Parametric equations

Dean54321

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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)
 

Prove It

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Your post isn't clear. Is it this?

\begin{align*} x &= \frac{1}{2}\,\left( 2^t + 2^{-t} \right) \\
y &= \frac{1}{2}\,\left( 2^t - 2^{-t} \right) \end{align*}
 

Subhotosh Khan

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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)

Calculate [(x/y) + 1]

Calculate [(x+y)/y]

Calculate [(x - y)/y]

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

Continue.....
 

Jomo

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Why do you want to solve for t?? What if (for example) you had discovered that 3x = 2t^2 -11 and that y^2 = 2t^2-11? Wouldn't that tell you that 3x=y^2?? And you never solved for t! Stop thinking mechanically!

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

Prove It

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Why do you want to solve for t?? What if (for example) you had discovered that 3x = 2t^2 -11 and that y^2 = 2t^2-11? Wouldn't that tell you that 3x=y^2?? And you never solved for t! Stop thinking mechanically!

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).
 

Jomo

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

Subhotosh Khan

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

Prove It

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Yes of course I read it. I'm just pointing out that it is perfectly reasonable to solve for t (although as you pointed out, not the most concise way).

\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...
 

Jomo

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Yes of course I read it. I'm just pointing out that it is perfectly reasonable to solve for t (although as you pointed out, not the most concise way).

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

Jomo

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Prove It

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Sorry, but I don't see how you can solve for t yet.
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...
 
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