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Parametric equation - velocity

A curve is given by x=a(θ−sinθ),y=a(1+cosθ)x=a(\theta -sin\theta),y=a(1+cos\theta)

A small smooth bead is slides on the curve. Show that its velocity (dxdt,dydt)(\frac{dx}{dt},\frac{dy}{dt}) is 2adθdtcosθ2(cosθ2,sinθ2)2a\frac{d\theta}{dt}cos\frac{\theta}{2} (cos\frac{\theta}{2},sin\frac{\theta}{2})

When i differentiate x and y with respect to t, i don't get this. I seem to get 2adθdtsinθ2(sinθ2,−cosθ2)2a\frac{d\theta}{dt}sin\frac{\theta}{2} (sin\frac{\theta}{2},-cos\frac{\theta}{2})

Could someone give me a hand with the question please
lilman91
A curve is given by x=a(θ−sinθ),y=a(1+cosθ)x=a(\theta -sin\theta),y=a(1+cos\theta)

A small smooth bead is slides on the curve. Show that its velocity (dxdt,dydt)(\frac{dx}{dt},\frac{dy}{dt}) is 2adθdtcosθ2(cosθ2,sinθ2)2a\frac{d\theta}{dt}cos\frac{\theta}{2} (cos\frac{\theta}{2},sin\frac{\theta}{2})

When i differentiate x and y with respect to t, i don't get this. I seem to get 2adθdtsinθ2(sinθ2,−cosθ2)2a\frac{d\theta}{dt}sin\frac{\theta}{2} (sin\frac{\theta}{2},-cos\frac{\theta}{2})

Could someone give me a hand with the question please


With the information you have given, I agree with your answer.

So, we've both made the same mistake, or there is an error in the given question or answer.
Reply 2
Avatar for lilman91
lilman91
OP
4
ghostwalker
With the information you have given, I agree with your answer.

So, we've both made the same mistake, or there is an error in the given question or answer.


its from an introductory worksheet that cambridge sent me. Here's what the question is - i don't think i missed anything out when i posetd
question.jpg30.0KB
I don't even agree with the answer to the first part as I get -cot(theta/2)
Reply 4
Avatar for james.h
james.h
11
ghostwalker
I don't even agree with the answer to the first part as I get -cot(theta/2)

:no: I don't agree either.

I got down to dydx=−sinθ1−cosθ\dfrac{dy}{dx}=-\dfrac{sin\theta}{1-cos\theta}

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