Curvature proof problem

• Nov 17th 2009, 10:11 PM
Durgoth
Curvature proof problem
Consider a curve given in parametric form x(t), y(t). Starting from the definition of the tangent angle (top equation set in picture).

http://img132.imageshack.us/img132/2267/c3cprob1.jpg

prove the bottom half of the picture (The "dots" refer to differentiation with respect to the parameter t.

So far all I have for this problem is that since tanθ=sinθ/cosθ, that y'=sinθ and x'=cosθ, but I have no idea on how to proceed with this, and I am completely in the dark. Any help would be greatly appreciated.
• Nov 17th 2009, 11:21 PM
Chris L T521
Quote:

Originally Posted by Durgoth
Consider a curve given in parametric form x(t), y(t). Starting from the definition of the tangent angle (top equation set in picture).

http://img132.imageshack.us/img132/2267/c3cprob1.jpg

prove the bottom half of the picture (The "dots" refer to differentiation with respect to the parameter t.

So far all I have for this problem is that since tanθ=sinθ/cosθ, that y'=sinθ and x'=cosθ, but I have no idea on how to proceed with this, and I am completely in the dark. Any help would be greatly appreciated.

Note that $\displaystyle \tan\theta=\frac{\dot{y}}{\dot{x}}\implies\theta=\ arctan\!\left(\frac{\dot{y}}{\dot{x}}\right)$.

So when you differentiate both sides with respect to t, we see that

$\displaystyle \frac{\,d\theta}{\,dt}=\frac{1}{1+\left(\frac{\dot {y}}{\dot{x}}\right)^2}\cdot\frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{\dot{x}^2}=\frac{\ddot{y}\dot{x}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}$

Does this make sense?
• Nov 17th 2009, 11:28 PM
TwistedOne151
Take your equation $\displaystyle \tan\theta=\frac{\dot{y}}{\dot{x}}$ and take the derivative of both sides with respect to t. By the chain rule, $\displaystyle \frac{d}{dt}(\tan\theta)=(\sec^2\theta)\left(\frac {d\theta}{dt}\right)=(1+\tan^2\theta)\frac{d\theta }{dt}$; and you can find $\displaystyle \frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)$ via the quotient rule. Then replace $\displaystyle \tan\theta$ on the left hand side with $\displaystyle \frac{\dot{y}}{\dot{x}}$ (since those are equal), then solve the equation for $\displaystyle \frac{d\theta}{dt}$ in terms of $\displaystyle \dot{x}$, $\displaystyle \dot{y}$, $\displaystyle \ddot{x}$, and $\displaystyle \ddot{y}$.

--Kevin C.
• Nov 19th 2009, 08:02 PM
Durgoth
Thank you for the help!