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Short question:

$z(x,y)$ is a function that is implicitly defined by the equation

$$4x+3y+5z+4\cos(4z)+3=0$$

in the neigbourhood of the point $p=(-5\pi/16, 1/3, \pi/4)$

I am going to calculate $\partial$ in $p$.


$F$ differentiated with respect to $x$ is $4$ and with respect to $y$ is $3$.

Differentiated with respect to $z$ is $5-16\sin(4z)$

Then we get

$$\frac{\partial z}{\partial x}=-4/5$$

and

$$\frac{\partial z}{\partial y}=-3/5 $$

by plugging in $z=\pi/4$

But now I don't know how to calculate $\frac{\partial^2z}{\partial x \partial y}$.

Can somebody help me?

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  • $\begingroup$ Hi and welcome to the site, it is great that you show your own attempts on questions. I tried helping you typeset with MathJax LaTeX which is the preferred typesetting. $\endgroup$ – mathreadler Apr 24 '17 at 18:14
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In computing these partials, we will assume $F_z \not = 0$. Thus, since $z = g(x,y)$, for a small neighborhood of the points $(x,y,g(x,y))$ we can choose a curve $c(t) \in \mathbb{R}^2$ such that:

\begin{align*} (x,y,g(x,y)) \in F( c(t)) \Rightarrow 0 &= \frac{d}{dt}(F \circ c)(t)\\&= \nabla F \cdot \frac{\partial}{\partial x} (x,y,g(x,y) \\&= (F_x, F_y, F_z) \cdot (1,0, g_x) \\& \Rightarrow F_x + F_z g_x =0 \\& \frac{\partial z}{\partial x} = - \frac{F_x}{F_z} \end{align*}

Here $F(x,y,z) = 4x+3y+5z+4\cos(4z)+3$. I'll leave you to the rest. Just know you should get zero.

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  • $\begingroup$ The point is that you have to differentiate that final expression again (with respect to $y$) and use the chain rule wherever you see $z$'s. Second derivatives always cause people problems here. $\endgroup$ – Ted Shifrin Apr 24 '17 at 18:24
  • $\begingroup$ @TedShifrin: That's correct. I didn't want to get a formula for the second derivative since it would of took longer, plus the OP has an explicit expression for $F$. I didn't even remember this formula for the first derivative and so I had to derive it. $\endgroup$ – Faraad Armwood Apr 24 '17 at 18:28
  • $\begingroup$ Okay thanku! I think i understand what to do now $\endgroup$ – anne Apr 24 '17 at 18:43
  • $\begingroup$ But the OP has the correct value for $\partial z/\partial x$ at the point (although not as a function of $(x,y,z)$), so the intent of the question is clearly how to do the second derivative. $\endgroup$ – Ted Shifrin Apr 24 '17 at 18:44
  • $\begingroup$ @TedShifrin. It is not clear that the OP needs to compute the second derivative explicitly. I understood it to mean what is the second derivative evaluated at $p$. If the first part didn't require them to derive a formula for the first derivative and then evaluate, then why should I think the second part means, get a formula for the second derivative and then evaluate at $p$? $\endgroup$ – Faraad Armwood Apr 24 '17 at 18:48
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For the higher derivatives you first note that $\frac{\partial z}{\partial x}(x,y) = -\frac{4}{5-16\sin(4z(x,y))}$. Now let's differentiate this expression with respect to $y$. After some work we should get that: \begin{align*} \frac{\partial^2z}{\partial x \partial y}(x,y) = -\frac{256 \frac{\partial z}{\partial y}(x,y) \cos(4z(x,y))}{(5-16\sin(4z(x,y)))^2}. \end{align*} Now remember from the first derivatives you already calculated that $\frac{\partial z}{\partial y}(x,y) = -\frac{3}{5-16\sin(4z(x,y))}$. You can now plug that into the expression for $\frac{\partial^2z}{\partial x \partial y}(x,y)$.

Hope that helped.

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