Curvature proof problem

**Durgoth** · November 17th 2009, 10:11 PM

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

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.

**Chris L T521** · November 17th 2009, 11:21 PM

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).

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

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

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

Does this make sense?

**TwistedOne151** · November 17th 2009, 11:28 PM

Take your equation $\tan\theta=\frac{\dot{y}}{\dot{x}}$ and take the derivative of both sides with respect to t. By the chain rule, $\frac{d}{dt}(\tan\theta)=(\sec^2\theta)\left(\frac {d\theta}{dt}\right)=(1+\tan^2\theta)\frac{d\theta }{dt}$ ; and you can find $\frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)$ via the quotient rule. Then replace $\tan\theta$ on the left hand side with $\frac{\dot{y}}{\dot{x}}$ (since those are equal), then solve the equation for $\frac{d\theta}{dt}$ in terms of $\dot{x}$ , $\dot{y}$ , $\ddot{x}$ , and $\ddot{y}$ .

--Kevin C.

**Durgoth** · November 19th 2009, 08:02 PM

Thank you for the help!

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