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

<br /> x = e^{-t} + t<br />

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.