A particle moves along the curve defined by the parametric equations $x\left(t\right)=2t$ and $y\left(t\right)=36-t^2$ for time $0\le t\le 6$. A laser light on the particle points in the direction of motion and shines on the x-axis.
a) What is the velocity vector of the particle?
$s=<2t,\:36-t^2>$, so $v=<2,\:-2t>$
My answer: $v=<2,\:-2t>$
b) Write an equation of the line tangent to the graph at $\left(2t,\:36-t^2\right)$ in terms of t and x.
$\frac{dy}{dt}=-2t$ and $\frac{dx}{dt}=2$, so $\frac{dy}{dx}=\frac{-2t}{2}=-t$
My answer: $y-\left(36-t^2\right)=-t\left(x-2t\right)$ or $y=t^2-xt+36$
c) Express the x-coordinate of the point on the x-axis that the light hits as a function of t.
Would I set $y=t^2-xt+36$ to 0 in order to get $x=\frac{t^2+36}{t}$?
d) At what time t is the light moving along the x-axis with the slowest speed? Justify your answer.
I'm not sure what to do here?
From your answer to Part (c), the light beam's position (as time varies) along the $x$-axis is$$x=\frac{t^2+36}{t}.$$ So, its velocity along the $x$-axis is $$\frac{\mathrm dx}{\mathrm dt}.$$ So, its speed along the $x$-axis is $$\left|\frac{\mathrm dx}{\mathrm dt}\right|=\frac{\mathrm dx}{\mathrm dt}.$$ This speed is least when its derivative with respect to $t$ equals $0,$ i.e., when $$\frac{\mathrm d\left|\frac{\mathrm dx}{\mathrm dt}\right|}{\mathrm dt}=0,$$ i.e., when $\frac{\mathrm d^2x}{\mathrm dt^2},$ the light beam's acceleration along the $x$-axis, equals $0.$