X=a (1+sint),y=a (1-cost) p.t. d2y/dx2=1/a at x=90°
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Taking x = a (1 + sin t)
differentiating with respect to t we get,
dx/dt = acos t ------------> (1)
Taking
y = a (1 − cos t)
differentiating with respect to t we get,
dy/dt = a (0 + sin t)
dy/dt = a sin t .............(2)
Dividing (2) by (1)
we get,
(dy/dt) /(dx/dt) = a sin t/acos t
⇒ dy/dx = tant
= tan t
d^2 y / dt^2 = acost
d^2 x / dt^2 = -asint
d^2 y/dx^2 = - tant
t = π/2
d^2 y / dx^2 = 0
differentiating with respect to t we get,
dx/dt = acos t ------------> (1)
Taking
y = a (1 − cos t)
differentiating with respect to t we get,
dy/dt = a (0 + sin t)
dy/dt = a sin t .............(2)
Dividing (2) by (1)
we get,
(dy/dt) /(dx/dt) = a sin t/acos t
⇒ dy/dx = tant
= tan t
d^2 y / dt^2 = acost
d^2 x / dt^2 = -asint
d^2 y/dx^2 = - tant
t = π/2
d^2 y / dx^2 = 0
arshdeeplaurgill365:
Yes
After getting dy/dx; find d2y/dx2 w.r.t. t
It will be prover
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