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For the function $f(x,y,z) = \frac{1 }{ x^2+y^2+z^2}$ what is the direction of the fastest increase at $(1,1,1)$?

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2 Answers

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In the direction of the gradient of $f$ evaluated at $(1,1,1)$.

This is a general fact concerning functions $f=f(x,y,z)$: The direction in which a differentiable function $f$ increases most rapidly at the point $(a,b,c)$ is in the direction of $\nabla f(a,b,c)$.

You'll need to find the general formula for $\nabla f(x,y,z)$ first. Then simply evaluate $\nabla f(1,1,1)$. The solution is below, but it would be to your benefit to try the computations before looking at it.

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Solution:

The first thing you have to do is compute the gradient of $f$ in general; that is compute $\nabla f(x,y,z)$:

$$\eqalign{ \nabla f(x,y,z) &=f_x(x,y,z)\,{\bf i} +f_y(x,y,z)\,{\bf j} + f_z(x,y,z)\,{\bf k} \cr &={\partial \over\partial x} {1\over x^2+y^2+z^2}\,{\bf i} +{\partial \over\partial y} {1\over x^2+y^2+z^2} \,{\bf j} +{\partial \over\partial z} {1\over x^2+y^2+z^2}\,{\bf k} \cr &= -(x^2+y^2+z^2)^{-2}\cdot 2x\,{\bf i} -(x^2+y^2+z^2)^{-2}\cdot 2y\,{\bf j} -(x^2+y^2+z^2)^{-2}\cdot 2z\,{\bf k}\cr &={ -2x\over (x^2+y^2+z^2)^{2}}\,{\bf i} + { -2y\over (x^2+y^2+z^2)^{2}}\,{\bf j} + { -2z\over (x^2+y^2+z^2)^{2}}\,{\bf k}. } $$

Now evaluate the gradient at the point $(1,1,1)$: $$\eqalign{ \nabla f(1,1,1) &={ -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf i} +{ -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf j} + { -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf k}\cr &={-2\over9}\,{\bf i}+{-2\over9}\,{\bf j}+{-2\over9}\,{\bf k} . } $$

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The direction of the fastest increase at a point $(x_0, y_0, z_0)$ is the direction of the gradient $\nabla f (x_0, y_0 , z_0)$. So you have $$\begin{align} \nabla f &= (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) \\ &= ... \end{align}$$ This expression you evaluate at $(1,1,1)$.

Note, however, that because of the symmetry you only have to find one of the partial derivatives and evaluate that at $1$. The other coordinates are going to be the same value. And since it is clear that the derivative $\frac{\partial f}{\partial x}$ is negative at something positive, if you are just interested in the direction, you get the same as the direction $(-1,-1,-1)$.

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