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I am new to partial derivative and I need some help in understanding if what I have done so far is correct.

Let $S$ be the surface given by $x^2 + y^2 - 3z^2 = 5$

I want to calculate the partial derivative:

$\frac{\partial z}{\partial x}$ at the point $(2,2,1)$ and $(2,2,-1)$

This is what I have done:

$x^2 + y^2 - 3z^2 = 5$

$z^2 = \frac{x^2 + y^2 - 5}{3}$

$z = \pm \sqrt\frac{x^2 + y^2 - 5}{3}$

$\frac{\partial z}{\partial x} = \frac{\frac{1}{2}(x^2 + y^2 - 5)^{-\frac12}(2x)}{\sqrt3}$

$\frac{\partial z}{\partial x} = \frac{2x}{2\sqrt{3}\sqrt{x^2 + y^2 - 5}}$

$\frac{\partial z}{\partial x} = \frac{x}{\sqrt{3}\sqrt{x^2 + y^2 - 5}}$

But I am unsure of how to continue after this, and how to use the points (2,2,1) and (2,2,-1).

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  • $\begingroup$ There is sign error at $z^2=\dots$ $\endgroup$
    – Bernard
    Oct 3, 2018 at 22:09
  • $\begingroup$ @Bernard edited! $\endgroup$ Oct 3, 2018 at 22:16

2 Answers 2

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Use differential calculus with the equation of the surface: differentiating both sides yields $$2x\,\mathrm dx+2y\,\mathrm dy-6z\,\mathrm dz=0,$$ whence $$\mathrm dz=\frac{x\,\mathrm dx+y\,\mathrm dy}{3z}.$$ Now $\;\dfrac{\partial z}{\partial x}$ is the coefficient of $\mathrm dx$, and similarly for $\;\dfrac{\partial z}{\partial y}$.

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  • $\begingroup$ I am rather new to this so I don't really follow your train of thought. Does this mean that I can simply substitute the values $(2,2,1)$ and $(2,2,-1)$ into the equation as stated, such that for point $(2,2,1)$, the partial derivative is $\frac{4}{3}$ and for $(2,2,-1)$, it is $-\frac{4}{3}$? $\endgroup$ Oct 3, 2018 at 22:41
  • $\begingroup$ Yes, absolutely. $\endgroup$
    – Bernard
    Oct 3, 2018 at 22:43
  • $\begingroup$ Isn't $\frac{\partial z}{\partial x} = \frac{2}{3}$ at $(2, 2, 1)$ and $\frac{\partial z}{\partial x} = -\frac{2}{3}$ at $(2, 2, -1)$ based on your own calculations? $\endgroup$ Oct 4, 2018 at 9:45
  • $\begingroup$ @N.F.Taussig: You're absolutely right. Actually, I didn't check the numeric values given in the O.P.'s comment. $\endgroup$
    – Bernard
    Oct 4, 2018 at 16:44
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I'm also not expert partial derivative but I have thought like that: your final result is a global result. So you can find desired partial derivative just put x and y values into the found equations. Z values are only affect the sign of the result. For a positive z values, your solution is fully correct but for a negative z value, you should add a minus sign to the equation when square rooting z value in your solution.

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  • $\begingroup$ Do you mean For $\frac{\partial z}{\partial x} = \frac{x}{\sqrt{3}\sqrt{x^2 + y^2 - 5}}$ When $z = 1: \frac{\partial z}{\partial x} = \sqrt{\frac23}$ When $z = -1, \frac{\partial z}{\partial x} = -\sqrt{\frac23}$ Is this what you meant? $\endgroup$ Oct 3, 2018 at 22:21
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    $\begingroup$ Actually, result should be 2/3 for z=1 and -2/3 for z=-1. $\endgroup$
    – user600181
    Oct 4, 2018 at 3:46

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