Maximum Acceleration along Parabolic Curve

AI Thread Summary
The discussion focuses on deriving the maximum acceleration along a parabolic curve defined by the equation y^2 = 4ax. Participants explore implicit differentiation to find the first and second derivatives, leading to the radius of curvature. The maximum acceleration occurs where the radius of curvature is smallest, specifically at the apex of the parabola. A key point is that substituting the apex coordinates into the acceleration formula yields the maximum acceleration value. The conversation highlights the importance of careful differentiation and substitution in solving the problem effectively.
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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.
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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.
 
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Try this:

for a in a=v^2/ρ to be min or max,
da/dρ = 0
 
Isn't the radius of curvature a minimum at the apex of the parabola?
 
cupid.callin said:
Try this:

for a in a=v^2/ρ to be min or max,
da/dρ = 0

yeah i did that and got it down to (1+a/x)^(9/4) = 0 which i assume is wrong
 
Simon Bridge said:
Isn't the radius of curvature a minimum at the apex of the parabola?

yep but i don't know how this would help me to find the max acceleration or how to go about it
 
ok i went about it a completely different way. this time rearranging it to x= y^2/4a now i will differentiate with respect to y instead of x.

so dx/dy = 2y/4a → y/2a

d^2x/dy^2 = 1/2a

therefore curvature p = ((1+(y/2a)^2)3/2)/(1/2a)

at the vertex which is (0,0)the curvature is 1*1/2a therefore curvature = 2a

and max acceleration is v^2(1/2a)

does this seem legitimate?
 
Last edited:
yep but i don't know how this would help me to find the max acceleration or how to go about it
You have derived an equation for the acceleration as a function of the (x,y) position.
The max acceleration is where the radius of curvature is smallest.
The radius of curvature is smallest at the apex of the parabola.
You have the equation of the shape of the parabola.
Therefore, you know the position of the apex of the parabola...
What happens when you plug that into your equation for acceleration?

Oh I see you've done something like that.
 
Simon Bridge said:
You have derived an equation for the acceleration as a function of the (x,y) position.
The max acceleration is where the radius of curvature is smallest.
The radius of curvature is smallest at the apex of the parabola.
You have the equation of the shape of the parabola.
Therefore, you know the position of the apex of the parabola...
What happens when you plug that into your equation for acceleration?

Oh I see you've done something like that.
the problem was i was dividing by 0 when i differentiated with respect to x
 
That doesn't make sense - let's see how I do:

\frac{d}{dx}\big [y^2=4ax\big ] \Rightarrow 2y\frac{dy}{dx} = 4awhich tells us that \bigg ( \frac{dy}{dx} \bigg )^2 = \frac{4a^2}{y^2}
\frac{d}{dx}\big [y\frac{dy}{dx} = 2a \big ] \Rightarrow \bigg ( \frac{dy}{dx} \bigg )^2 + y\frac{d^2y}{dx^2} = 0which tells us that\frac{d^2y}{dx^2}=-\frac{4a^2}{y^3}
So:
\frac{1}{\rho} = -\frac{4a^2}{y^3(1+\frac{4a^2}{y^2})^{3/2}} = -\frac{4a^2}{(y^2+4a^2)^{3/2}}

We can put y=0 safely here. Agrees with your v^2/2a.
 
  • #10
very nice! good answer
 
  • #11
You did pretty well yourself.
I should have spotted the problem with the final stage in your calc in post #1 earlier - you just needed to substitute for x at the end (or not substitute for y earlier). That's the only difference between mine and yours.

Since you'd already worked it out I figured it'd do no harm to model the direct approach.

Notice how LaTeX makes the equations really clear?
It is very much worth learning.
 
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