0
$\begingroup$

I am learning the partial derivative and I am stuck on how to proceed with the following expression.

z = √(4 - x^2 - y^2)

the expressions I need to find are Δ_x z,Δ_y z and Δz, which are are intersections in the plane x,y,z, where
Δ_x z = f(x+Δx,y) - f(x,y) Δ_y z = f(x,y+Δy) - f(x,y) Δz = f(x+Δx,y+Δy) - f(x,y)

I have been out of school for a few years now and I can't find any explanations on how to simplify addition expression in a radical such as the expresion z above and this is blocking me.

My first instinct would be to leave the expression complete (for example, Δ_x z = √(4 - (x+Δx)^2 - y^2) - √(4 - x^2 - y^2), but in my previous examples, the -f(x,y) part of the expressions would simplify the non-delta expressions and that confuses me.

Thank you for your help.

$\endgroup$
1
$\begingroup$

If you are trying to compute the partial derivative wrt $x$ of $z = \sqrt{4-x^2-y^2}$, given by $$\frac{\partial z}{\partial x} = \lim_{\Delta x \rightarrow 0} \frac{\sqrt{4-(x+\Delta x)^2-y^2} - \sqrt{4-x^2-y^2}}{\Delta x}$$, a good trick is to multiply numerator and denominator by the conjugate expression $\sqrt{4-(x+\Delta x)^2-y^2} + \sqrt{4-x^2-y^2}$ and simplify. This will eliminate the square roots.

$\endgroup$
  • $\begingroup$ Unless I am missing something, to multiply by the conjugate expression, I need to multiply it on denominator and numerator (to multiply by 1), so I will be left with the conjugate at the denominator, no? edit : I misread a little bit here. But still, the conjugate expression will still be present at the denominator and I am not sure how to go from there neither. $\endgroup$ – Jeph Gagnon Apr 30 '18 at 18:26
  • $\begingroup$ After multiplying numerator and denominator, you can distribute and should be able to cancel out a $\Delta x$. $\endgroup$ – D.B. Apr 30 '18 at 18:40
  • $\begingroup$ I will still be stuck with ${2x}{/}{(}{\sqrt{4-(x+\Delta x)^2-y^2} +\sqrt{4-x^2-y^2}}{)}$ and I still have the same issue as when I started. edit: sorry for the edits, begginer mathjax $\endgroup$ – Jeph Gagnon Apr 30 '18 at 19:19
  • $\begingroup$ Exactly ! In the limit as $\Delta x$ goes to zero, this becomes $\frac{x}{\sqrt{4-x^2-y^2}}$ $\endgroup$ – D.B. Apr 30 '18 at 21:24
  • $\begingroup$ which is the answer $\endgroup$ – D.B. Apr 30 '18 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.