A particle moves along the curve $x=\ln y$ with a constant speed of $4$ units per second. Find the normal scalar component of acceleration as a function of $x$.
Honestly, what I don't understand is how to make it into a vector function. I know how to proceed if I had it all in terms of $t$. I would derive the acceleration and use the formula for the normal scalar. Essentially what I have is the positions of the particle, how do I take it and make $\vec r(t)$? Is this how? $$\vec{r}(t)= x \hat i + \ln y \hat j$$
The easiest way to to convert between a curve that is given explicitly is to let $x$ or $y$ be $t$, and then continue from there. So, if we have the equation: $$x=\ln y$$ This is the same as: $$y = e^x$$
Now, we can re-write this as a system of equations: $$\begin{cases} x=t\\ y=e^t \end{cases}$$ This system can be turned into a vector-valued-function: $$\vec{r} = (t, e^t)$$
Or, in $\hat i\hat j\hat k$ form: $$\vec{r} = t\hat{i}+e^t\hat j$$
If I've skipped over a step somewhere, let me know with a comment below.