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I know the $\surd$ sign refers to the positive square root. Does the exponent 1/2 mean the positive square root too by convention?

I ask because I'm converting from parametric to cartesian here...

$x=t^2$ and $y=t^3$

So $t=\pm \sqrt{x}$

Then $y=\pm x^{3/2}$

Yet the given answer is $y=x^{3/2}$ in the textbook. Can someone clarify please?

Thanks, Rob

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    $\begingroup$ Most likely in this problem $t\ge 0$, so there is no ambiguity to resolve. $\endgroup$ – vadim123 Apr 27 '14 at 19:31
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    $\begingroup$ I think you mean $t = \pm \sqrt{x}$ with the attendant change to $y$. $\endgroup$ – Eric Towers Apr 27 '14 at 19:32
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    $\begingroup$ $a^{1/2}=\sqrt{a}$. Solving the equation $x^2 = a$ is different, and doors give two possible answers. $\endgroup$ – vonbrand Apr 27 '14 at 19:51
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A positive number to a real power is always, by convention, positive. This is because $a^b$ is generally defined as $e^{b \ln a}$.

In particular, in your case $t^{3/2}$ refers to $e^{(3/2) \ln t}$, a positive value.

It gets slightly more complicated if you have a negative number as a base, in which case for instance $a^{1/3}$ could mean the negative cube root, but $a^x$ is generally undefined for real $x$.

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  • $\begingroup$ Happy to help @Saxobob. $\endgroup$ – 6005 Apr 27 '14 at 20:42
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Unless the problem explicitly states that $t\ge 0$, then the answer in the textbook is wrong. It should be $y=\pm x^{3/2}$, as you suspect. This can easily be seen by putting, for example, $t=-1$; then $x=1$ but $y=-1$.

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