mrkrgn
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Homework Statement
Hello,
So i was just doing a few past papers for an upcoming exam and i got stuck on this one.
i have got to the stage where i have 1/p all in term of x and y, but i can't see how you would go about getting the maximum acceleration just from x and y values.
Homework Equations
a=v^2/r
The Attempt at a Solution
so here is how i went about it:
y^2 = 4ax
therefore by implicit differentiation
2ydy/dx = 4a
thus dy/dx = 2a/y
for the equation for 1/p we need (dy/dx)^2
which is,
4a^2/y^2 →4a^2/4ax since y^2 = 4ax
thus we get (dy/dx)^2 = a/x
now we need d^2y/dx^2
so by implicit differentiation again 2a*y^-1 → -2a*y^-2*dy/dx
(-2a/y^2)(2a/y) → (-2a/4ax)(2a/y) → -a/xy
therefore 1/p = ((1+a/x)^(3/2))/(-a/xy)
This is where I found it difficult to continue. I tried by saying a is a maximum when p is a minimum (a=v^2/p) so da/dp = -v^2/p^2, but i couldn't really get any further any help would be greatly appreciated because it is just annoying me now haha.