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Converting from parametric

How would you find the intrinsic equation of the curve defined by the parametric equations x=3ap2,y=2ap3x=3ap^2, y=2ap^3 , a is constant.
I have tanψ=p \tan \psi =p but don't really know where to go from there.
(edited 7 years ago)
Original post by Ano123
How would you find the intrinsic equation of the curve defined by the parametric equations x=3ap2,y=2ap3x=3ap^2, y=2ap^3 , a is constant.
I have tanψ=p \tan \psi =p but don't really know where to go from there.

So the task is to determine the specified arclength s(p)s(p) satisfying dsdx=1+(dydx)2\dfrac{ds}{dx} = \sqrt{1+ \left( \dfrac{dy}{dx} \right) ^2 } and an initial condition of your convenience.

By two uses of the chain rule, observe that this can be rewritten in the form:

dsdp=dxdp1+(dydx)2=(dxdp)2+(dydp)2\dfrac{ds}{dp} = \dfrac{dx}{dp}\sqrt{1+\left( \dfrac{dy}{dx} \right)^2} = \sqrt{\left( \dfrac{dx}{dp} \right)^2 + \left(\dfrac{dy}{dp}\right)^2}

Can you finish it from there?
Reply 2
Original post by Farhan.Hanif93
So the task is to determine the specified arclength s(p)s(p) satisfying dsdx=1+(dydx)2\dfrac{ds}{dx} = \sqrt{1+ \left( \dfrac{dy}{dx} \right) ^2 } and an initial condition of your convenience.

By two uses of the chain rule, observe that this can be rewritten in the form:

dsdp=dxdp1+(dydx)2=(dxdp)2+(dydp)2\dfrac{ds}{dp} = \dfrac{dx}{dp}\sqrt{1+\left( \dfrac{dy}{dx} \right)^2} = \sqrt{\left( \dfrac{dx}{dp} \right)^2 + \left(\dfrac{dy}{dp}\right)^2}

Can you finish it from there?


The task is to find the arc length s s in terms of the angle of the tangent ψ \psi from whatever direction.
Original post by Ano123
The task is to find the arc length s s in terms of the angle of the tangent ψ \psi from whatever direction.

Right, so what's the issue? If you follow through the above post, you'll see that this is easily obtained by plugging in tanψ=p\tan \psi = p after finding s(p)s(p).

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