# Curvilinear Motion n/t coordinates

 P: 23 1. The problem statement, all variables and given/known data Given, y2 = 9x3 + 6x, where x and y are in metres and y is positive. What is the normal component of the acceleration when x= 3m, $$\dot{x}$$ = 7ms-1 and $$\ddot{x}$$ = 8 ms-2? 2. Relevant equations V = Vxi + Vyj Vx = $$\dot{x}$$ Vy = $$\dot{y}$$ a = axi + ayj ax = $$\ddot{x}$$ ay = $$\ddot{y}$$ 3. The attempt at a solution I was fairly confident that my following attempt would yield the correct solution, however it did not. I still fail to see any problems with my attempt, so I thought I'd post it. Any feedback on my work would be greatly appreciated.
 HW Helper P: 5,003 Hmmm.... it's difficult to spot the error in your calculations, but you might have an easier time if you don't mess about with too many angles... Everything up to and including your calculation of $\theta$ looks correct to me. From that point, you should be able to basically look at diagram and read off the vector expression for the unit tangent and unit normal; $\textbf{T}=\cos\theta\textbf{i}+\sin\theta\textbf{j}$ and $\textbf{N}=\cos(\frac{\pi}{2}-\theta)\textbf{i}-\sin(\frac{\pi}{2}-\theta)\textbf{j}=\sin\theta\textbf{i}-\cos\theta\textbf{j}$...The advantage of expressing the unit normal like this is that once you calculate $\textbf{a}=\ddot{x}\textbf{i}+\ddot{y}\textbf{j}$, all you need to do to find its normal component is take the dot product $a_N=\textbf{a}\cdot\textbf{N}$.