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The Time of Sliding Points Down a Curve and Parametric Equations


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I am taking Calculus III for one of my classes and Im learning about solving for Tangent Vectors and Normal Vectors for a curve. I figured out a way to determine the g-force exerted on a "particle" at a particular point on the curve by using vector calculus.

So, just for laughs, I wanted to find out what g-forces a particle would experiance on the curve f(x) = x2. However, in order to solve it, I need the parametric equations for f(x) = x2. Im trying to make the parametric equations work so f(x) is like a rollercoster. However, everytime I try and solve for the parametric equations, I get gravity working "upside down" and my "particle" slows down as it approaches the origin (the bottom of my "rollercoster", whereas it should be speeding up.) Can anyone help? Thanks!

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You can parametrize the curve as simply [math]\mathbf{r}(t)=(t,t^2,0)[/math]. The tangent vector to the curve is then [math]\mathbf{r}'(t)=(1,2t,0)[/math], and the acceleration vector is [math]\mathbf{r}''(t)=(0,2,0)[/math]. The tangent component of the acceleration to the curve is the dot product of the acceleration vector with the unit tangent, the normal component is therefore:


[math]a_\perp = \frac{|\mathbf{r}'\times \mathbf{r}''|}{|\mathbf{r}'|}[/math]


So we have:


[math]a_\perp (t) = \frac{|(1,2t,0)\times (0,2,0)|}{|(1,2t,0)|}=\frac{|(0,0,2)|}{\sqrt{1+4t^2}}=\frac{1}{\sqrt{t^2+1/4}}[/math]


Plotting a graph:




This looks as we'd expect. The maximum acceleration occurs at t=0, when the point is at the minimum of the y=x2 graph, and it tapers off in either direction.

Edited by elfmotat
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