# *solved*Particle moving along a parabola

• rxh140630
In summary: 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.
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?

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.

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.

SammyS and vela

## 1. What is a parabola?

A parabola is a U-shaped curve that is formed by the graph of a quadratic function. It is defined by the equation y = ax^2 + bx + c, where a, b, and c are constants.

## 2. How is a particle moving along a parabola solved?

The motion of a particle along a parabola can be solved using the principles of calculus. By finding the derivative of the equation of the parabola, the velocity and acceleration of the particle can be determined. The position of the particle at any given time can also be found by integrating the velocity function.

## 3. What factors affect the motion of a particle along a parabola?

The motion of a particle along a parabola is affected by several factors, including the initial velocity of the particle, the angle at which it is launched, and the force of gravity. Other external forces, such as air resistance, can also affect the motion of the particle.

## 4. Can a particle move along a parabola without any external forces?

No, a particle cannot move along a parabola without any external forces. In order for the particle to move along a curved path, there must be a force acting upon it. In the case of a parabola, the force of gravity is typically the dominant force affecting the motion of the particle.

## 5. What real-life applications involve a particle moving along a parabola?

The motion of a projectile, such as a ball thrown into the air, can be modeled by a parabola. This is also true for the motion of objects launched from a slingshot or roller coaster. In addition, the trajectory of a satellite orbiting the Earth can also be described by a parabola.

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