*solved*Particle moving along a parabola

[rxh140630]

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?

 

[PeroK]

Don't you need something like a force of gravity here?

 

[rxh140630]

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.

 

[PeroK]

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.

 

[rxh140630]

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.

 

[PeroK]

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 $t$.

 

[rxh140630]

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

 

[Lnewqban]

... 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.