- Thread starter Dean54321
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calculate (x/y)Solve the parametric equations whose parameters are:

x = 1/2(2^t+2^-t)

y = 1/2(2^t-2^-t)

I can't seem to isolate the index t. So far, I have simplified equation1 to be x = 2^(t-1+2)^(-t-1)

Calculate [(x/y) + 1]

Calculate [(x+y)/y]

Calculate [(x - y)/y]

Calculate [(x+ y)/ (x - y)]

Continue.....

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Can you calculate x+y? How about x-y?

I assume the OP is trying to get a relationship between x and y (so without any t terms).

Can you calculate x+y? How about x-y?

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That is correct. Even so, there is not need to solve for t. There is just a need to eliminate t. Did you read my example above?I assume the OP is trying to get a relationship between x and y (so without any t terms).

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And Jomo is showing the OP one of the ways!I assume the OP is trying to get a relationship between x and y (so without any t terms).

\begin{align*}

x &= \frac{1}{2} \left( 2^t + 2^{-t} \right)\\

y &= \frac{1}{2} \left( 2^t - 2^{-t} \right) \\

\\

2\,x &= 2^t + \frac{1}{2^t} \\

2\,y &= 2^t - \frac{1}{2^t} \\

\\

2\,x &= \frac{\left( 2^t \right) ^2 + 1}{2^t} \\

2\,y &= \frac{\left( 2^t \right) ^2 - 1 }{2^t} \\

\\

2\,x \left( 2^t \right) &= \left( 2^t \right) ^2 + 1 \\

2\,y \left( 2^t \right) &= \left( 2^t \right) ^2 - 1 \\

\\

0 &= \left( 2^t \right) ^2 - 2\,x \left( 2^t \right) + 1 \\

0 &= \left( 2^t \right) ^2 - 2\,y \left( 2^t \right) - 1 \\

\\

0 &= \left( 2^t \right) ^2 - 2\,x \left( 2^t \right) + \left( -x \right) ^2 - \left( -x \right) ^2 + 1 \\

0 &= \left( 2^t \right) ^2 - 2\,y \left( 2^t \right) + \left( -y \right) ^2 - \left( -y \right) ^2 - 1 \\

\\

0 &= \left( 2^t - x \right) ^2 - x^2 + 1 \\

0 &= \left( 2^t - y \right) ^2 - y^2 - 1

\end{align*}

You should be able to solve for t now...

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Is it always possible to solve for t? That is a serious question. I suspect maybe not as you might run into domain problems(?)

\begin{align*}

x &= \frac{1}{2} \left( 2^t + 2^{-t} \right)\\

y &= \frac{1}{2} \left( 2^t - 2^{-t} \right) \\

\\

2\,x &= 2^t + \frac{1}{2^t} \\

2\,y &= 2^t - \frac{1}{2^t} \\

\\

2\,x &= \frac{\left( 2^t \right) ^2 + 1}{2^t} \\

2\,y &= \frac{\left( 2^t \right) ^2 - 1 }{2^t} \\

\\

2\,x \left( 2^t \right) &= \left( 2^t \right) ^2 + 1 \\

2\,y \left( 2^t \right) &= \left( 2^t \right) ^2 - 1 \\

\\

0 &= \left( 2^t \right) ^2 - 2\,x \left( 2^t \right) + 1 \\

0 &= \left( 2^t \right) ^2 - 2\,y \left( 2^t \right) - 1 \\

\\

0 &= \left( 2^t \right) ^2 - 2\,x \left( 2^t \right) + \left( -x \right) ^2 - \left( -x \right) ^2 + 1 \\

0 &= \left( 2^t \right) ^2 - 2\,y \left( 2^t \right) + \left( -y \right) ^2 - \left( -y \right) ^2 - 1 \\

\\

0 &= \left( 2^t - x \right) ^2 - x^2 + 1 \\

0 &= \left( 2^t - y \right) ^2 - y^2 - 1

\end{align*}

You should be able to solve for t now...

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Sorry, but I don't see how you can solve for t yet.You should be able to solve for t now...

It's literally just a case of reversing all the operations and isolating the t in each equation. There's only one term with t in it in each now...Sorry, but I don't see how you can solve for t yet.