rxh140630
- 60
- 11
- Homework Statement
- A particle is constrained to move along a parabola whose equation is y=x^2. At what point on the curve are the abscissa and the ordinate changing at the same rate?
- Relevant Equations
- x^2=y, dx/dy= 2x
Would the trivial solution be x=0,y=0?
Non trivial:
let y=x^2
\frac{dy}{dx}=2x, \frac{dx}{dy} = \frac12y^{-\frac12}
x=\frac14 y^{-\frac12}
here x=1 and y = 1/16 is a solution
but my book says the answer is x=1/2 and y=1/4
this is one answer that you get with the equation I derived, but I feel like it's not the only answer. Am I missing something?
Non trivial:
let y=x^2
\frac{dy}{dx}=2x, \frac{dx}{dy} = \frac12y^{-\frac12}
x=\frac14 y^{-\frac12}
here x=1 and y = 1/16 is a solution
but my book says the answer is x=1/2 and y=1/4
this is one answer that you get with the equation I derived, but I feel like it's not the only answer. Am I missing something?