Acceleration on an Elliptical Path

Archit ur finding tangential acc. which is const.(actually zero) but u need to find normal acc.....

Well I got the answer as $\frac{b}{a}$ $cosec^{2}$ t... $\frac{dx}{dt}$ = -asin(t) and $\frac{dy}{dt}$ = bcos(t). Now using the chain rule: $\frac{dy}{dx}$ = $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ = - $\frac{b}{a}$ . $\frac{cos(t)}{sin(t)}$ . Now we need to be more careful when working out the 2nd derivative (also I'm going to ignore the constant for now): " $\frac{d^2 y}{dx^2}$ " = $\frac{d}{dx}$ [ $\frac{dy}{dx}$ ] = $\frac{d}{dt}$ [ $\frac{cos(t)}{sin(t)}$ ] $\frac{dt}{dx}$ (by the chain rule). Now using the quotient rule: $\frac{d^2 y}{dx^2}$ = $\frac{1}{v}$ $\frac{du}{dx}$ - $\frac{u}{v^2}$ $\frac{dv}{dx}$ = -[ $\frac{sin^2t +cos^2t}{sin^2t}$ ] = - $cosec^{2}$ (t). Now multiplying by the constant that we neglected earlier gives: $\frac{d^2 y}{dx^2}$ = $\frac{b}{a}$ $cosec^{2}$ (t)

It is not tangential acceleration it is radial.It is requires to keep body on given curve.Example when u sit on merry-go-round u experience an outward force so to balance that an inward force is required(equal in magnitude to outward force).Hence this force is not increasig speed.

I am from Jaipur and sorry for replying late..