Two-Dimensional Motion of a Particle: Velocity & Acceleration

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Homework Statement



The two-dimensional motion of a particle is defined by the relationship [tex]r = \frac {1}{sin\theta - cos\theta}[/tex] and [tex]tan\theta = 1 + \frac {1}{t^2}[/tex], where [tex]r[/tex] and [tex]\theta[/tex] are expressed in meters and radians, respectively, and [tex]t[/tex] is expressed in seconds. Determine (a) the magnitudes of velocity and acceleration at any instant, (b) the radius of curvature of the path.


Homework Equations



[tex]r = \frac {1}{sin\theta - cos\theta}[/tex]


[tex]tan\theta = 1 + \frac {1}{t^2}[/tex]



The Attempt at a Solution



I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex]tan\theta[/tex] for [tex]\theta[/tex]. Once I've done that I figure I'd need to differentiate both [tex]r[/tex] and [tex]\theta[/tex] to find [tex]\dot{r}, \ddot{r}, \dot{\theta}, \ddot{\theta}[/tex].

I don't know if I'm on the correct route but any help would be appreciated. thanks!
 
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chart2006 said:
I've made a few attempts but they seem way more complicated than the problem should be I think. I'm assuming I need to solve [tex]tan\theta[/tex] for [tex]\theta[/tex].

Tangent is not a one-to-one function, so that's a bad idea.

Instead, draw a picture!:smile: I think you can find expressions for [itex]\sin\theta[/itex] and [itex]\cos\theta[/itex] in terms of [itex]t[/itex] without actually solving for [itex]\theta[/itex] first...think 'right triangle':wink: