if x=a(t-sin t),(y=a(1+cos t) find dy÷dx
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Step-by-step explanation:
We can write
dydx=dydtdxdt
Since, x=a(t+sint)
⟹dxdt=a(1+cost)
And y=a(1−cost)
⟹dydt=a[0−(−sint)]=asint
Therefore,
dydx=asinta(1+cost)=2sint2 cost22cos2 t2
=tan(t2)
Thus our first answer:
dydx=tan(t2)
Now for the second
d2ydx2=ddx(dydx)
Thus,
d2ydx2=sec2 (t2)⋅12⋅dtdx
Since,
dtdx=1dxdt=1a(1+cost)
Therefore,
d2ydx2=12sec2(t2)1a(1+cost)
d2ydx2=sec2 t22a⋅2cos2 t2
Hence the final result,
d2ydx2=sec4 t24a
Hope that helps :)
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