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If $\frac{x^2}{36} + \frac{y^2}{64} = 1$ and $y(3) = 6.92820323$ find $y'(3)$

Okay so, we get:

$\frac{36\cdot2x}{36^2} + \frac{64\cdot 2y(y')}{64^2} = 0 $

$\frac{72x}{1296} + \frac{128y(y')}{4096} = 0 $

Where the heck do we go from here?


Okay: Here's what I did. I got really confused with the y(3) and y'(3) stuff. I sometimes don't know what goes where with this added notation (that I'm not used to yet).

$\frac{72\cdot3}{1296} + \frac{128\cdot 6.92820323z}{4096} = 0 $

We have to substitute the 3 for the x, and the 6.92820323 for the y, and I'm just going to use z to represent y'.

Then:

$z = -0.769802$

This works because our $x$, $3$ is the same in both cases. When we plug in our y, which we know to be $6.928...$, it is synonymous to plugging the original equation back in (like we'd do if we didn't know what y was).

Thanks!

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How about solving for $y'(x)$ and then substituting in $x=3$ to get the value? –  Amzoti Oct 3 '13 at 4:01
    
Set $x=3$ and solve for $y'(3)$. –  lhf Oct 3 '13 at 4:02
    
I think you have an extra 36 and an extra 64 in your implicit derivative. –  DanielV Oct 3 '13 at 4:09
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The $y(3)\approx \text{mess}$ does carry some information. There are two values of $y$ corresponding to $x-3$, –  André Nicolas Oct 3 '13 at 4:54
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Okay: Here's what I did. I got really confused with the y(3) and y'(3) stuff. I sometimes don't know what goes where with this added notation (that I'm not used to yet).

72⋅31296+128⋅6.92820323z4096=0 We have to substitute the 3 for the x, and the 6.92820323 for the y, and I'm just going to use z to represent y'.

Then:

z=−0.769802 This works because our x, 3 is the same in both cases. When we plug in our y, which we know to be 6.928..., it is synonymous to plugging the original equation back in (like we'd do if we didn't know what y was).

Thanks!

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