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## Homework Statement

There is an elliptical path and pegs A and B are restricted to move around it. If the link moves with a constant speed of 10m/s, determine the magnitude of velocity when x=0.6m

[PLAIN]http://users.adam.com.au/shortround/Prob.12-78.jpg [Broken]

## Homework Equations

[tex]\frac{x^2}{4}[/tex]+y

^{2}=1

[tex]\rho[/tex]=[tex]\frac{(1+(\frac{dy}{dx})^2)^\frac{3}{2}}{\left|\frac{dy}{dx}\right|}[/tex]

a=[tex]\frac{dv}{dt}[/tex] [tex]\vec{e}[/tex]

_{t}+[tex]\frac{v^2}{\rho}[/tex] [tex]\vec{e}[/tex]

_{n}

Where [tex]\rho[/tex] is the radius of curvature.

## The Attempt at a Solution

I rearranged [tex]\frac{x^2}{4}[/tex]+y

^{2}=1 to get x

^{2}+4y

^{2}=4

I then differentiated this to get: [tex]\frac{dy}{dx}[/tex]=[tex]\frac{-x}{4y}[/tex] and [tex]\frac{d^2y}{dx^2}[/tex]=[tex]\frac{(x/y)-1}{4y}[/tex]

Using x=0.6m, y=[tex]\sqrt{0.91}[/tex]=0.9539

By substituting this into the derivative and second derivatives and then putting these into the radius of curvature equation, I found the radius of curvature. However I am not sure if this is the correct way to do it. Also once I have the radius of curvature, how do I find the velocity?

Thanks!!

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