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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? $$\vec{r}(t)= x \hat i + \ln y \hat j$$

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    $\begingroup$ 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 r(t)= xi + lnyj is that how? $\endgroup$
    – The Doctor
    Dec 10 '13 at 3:03
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    $\begingroup$ Thanks! Knowing that, your question is much more specific and answerable. (I'll write up an answer now...) $\endgroup$
    – apnorton
    Dec 10 '13 at 3:04
  • $\begingroup$ Hey, anorton is there anyway you could help me out a little more? $\endgroup$
    – The Doctor
    Dec 10 '13 at 3:20
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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.

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  • $\begingroup$ That's consider a parametric correct? Thank you so much, I've been killing myself over how to make it into a position vector. THANK YOU! I just have one question. The next step is to get v(t) and then get a(t) by deriving, I'm not entirely sure as to what to do with the speed. I know the speed is the llvll. Is that to calculate the normal scalar? $\endgroup$
    – The Doctor
    Dec 10 '13 at 3:14
  • $\begingroup$ @Jose Yes, that is a parametric equation. To find the normal scalar component of acceleration, I suggest this paper: math.vt.edu/people/alattime/1224/1224-recitation10.pdf It shows the method to an easy(ish) way to find the normal scalar component of acceleration. $\endgroup$
    – apnorton
    Dec 10 '13 at 3:21
  • $\begingroup$ I see. Well the equation that I learned to find the normal scalar component involves using the cross product of the vector for velocity and acceleration and dividing by the llvll which in this case is 4 units per second. Is that more or less in the right track? $\endgroup$
    – The Doctor
    Dec 10 '13 at 3:28
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$
    – apnorton
    Dec 10 '13 at 3:29
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    $\begingroup$ The velocity is not constant, but it is determined by the slope of the curve. There is no freedom. $(dy/dt)/(dx/dt)$ is the slope of the curve, and $(dx/dt)^2+(dy/dt)^2$ must be $4^2$. Together, these determine both components of the velocity. $\endgroup$
    – dfeuer
    Dec 10 '13 at 4:18

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