The standard helix was parameterized in the lecture: $[cost,sint,t{]}^T.$

$($a$,$ b$)$ Create a function hinR$R^3$ whose path will be an helix that rises exactly $6$ units as it turns one full revolution. $($Hint: you want this new helix to be at the point $[1,0,0{]}^T$ at time $t=0$ and at the point $[1,0,6{]}^T$ after one revolution.$)$ By simply using a felicitous placement of a scalar or two in the parameterization of the standard helix, you can create this "special" helix. $($Muttiple right answers are possible.$)$

$h(t),=$

$($b$)$ If you got problem $($a$)$ right, then find the function's first and second derivative. If you did not get it right, still answer and hope for partial credit:

$h^{\text{'}}(t)=$

$h^{\text{'}\text{'}}(t)=$

$($c$,$ d$)$ Find the length of one turn of this special helix. Two methods of solution are possible: $($a$)$ using "brute force" with your function $h($it$\text{'}$s not too hard$),$ and $($b$)$ using a clever argument in the manner of the solution of a competition problem. In any case, get me a good, satisfying answer. Only half credit will be awarded for an answer that is an integral that has not been evaluated.

Arclength of one turn $=$