What is the gradient of $f (x, (y^2+z^2)^{1/2})$?

Question

Consider a $c \in \mathbb R$ and a function $f: U \subset \mathbb R^2\rightarrow \mathbb R $ with $ \nabla f (p) \neq 0, \forall p \in f^{-1}(c)$, where $U$ is contained in the upper half plane y > 0.

Now define $g: U \times \mathbb R \rightarrow \mathbb R$ by $g(x, y, z) = f (x, (y^2+z^2)^{1/2})$. Then it is to be shown that $\nabla g(q) \neq 0, \forall q \in g^{-1}(c)$.

But I am not able to figure out how to write $\nabla f$ in way that would help me solve the problem.

Answer 1

I assume that what you want is to find $\nabla g$. In order to do so, we have to apply the chain rule. First of all, notice that the function $g$ can be interpreted as the composition of $f: U \subset \mathbb{R}^2\to\mathbb{R}$ and $h: \mathbb{R}^3\to\mathbb{R}^2$, where $h(x,y,z) = (x, (y^2+z^2)^\frac{1}{2})$. The composition is well defined and $g = f \:\circ h $. Then the partial derivatives of $g$ are:

• $\frac{\partial g}{\partial x}(x,y,z) = f^{(1,0)}(x, (y^2+z^2)^\frac{1}{2})$

• $\frac{\partial g}{\partial y}(x,y,z) = \frac{y}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2})$

• $\frac{\partial g}{\partial z}(x,y,z) = \frac{z}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2})$

The easiest way to compute the partial derivatives using the chain rule is, for me at least, using Jacobian matrices. More in particular: $$ J_f(x,y) = \begin{bmatrix} f^{(1,0)} & f^{(0,1)} \end{bmatrix} $$ $$ J_h(x,y,z)= \begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{y}{(y^2+z^2)^\frac{1}{2}} & \frac{z}{(y^2+z^2)^\frac{1}{2}} \end{bmatrix} $$ The chain rule tell us that $J_g(x,y,z) = J_f(h(x,y,z))J_h(x,y,z)$. Therefore: $$ J_g(x,y,z) = \begin{bmatrix} f^{(1,0)}(x, (y^2+z^2)^\frac{1}{2}) & \frac{y}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2}) & \frac{z}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2}) \end{bmatrix} $$ Therefore its gradient is: $$\nabla g (x,y,z) = (f^{(1,0)}(x, (y^2+z^2)^\frac{1}{2}), \frac{y}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2}), \frac{z}{(y^2+z^2)^\frac{1}{2}}f^{(0,1)}(x, (y^2+z^2)^\frac{1}{2}))$$

Answer 2

If memory serves, and that's a big if, the gradient is defined as $\nabla f(x,y)=(f_x,f_y)$.

Here we get $\nabla g=\nabla (f\circ h)$, where $h(x,y,z)=(x,(y^2+z^2)^{1/2})$. So $h:\Bbb R^3\to\Bbb R^2$.

Now before I try to remember the chain rule in this context, do you remember it?

We should get that $\rm dh $ is a $2×3$ matrix of partials. Each column a gradient of coordinate functions. If that isn't confusing enough, let's try. I get $\begin{pmatrix}1\quad\quad 0\quad \quad 0\\0\frac y{\sqrt{y^2+z^2}} \frac z{\sqrt{y^2+z^2}}\end{pmatrix}$.

Now, when you dot this with $\nabla f$ you get $(f_x,\frac{f_yy}{\sqrt{y^2+z^2}},\frac{f_yz}{\sqrt{y^2+z^2}})$.

Now we need to reason why this $3$-tuple is nonzero.

Well, one of $f_x,f_y$ is nonzero. Now, if it's $f_x$ we're done. And since $y>0$, we are also done if it's $f_y$.