Solving Parametric Equations: An Impossible Challenge?

[Qbit42]

Homework Statement

Solve the following parametric equations for y = f(x)

x(t) = e$$^{-t}$$ + t
y(t) = e$$^{t}$$ - t

The Attempt at a Solution

I'm completely stuck on this one. I've noticed x + y = e$$^{-t}$$ + e$$^{t}$$ and also

y = $$\frac{1}{x-t}$$ $$- t$$

but that's about as far as I've gotten.

 

[MisterMan]

Plug -t into the y equation. Like this:

y(-t) instead of y(t).

I hope that's what you're looking for, I assume, by reading the problem statement, that you want y as a function of x?

 

[Qbit42]

Well I've never seen anything like that, every example I've seen involved elimintating t from the system of equations using algebraic manipulations. But yes I am trying to find y as a function of only x

 

[gabbagabbahey]

Plug -t into the y equation. Like this:

y(-t) instead of y(t).

I hope that's what you're looking for, I assume, by reading the problem statement, that you want y as a function of x?

I don't see how that works. Sure, you can say $y(-t)=x(t)$, but that equation still makes explicit reference to $t$.

 

[gabbagabbahey]

I'm completely stuck on this one. I've noticed x + y = e$$^{-t}$$ + e$$^{t}$$ and also

y = $$\frac{1}{x-t}$$ $$- t$$

but that's about as far as I've gotten.

Take a look at $x(t)y(t)$ and $x(t)-y(t)$

 

[Mark44]

And if I understand what gabbagabbahey is driving at, sinh(t) = (1/2)(e^t - e^(-t)) and cosh(t) = (1/2)(e^t + e^(-t)) might be helpful.

 

[Qbit42]

Take a look at x(t)y(t) and x(t)-y(t)

I'm sorry but I'm just not seeing it.

x(t) - y(t) = e$$^{-t}$$ -e$$^{t}$$ + 2t

x(t)y(t) = 1 + t(-e$$^{-t}$$ + e$$^{t}$$ - t) = 1 - t(e$$^{-t}$$ - e$$^{t}$$ + t) = 1 - t(x - y - t)

I've tried combining both equations in a couple of different ways but I always end up with this equality. Any more help?

 

[gabbagabbahey]

I'm sorry but I'm just not seeing it.

x(t) - y(t) = e$$^{-t}$$ -e$$^{t}$$ + 2t

x(t)y(t) = 1 + t(-e$$^{-t}$$ + e$$^{t}$$ - t) = 1 - t(e$$^{-t}$$ - e$$^{t}$$ + t) = 1 - t(x - y - t)

I've tried combining both equations in a couple of different ways but I always end up with this equality. Any more help?

Oops! When I quickly calculated it out on a scrap of paper, I had incorrectly written down x-y=2t...which would have made things nice and easy.

The only way that I can think of is to find $y'(x)$ and $y''(x)$ using the chain rule, in order to get an ODE that doesn't depend on $t$, and then solve that...but it gets messy pretty quick.

 

[Qbit42]

And if I understand what gabbagabbahey is driving at, sinh(t) = (1/2)(e^t - e^(-t)) and cosh(t) = (1/2)(e^t + e^(-t)) might be helpful.

I talked to my professor today and he said that we are supposed to use the inverse cosh function.

x + y = e$$^{-t}$$ + e$$^{t}$$ = 2cosh(t). Therefore t = arccosh((x + y)/2)

then subsitute t into y = $$\frac{1}{x-t}$$-t

I never would have thought that inverse cosh would be an acceptable answer, in my experience if you have to rely on inverse functions you've got the wrong answer lol.

 

[Char. Limit]

I suppose Mark44 was on the right track then.

 

[Mark44]

I suppose Mark44 was on the right track then.

Actually, I don't think I was. You can get to x + y = 2cosh(t), and then use cosh^(-1) to solve for t, but I don't see any simplification that ends with y as a function of x.

 

[Qbit42]

The answer he is expecting is y = $$\frac{1}{x - arccosh((x+y)/2)}$$ - arccosh((x+y)/2). I know that its not an explict function of x but y is implicitly defined by the previous equation.

