How to find the Jacobian for implicit functions?

Question

I have to find the Jacobian for $$\begin{align} u&= x/ (1-r^2)^{1/2}\\ v&= y/ (1-r^2)^{1/2}\\ w &= z/ ( 1-r^2) \end{align}$$ where $r^2 = x^2 + y^2 + z^2$ but I am not able to solve it without making it more complicated. I think that maybe it can be done by using implicit functions but maybe I am wrong.

Answer 1

I assume that when you say Jacobian, you want the matrix

$$ \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{bmatrix}.$$

If so, then you can find the partial derivative using a combination of the Quotient Rule, $$\frac{d}{dx} \left( \frac{f}{g }\right) = \frac{f^\prime g - f g^\prime}{g^2}$$ and the Chain Rule $$ \frac{d}{dx} \left(f(g(x)) \right) = f^\prime (g(x)) g^\prime (x).$$ Let me know if you need more direction!

Answer 2

You're right about using implicit differentiation, but it's only a minor part of it. I'll use $u_x$ to mean ${\partial u\over\partial x}$ and similar notation for other partial derivatives.

We have $$u={x\over(1-r^2)^{1/2}}$$

By the quotient rule, $$ \begin{align} u_x &= {(1-r^2)^{1/2}-x\left(\frac12\right)(1-r^2)^{-1/2}(-2r)r_x\over1-r^2}\\ &={(1-r^2)+xrr_x\over(1-r^2)^{3/2}}\\ \end{align} $$ Implicitly differentiating $r^2=x^2+y^2+z^2$ with respect to $x$, we have $rr_x=x,$ so that $$u_x=1-y^2-z^2$$

Similarly, $$\begin{align} u_y &= {-x\left(\frac12\right)(1-r^2)^{-1/2}(-2r)r_y\over1-r^2}\\ &={xrr_y\over(1-r^2)^{3/2}}\\ &={xy\over(1-r^2)^{3/2}} \end{align}$$

Similar formulas hold for all the other partial derivatives.

When you say "Jacobian" I'm not sure if you mean the Jacobian determinant or the Jacobian matrix. The matrix is just $${1\over(1-r^2)^{3/2}}\begin{bmatrix} 1-y^2-z^2&xy&xz\\ xy&1-x^2-z^2&yz\\ xz&yz&1-x^2-y^2 \end{bmatrix}$$