Caesius 0 Posted March 29, 2013 Share Posted March 29, 2013 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) = x^{2}. However, in order to solve it, I need the parametric equations for f(x) = x^{2}. 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! Link to post Share on other sites

swansont 7563 Posted March 29, 2013 Share Posted March 29, 2013 You are dealing with forces, which are vectors. If up is +, down will be -, i.e. the gravitational acceleration is -9.8 m/s^2 Link to post Share on other sites

elfmotat 324 Posted March 29, 2013 Share Posted March 29, 2013 (edited) 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=x^{2} graph, and it tapers off in either direction. Edited March 29, 2013 by elfmotat 1 Link to post Share on other sites

Caesius 0 Posted April 3, 2013 Author Share Posted April 3, 2013 That actually helps a lot. Thanks! Link to post Share on other sites

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