1
$\begingroup$

Given is the function $F(x,y,z)=x^2+y^3-z$.

Determine the Jacobian matrix $Dz$ in $P=(1,1,2)$ using implicit differentiation.

My idea is to calculate $\frac{∂z}{∂x}$ in $P(1,1,2)$ and $\frac{∂z}{∂y}$ in $P(1,1,2)$ and then just write it in matrix form.

So,

$F(x,y,z)=x^2+y^3-z=0$

$\frac{∂z}{∂x}=-\frac{\frac{∂F}{∂x}}{\frac{∂F}{∂z}}=-{\frac{2x}{-1}}=2x$

$\frac{∂z}{∂x} $ in $ P(1,1,2)=2$

$\frac{∂z}{∂y}=-\frac{\frac{∂F}{∂y}}{\frac{∂F}{∂z}}=-{\frac{3y^2}{-1}}=3y^2$

$\frac{∂z}{∂y} $ in $ P(1,1,2)=3$

$Dz(1,1,2)=(\frac{∂z}{∂x}(1,1,2) \qquad\frac{∂z}{∂y} (1,1,2)) $

$Dz(1,1,2)=(2 \qquad 3) $

I checked the result using explicit differentiation and I obtained the same.

But in the book that I use I saw another approach. Namely, as a hint was given this formula:

$D_{\underline x} f({\underline x°}) = -[D_{\underline y}F(\underline x°,\underline y°)]^{-1}D_\underline xF(\underline x°,\underline y°)$.

I don’t understand how this formula can be used in order to calculate $Dz$.

Any help is appreciated.

Thanks in advance.

$\endgroup$
0
$\begingroup$

It seems that both approaches are indeed equivalent.

Since $Dz$ is requested,

$D_{\underline x} f({\underline x°}) = -[D_{\underline y}F(\underline x°,\underline y°)]^{-1}D_\underline xF(\underline x°,\underline y°)$ becomes

$D_{\underline x, \underline y}f({\underline x°},{\underline y°}) = -[D_{\underline z}F(\underline x°,\underline y°\underline z°)]^{-1}D_{\underline x \underline y} F(\underline x°,\underline y°,\underline z°)$.

It follows that:

$ -[D_{\underline z}F(\underline x°,\underline y°\underline z°)]^{-1}=-[-1]^{-1}$

$D_{\underline x \underline y} F(\underline x°,\underline y°,\underline z°)= (2x\qquad 3x^2) $.

$D_{\underline x \underline y} F(1, 1, 2)= (2\qquad 3) $.

So,

$Dz = -[-1]^{-1}(2\qquad 3)=(2\qquad 3)$.

$\endgroup$

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.