*solved*Particle moving along a parabola

  • Thread starter Thread starter rxh140630
  • Start date Start date
  • Tags Tags
    Parabola
rxh140630
Messages
60
Reaction score
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?
 
Physics news on Phys.org
Don't you need something like a force of gravity here?
 
PeroK said:
Don't you need something like a force of gravity here?

No clue, it's a calculus book, I'm guessing they decided to remove gravity from the equation.
 
  • Sad
Likes PeroK
rxh140630 said:
No clue, it's a calculus book, I'm guessing they decided to remove gravity from the equation.
Perhaps it doesn't matter whether the particle is accelerating or not?

What about parameterising the curve? I assume "changing at the same rate" means with respect to time.
 
PeroK said:
Perhaps it doesn't matter whether the particle is accelerating or not?

What about parameterising the curve? I assume "changing at the same rate" means with respect to time.

Yeap, no mention of accelerating, I don't really know what that is. That's just the derivative of velocity right? It doesn't mention velocity or acceleration. Just how the two coordinates change with respect to each other, which I guess implies velocity? I need to start reading a physics textbook.

Yes, I believe changing at the same rate means with respect to time. I wrote out the whole question word for word as per the book. I wish the author was a tad more detailed and made less assumptions in his questions.
 
rxh140630 said:
Yeap, no mention of accelerating, I don't really know what that is. That's just the derivative of velocity right? It doesn't mention velocity or acceleration. Just how the two coordinates change with respect to each other, which I guess implies velocity? I need to start reading a physics textbook.

Yes, I believe changing at the same rate means with respect to time. I wrote out the whole question word for word as per the book. I wish the author was a tad more detailed and made less assumptions in his questions.
I had to look up "abscissa" and "ordinate" (and found they mean ##x## and ##y##), so perhaps this is a book from a bygone era. And I promptly forgot which one is which!

That said, you should parameterise the curve in terms of ##t##.
 
Last edited:
PeroK said:
I had to look up "abscissa" and "ordinate" (and found they mean ##x## and ##y##), so perhaps this is a book from a bygone era. And I prompty forgot which one is which!

That said, you should parameterise the curve in terms of ##t##.
Hmm actually I think implicit differentiation is the easiest way to do this problem.

it gives you the result x=1/2 and y=1/4, the books answer
 
rxh140630 said:
... Yes, I believe changing at the same rate means with respect to time. I wrote out the whole question word for word as per the book. I wish the author was a tad more detailed and made less assumptions in his questions.
Could it simply be the point of the curve around which Δx≈Δy?
If so, that would the point at which the slope of a straight line tangent to the curve equals 1, I believe.
 
  • Like
Likes SammyS and vela
Back
Top