 

[Mark44]

Well, then maybe I was on the right track.

x + y = e^t + e^-t = 2cosh(t)
==> (x + y)/2 = cosh(t)
==> cosh^-1((x + y)/2) = t
or t = arccosh((x + y)/2)

Substituting for t in y = e^t - t, we have
$$y = e^{arccosh((x + y)/2)} - arccosh((x + y)/2)$$

This seems to be going in the right direction, but I don't know how to deal with the exponential term. If this involved triangle trig inverses, I would draw a right triangle and see if I could rewrite the first term above, but I'm not as familiar with the geometry in the hyperbolic trig functions.

 

[Qbit42]

Well, then maybe I was on the right track.

x + y = et + e-t = 2cosh(t)
==> (x + y)/2 = cosh(t)
==> cosh-1((x + y)/2) = t
or t = arccosh((x + y)/2)

Substituting for t in y = et - t, we have
LaTeX Code: y = e^{arccosh((x + y)/2)} - arccosh((x + y)/2)

This seems to be going in the right direction, but I don't know how to deal with the exponential term. If this involved triangle trig inverses, I would draw a right triangle and see if I could rewrite the first term above, but I'm not as familiar with the geometry in the hyperbolic trig functions.

Once I'd found t I did some rearranging before I substituted it back into the equations.

x = e$$^{-t}$$ + t ==> x-t = e$$^{-t}$$

y = e$$^{t}$$ - t ==> y = $$\frac{1}{x-t}$$-t

then I substituted for t. It may not be any prettier but at least it doesn't have an exponential in it

 

[Unit]

I am taking a bit of a different approach:

x + y = exp(-t) + exp(t)

Multiply both sides by exp(t) to get:

(x + y) ext(t) = 1 + exp(t)^2

So 0 = exp(t)^2 - (x+y)exp(t) + 1

You could use the quadratic formula to solve for exp(t) = a big huge two-valued mess. Then take the ln of THAT and have t?

I also tried this:

(x - t)(y + t) = exp(-t) exp(t)
xy - ty + tx - t^2 = 1

Rearrange a little to get
0 = t^2 - t(x - y) + 1 - xy

And maybe you could solve this quadratic?

 

[Dickfore]

It's impossible to solve

$$x = e^{-t} + t$$

for t in terms of elementary functions.

 

[hgfalling]

Well, then maybe I was on the right track.

x + y = e^t + e^-t = 2cosh(t)
==> (x + y)/2 = cosh(t)
==> cosh^-1((x + y)/2) = t
or t = arccosh((x + y)/2)

Substituting for t in y = e^t - t, we have
$$y = e^{arccosh((x + y)/2)} - arccosh((x + y)/2)$$

This seems to be going in the right direction, but I don't know how to deal with the exponential term. If this involved triangle trig inverses, I would draw a right triangle and see if I could rewrite the first term above, but I'm not as familiar with the geometry in the hyperbolic trig functions.

$$\cosh^{-1} x = \ln (x - \sqrt{x^2-1})$$ (for real-valued functions) according to Wikipedia, which gets rid of the exponential term but doesn't obviously simplify to the term in the answer given.

http://en.wikipedia.org/wiki/Inverse_hyperbolic_function

 

[Unit]

$$\cosh^{-1} x = \ln (x - \sqrt{x^2-1})$$

x + y = exp(-t) + exp(t)
(x + y) ext(t) = 1 + exp(t)^2
0 = exp(t)^2 - (x + y) exp(t) + 1

Quadratic formula to solve for exp(t).

Not much help, but I think it's cool that we can go around in a circle: you can simplify the arcosh expression to be equivalent to the "minus" root of the radical.

$$\cosh^{-1} \frac{x+y}{2} = \ln \left( \frac{x+y}{2} - \sqrt{ \frac{(x+y)^2}{4} - 1} \right)$$

And as Mark44 said,

$$\cosh^{-1} \frac{x+y}{2} = t$$

$$t = \ln \left( \frac{x+y}{2} - \sqrt{ \frac{(x+y)^2}{4} - 1} \right)$$

$$e^t = \frac{x+y}{2} - \frac{\sqrt{(x+y)^2 - 4}}{4}$$

$$e^t = \frac{x+y - \sqrt{ (x+y)^2 - 4}}{2}$$, which is a solution to 0 = exp(t)^2 - (x + y) exp(t) + 1.