# Partial Derivative Q: continuity and directional deriv's

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1. Mar 16, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

a) Show that the function $$f(x,y)=\sqrt[3]{xy}$$ is continuous and the partial derivatives $$f_x$$ and $$f_y$$ exist at the origin but the directional derivatives in all other directions do not exist

b) Graph f near the origin and comment on how the graph confirms part (a).

2. The attempt at a solution

This is an odd question to me. First off, what do they mean by "directional derivatives in all other directions" ? They only specify the origin as a point at which # f_x # and $$f_y$$ should exist, but no direction is specified, so how can there be other directions? What other directions? All directions? Ie, $$f_x$$ and $$f_y$$ exist at (0,0,0) but in no direction do valid directional derivatives exist??

The other quandary for me is how to show that it is continuous at the origin. I have found the grad vector

$$\nabla f(x,y) = \frac{1}{3}\left<\frac{y^{\frac{1}{3}}}{x^{\frac{2}{3}}},\frac{x^{\frac{1}{3}}}{y^{\frac{2}{3}}}\right>$$

But there is no general L'Hopital rule for part. deriv's, so I can't decide these limits as $$(x,y) \rightarrow (0,0)$$. And if y is taken as a constant for $$f_x$$ then it just goes to infinity as x approaches zero and likewise for $$f_y$$.

So I decided to just go ahead and graph this function in Matlab. I used meshgrid and symbolics to get two different but similar graphs.

1) meshgrid

[x,y] = meshgrid(-1:0.01:1);
z = (x.*y).^(1/3);
z = real(z);
surf(x,y,z), axis equal, xlabel('x'), ylabel('y'), zlabel('z'), shading interp, colormap(jet)

2) using symbolics and ezsurf

syms x y
ezsurf((x*y)^(1/3)), axis equal, xlabel('x'), ylabel('y'), zlabel('z')

Why does using the symbolic ezsurf throw out half the surface? It must be related to taking only the real values for z in meshgrid yes??

So the meshgrid form interpolates points around the origin, suggesting that the surface exists at the origin, but the symbolic toolbox suggests that the surface doesn't exist at the origin.

2. Mar 16, 2015

### RUber

The question about directional derivatives is based on the definition of a derivative as the $\lim_{|h|\to0} \frac{f( x+h_x,y+h_y) - f(x,y) }{|h|}$ looking at it this way, you allow for h to be a vector in any direction on the x-y plane.
You should be able to show that this limit is continuous if h only has x components or y components, but as soon as you have both, the limit is not the same for h as it is for -h and therefore the derivative does not exist in that direction.
Considering that explanation, look again at your plot, and you should see what I am saying.

3. Mar 16, 2015

### Ray Vickson

A directional derivative at $\vec{p}_0 = (x_0,y_0)$ in direction $\vec{d}$---YOU pick whatever $\vec{d}$ YOU want!---is defined as
$$D_{\vec{d}} f(\vec{p}_0) = \lim_{t \to 0+} \frac{ f(\vec{p}_0 + t \vec{d}) - f(\vec{p}_0)}{t}$$
In other words, for small $t > 0$ we should have
$$f(\vec{p}_0 + t \vec{d}) \approx f(\vec{p}_0) + t \cdot D_{\vec{d}} f(\vec{p}_0)$$

So, pick any direction non-axial direction $\vec{d}$ that you want, and see if the required limit exists.