabla f\bigl({\bf c}(t)\bigr)\cdot{\bf c}'(t)=(\cos t,\sin t, t^2)(-\sin t,\cos t,1)=t^2\ .$$ This shows that $g'(t)\geq0$ for all $t$, and $=0$ only at $t=0$. It follows that $g$ is monotonically increasing. In fact $g(t)={1\over3}t^3+C$ for some constant $C$. When you draw the graph of $t$ you have an inflection point at $t=0$.
By the way: The function $f(x,y,z):={1\over2}(x^2+y^2)+{1\over3}z^3$ is an example for an $f$ described in the story.