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Find the equation of the line that is tangent to the curve at the point $(0,\sqrt{\frac{\pi}{2}})$. Given your answer in slope-intercept form.

I don't know how can I get the tangent line, without a given equation!!, this is part of cal1 classes.

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  • $\begingroup$ Do you mean $\frac{\sqrt{\pi}}{2}$ or $\sqrt{\frac{\pi}{2}}$? You could just respond with "the first one" or "the second one" and I will edit your question. $\endgroup$ – layman Apr 8 '15 at 15:50
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    $\begingroup$ Do you have anything? Like a picture, or a verbal description of the curve? $\endgroup$ – Arthur Apr 8 '15 at 15:51
  • $\begingroup$ the second one , thank you :) $\endgroup$ – Mioosh Apr 8 '15 at 15:52
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    $\begingroup$ no, that's the problem the question kind of missing something, $\endgroup$ – Mioosh Apr 8 '15 at 16:00
  • $\begingroup$ What is the dimension of the space where the curve lays? $\endgroup$ – demitau Apr 8 '15 at 16:00
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If we suppose that your curve is the graph of a function $y=f(x)$ such that $f(0) = \sqrt{\pi/2}$, than the equation of the tangent at $x=0$ is:

$ y-\sqrt{\pi/2}=f'(0)(x-0) $

i.e.

$y=f'(0) x+\sqrt{\pi/2}$

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So far we can infer $$ T(x) = m x + n $$ with $T(0) = \sqrt{\pi/2}$. Thus $$ T(x) = m x + \sqrt{\pi/2} $$ To determine the slope $m$ we need more information about the given curve.

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  • $\begingroup$ thank you so much, for your help. and i'll try t see my teacher if there is a farther info about it. THX $\endgroup$ – Mioosh Apr 8 '15 at 16:11

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