# Can we solve for x(t) and y(t) with respect to time?

• Sotiris Michos
In summary, the conversation discusses the attempt to find the exact position of a point mass sliding on a friction-free curve with zero initial velocity. The conservation of energy is used to find the velocity, but the resulting integral is not solvable in closed form. Alternative approaches, such as using a cycloid instead of an exponential, may lead to a more elegant solution. However, the conversation concludes that numerical integration is likely the best approach for finding the exact position with respect to time.
Sotiris Michos

## Homework Statement

I would like to ask a question I was discussing the other day with a friend of mine. Suppose you have a point mass m, sliding on the friction-free curve $y = e^{-x}$ starting from position $x(0) = 0$ and $y(0) = 1$ with zero initial velocity. Can we find in closed-form the exact position $(x,y)$ with respect to time?

2. The attempt at a solution
Going down the usual path, from conservation of energy I found that $v = \sqrt{2g(1-y)}$. Assuming $φ$ is the angle between $\vec{v}$ and the downward vertical direction (the direction of negative ordinates), and using that $\dot{x} = v \sin(φ)$, $\dot{y} = v \cos(φ)$, $φ = \tan^{-1}(e^{x})$, I reached

$$\dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-e^{x}}{e^{2x}+1}}$$
$$\dot{y} = -y\sqrt{2g}\sqrt{\frac{1-y}{1+y^2}}$$

From these two, I need to solve only one. Choosing the first, separating $dx$ and $dy$ and integrating, yields

$$\int_0^{x(t)}\sqrt{\frac{e^{2x} + 1}{e^{2x} - e^{x} }}dx = \sqrt{2g}t$$

However, plugging the integral in the LHS in Mathematica doesn't compute anything, most likely meaning that this integral is not solvable in closed form, let alone invertible in closed-form.

Choosing now the second equation and doing the same yields:
$$\int_1^{y(t)}\sqrt{\frac{1+y^2}{1 - y}}\frac{dy}{y} = - \sqrt{2g}t$$

Again, trying to compute the integral in the LHS, gives an assortment of complex elliptic integrals that are gain not invertible in some closed-form. Any thoughts about what can be done?

Last edited:
You reached $$\dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-1}{e^{2x}+1}}$$
but I reach something else ( from ##\ \ e^{2x}(1-e^{-x})\ \ ## under the square root in the numerator ).

Sotiris Michos
BvU said:
You reached $$\dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-1}{e^{2x}+1}}$$
but I reach something else ( from ##\ \ e^{2x}(1-e^{-x})\ \ ## under the square root in the numerator ).

Thank you for replying BvU! You are absolutely correct ##\dot{x}## is not the one I wrote but
$$\dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-e^{x}}{e^{2x}+1}} = \sqrt{2g}\sqrt{\frac{1-e^{-x}}{1+e^{-2x}}}$$
I also noted that ##\dot{y}## needs a minus in order for ##\dot{y} = -e^{-x}\dot{x} = -y \dot{x}## to hold. I will correct the initial post! However, after this correction, things get even more challenging, since the integral now is not analytically computable!

Did someone promise you there would be an elegant looking analytical solution ?
If you want that, you might look at a cycloid instead of an exponential

Sotiris Michos
BvU said:

Did someone promise you there would be an elegant looking analytical solution ?
If you want that, you might look at a cycloid instead of an exponential

Haha! None indeed BvU! As I said, it is just an inquiry I had with a friend of mine the other day. It surprised me that the only method I could find involved integration and then inversion of the result so I thought that I might be going down the wrong path. At best, there should be a way to acquire the solution in some integral form and then let numerical integration take over!

Thank you for your warm welcoming message!

## 1. Can we solve for x(t) and y(t) at the same time?

Yes, it is possible to solve for both x(t) and y(t) simultaneously with respect to time. This is often done in systems of equations where both variables are dependent on time.

## 2. What is the process for solving x(t) and y(t) with respect to time?

The process for solving x(t) and y(t) with respect to time involves using the given equations and manipulating them to isolate the variables on one side and all other terms on the other side. Then, by using algebraic techniques such as substitution or elimination, the variables can be solved for.

## 3. Can we solve for x(t) and y(t) if there are multiple equations?

Yes, if there are multiple equations involving x(t) and y(t), it is still possible to solve for both variables with respect to time. This may require using methods such as substitution or elimination to reduce the number of equations and variables.

## 4. Is there a specific method for solving x(t) and y(t) with respect to time?

There is no one specific method for solving x(t) and y(t) with respect to time. The approach may vary depending on the given equations and the specific problem. However, commonly used methods include substitution, elimination, and graphing.

## 5. Can we solve for x(t) and y(t) if the equations involve other variables?

Yes, the presence of other variables in the equations does not prevent us from solving for x(t) and y(t) with respect to time. The process may be more complex, but it is still possible to solve for the variables by manipulating the equations and using algebraic techniques.

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